In each case we split the interval $[0,\pi]$ into two parts, $[0,\epsilon]$
and $[\epsilon,\pi]$, with $\epsilon$ varying for the different cases, and
threat them separately. For the interval $[0,\epsilon]$ we use the asymptotic
bounds for the different functions that were introduced in the previous
sections. For the interval $[\epsilon,\pi]$ we evaluate the functions directly
using interval arithmetic. The method we use for bounding the supremum is the
one described in Section 6.1.

In most cases the subintervals in the computations are bisected at the
midpoint, meaning that the interval $[\underline{x},\overline{x}]$ would be
bisected into the two intervals
$[\underline{x},(\underline{x}+\overline{x})/2]$ and
$[(\underline{x}+\overline{x})/2,\overline{x}]$. However, when the magnitude
of $\underline{x}$ and $\overline{x}$ are very different it can be beneficial
to bisect at the geometric midpoint (see e.g. [27]), in that case we split the
interval into $\left[\underline{x},\sqrt{\underline{x}\overline{x}}\right]$
and $\left[\sqrt{\underline{x}\overline{x}},\overline{x}\right]$, where we
assume that $\underline{x}>0$.

The code [14] for the computer-assisted parts is implemented in Julia [5]. The
main tool for the rigorous numerics is Arb [40] which we use through the Julia
wrapper Arblib.jl 111https://github.com/kalmarek/Arblib.jl. Many of the basic
interval arithmetic algorithms, such as isolating roots or enclosing maximum
values, are implemented in a separate package, ArbExtras.jl
222https://github.com/Joel-Dahne/ArbExtras.jl. For finding the coefficients
$\\{a_{j}\\}$ with $j\geq 1$ and $\\{b_{n}\\}$ of $u_{\alpha}$ we make use of
non-linear solvers from NLsolve.jl [48].

For $\alpha\in I_{1}$ and $\alpha\in I_{3}$ the computations were done on an
AMD Ryzen 9 5900X processor with 32 GB of RAM using 12 threads. For $\alpha\in
I_{2}$ most of the computations where done on the Dardel HPC system at the PDC
Center for High Performance Computing, KTH Royal Institute of Technology. The
nodes are equipped with two AMD EPYC™ Zen2 2.25 GHz 64 core processors and 256
GB of RAM.

We handle the intervals $I_{1}$, $I_{2}$ and $I_{3}$ separately, the all
require slightly different approaches for computing the bounds.

###  12.1 Bounds for $I_{1}$

We here give bounds of $n_{\alpha}$, $\delta_{\alpha}$ and $D_{\alpha}$ for
$\alpha\in I_{1}=(-1,-1+\delta_{1})$, with $\delta_{1}=10^{-4}$. The bounds
are split into three lemmas. Recall that in this case the weight is given by
$w_{\alpha}(x)=|x|^{(1-\alpha)/2}\log(2e+1/|x|)$. We take $u_{\alpha}$ is in
(20) with $\hat{\alpha}=-0.9997$, $N_{\hat{\alpha},0}=1929$ and $N_{-1,1}=16$.

######  Lemma 12.1.

The constant $n_{\alpha}$ satisfies the inequality
$n_{\alpha}\leq\bar{n}_{1}=1.79$ for all $\alpha\in I_{1}$.

###### Proof.

A plot of $N_{\alpha}(x)$ on the interval $[0,\pi]$ is given in Figure 3(a).
It hints at a local maximum at $x=\pi$ but does not fully show what happens
near $x=0$, a plot closer to the origin is given in Figure 3(b), and we see
that the function flattens out around $\pi/2$ when $x$ approaches zero, which
indicates that the maximum indeed is attained at $x=\pi$.

For the interval $[\epsilon,\pi]$ we don’t compute with $N_{\alpha}(x)$
directly but instead with
$\bar{N}_{\alpha}(x)=\frac{w_{\alpha}(x)}{2u_{-1}(x)}$ which satisfies
$N_{\alpha}(x)\leq\bar{N}_{\alpha}(x)$, as mentioned in Section 9.2.

The value of $\epsilon$ is chosen dynamically. We don’t compute an enclosure
of the maximum for the interval $[0,\epsilon]$ but only prove that it is
bounded by $\bar{N}_{\alpha}(\pi)$. The value for $\epsilon$ is determined in
the following way. Starting with $\epsilon=0.45$ we compute an enclosure of
$N_{\alpha}([0,\epsilon])$. If this enclosure is bounded by
$\bar{N}_{\alpha}(\pi)$ we stop, otherwise we multiply $\epsilon$ by $0.8$ and
try again. Once the bound holds we stop, in practice this happens for
$\epsilon\approx 0.075$.

For the interval $[\epsilon,\pi]$ we compute an enclosure of the maximum of
$\bar{N}_{\alpha}(x)$. This gives us

$n_{\alpha}\leq[1.7870\pm 8.20\cdot 10^{-5}],$

which is upper bounded by $\bar{n}_{1}$. On this interval we are able to
compute Taylor expansions of $\bar{N}_{\alpha}(x)$ on the interval
$[\epsilon,\pi]$, allowing us to use the better version of the algorithm,
based on the Taylor polynomial, for enclosing the maximum. We use a Taylor
expansion of degree $0$, which is enough to pick up the monotonicity after
only a few bisections in most cases.

The runtime is about 6 seconds, most of it for handling the interval
$[\epsilon,\pi]$. ∎

######  Lemma 12.2.

The constant $\delta_{\alpha}$ satisfies the inequality
$\delta_{\alpha}\leq\bar{\delta}_{1}=0.0005$ for all $\alpha\in I_{1}$.

###### Proof.

A plot of $F_{\alpha}(x)$ on the interval $[0,\pi]$ is given in Figure 4(a), a
logarithmic plot on the interval $[10^{-5},10^{-1}]$ is given in Figure 4(b).

In this case we don’t try to compute a very accurate bound but instead only
prove that $\delta_{\alpha}$ is bounded by $\bar{\delta}_{1}$.

The value for $\epsilon$ is chosen such that $F_{\alpha}(\epsilon)$ evaluated
using the asymptotic version gives an enclosure that satisfies the bound. We
find that $\epsilon\approx 0.48$ works.

For the interval $[0,\epsilon]$ we first find $\epsilon_{1}$ such that
$F_{\alpha}([0,\epsilon_{1}])$ directly gives an enclosure that is bounded by
$\bar{\delta}_{1}$. We get that $\epsilon_{1}\approx 10^{-7000}$ works. For
the interval $[\epsilon_{1},\epsilon]$ we compute an enclosure of the maximum
by iteratively bisecting the interval. Since the endpoints are very different
in size we bisect at the geometric midpoint. We stop once the enclosure is
bounded by $\bar{\delta}_{1}$. The asymptotic version of $F_{\alpha}$ doesn’t
allow for evaluation with Taylor expansions, it therefore requires many
bisections to get a good enough enclosure, the maximum depth for the bisection
is $30$.

For the interval $[\epsilon,\pi]$ we also compute a bound by bisection,
stopping once the bound is less than $\bar{\delta}_{1}$. Similar to in the
previous lemma, and as mentioned in Section 10.2, we use $u_{-1}(x)$ in the
numerator of $F_{\alpha}(x)$, which gives an upper bound. In this case we use
Taylor polynomials of degree $4$ and maximum depth for the bisection is only
$4$.

The runtime is about $380$ seconds. ∎

######  Lemma 12.3.

The constant $D_{\alpha}$ satisfies the inequality
$D_{\alpha}\leq\bar{D}_{1}=0.938$ for all $\alpha\in I_{1}$.

###### Proof.

A plot of $\mathcal{T}_{\alpha}(x)$ on the interval $[0,\pi]$ is given in
Figure 3(c). It hints at the maximum being attained around $x\approx 1.4$.

We take $\epsilon=0.1$. We first bound the maximum on $[\epsilon,\pi]$, for
which we get that the maximum is bounded by $[0.93\pm 7.86\cdot 10^{-3}]$,
which in turn is bounded by $\bar{D}_{1}$.

For the interval $[0,\epsilon]$ we only prove that the maximum is bounded by
$[0.93\pm 7.86\cdot 10^{-3}]$. This is done by first computing
$\mathcal{T}_{\alpha}([0,10^{-10}])$ and verifying that this satisfies the
bound. The interval $[10^{-10},\epsilon]$ is then bisected until the bound can
be verified, in this case the bisection is done at the geometric midpoint.

For the interval $[0,\epsilon]$ we do not make use of Taylor expansions but
only use direct evaluation. For the interval $[\epsilon,\pi]$ we are not able
to compute Taylor expansion of $U_{\alpha}$, and hence not of
$\mathcal{T}_{\alpha}$. We do however make one optimization to get better
enclosures. Recall that

$\mathcal{T}_{\alpha}(x)\leq\frac{U_{\alpha}(x)}{\pi|w_{\alpha}(x)||u_{-1}(x)|}.$

For a given interval x both $U_{\alpha}(\textbf{{x}})$ and
$w_{\alpha}(\textbf{{x}})$ are evaluated directly. To enclose
$u_{-1}(\textbf{{x}})$ we first check if it is monotone by computing the
derivative and checking that it is non-zero. If it is monotone we compute an
enclosure by evaluating it at the endpoints of x.

The runtime for the computation is around 410 seconds, the majority for
handling the interval $[\epsilon,\pi]$. ∎

Combining the above three lemmas we get the following result.

######  Lemma 12.4.

For all $\alpha\in I_{1}$ the following inequality is satisfied

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}.$

###### Proof.

Follows immediately from the above lemmas, noticing that

$\delta_{\alpha}\leq\bar{\delta}_{-1}<\frac{(1-\bar{D}_{\alpha})^{2}}{4\bar{n}_{\alpha}}\leq\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}.$

∎

(a)

(b)

(c)

Figure 3: Plot of the functions $N_{\alpha}$ and $\mathcal{T}_{\alpha}$ for
$\alpha\in I_{1}$. The dashed green lines show the upper bounds as given in
Lemmas 12.1 and 12.3

(a)

(b)

Figure 4: Plot of the function $F_{\alpha}$ for $\alpha\in I_{1}$. The
function is plotted on the interval $[0,\pi]$ and $[10^{-5},10^{-1}]$. The
dashed green line shows the upper bound as given in Lemma 12.2. The dotted red
line shows $\frac{(1-\bar{D}_{1})^{2}}{4\bar{n}_{1}}$, which is the value we
want $\delta_{\alpha}$ to be smaller than.

###  12.2 Bounds for $I_{2}$

We here give bounds of $n_{\alpha}$, $\delta_{\alpha}$ and $D_{\alpha}$ for
$\alpha\in I_{2}=(-0.9999,-0.0012)$.

In this case it is not possible to give uniform bounds valid on the whole
interval $I_{2}$. Instead we split $I_{2}$ into many subintervals and for each
individual subinterval bounds are computed and they are checked to satisfy

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}},$

as required for Proposition 2.2.

When the midpoint of the subinterval is less than $-0.95$ the hybrid approach
corresponding to $\alpha$ near $-1$ is used. In this case the weight is given
by $w_{\alpha}(x)=|x|^{p}\log(2e+1/|x|)$. If the midpoint of the subinterval
is larger than $-0.16$ then the hybrid approach corresponding to $\alpha$ near
$0$ is used, in which case the weight is $w_{\alpha}(x)=|x|$. For the rest of
the interval the default approach is used, with the weight
$w_{\alpha}(x)=|x|^{p}$.

The value of $p$ is chosen based on the midpoint of the subinterval. If the
midpoint is above $-0.33$ we use $p=1$ and if it is between $-0.5$ and $-0.33$
we use $p=\frac{3}{4}$. If it is below $-0.5$ we take $p=\frac{1-\alpha}{2}$,
evaluated at the midpoint of the interval. Note that in all cases we have
$-\alpha<p\leq 1$.

The interval is split into $72000$ subintervals. The sizes of the subintervals
are smaller near $\alpha=-1$ and $\alpha=0$. The interval $(-0.9999,-0.0012)$
is first split non-uniformly into $32$ subintervals, each of these
subintervals are then further split uniformly into a given number of
subintervals, the number varying for the different parts, see Table 1. Of
these $72000$ subintervals there are a few for which the computed bounds are
too bad to satisfy the required inequality. This happens in $28$ cases, in all
of these cases it is enough to bisect the subinterval once for the bounds to
be good enough. The final result is therefore based on a total of $72028$
subintervals.

Interval | Subintervals | Interval | Subintervals
---|---|---|---
$[-0.9999,-0.999875]$ | $500$ | $[-0.7,-0.6]$ | $1000$
$[-0.999875,-0.99985]$ | $500$ | $[-0.6,-0.5]$ | $1500$
$[-0.99985,-0.9998]$ | $1000$ | $[-0.5,-0.45]$ | $1000$
$[-0.9998,-0.9996]$ | $1000$ | $[-0.45,-0.41]$ | $1000$
$[-0.9996,-0.9993]$ | $1000$ | $[-0.41,-0.37]$ | $1000$
$[-0.9993,-0.999]$ | $1000$ | $[-0.37,-0.33]$ | $1000$
$[-0.999,-0.998]$ | $1000$ | $[-0.33,-0.2]$ | $10000$
$[-0.998,-0.996]$ | $1000$ | $[-0.2,-0.16]$ | $3000$
$[-0.996,-0.993]$ | $1000$ | $[-0.16,-0.1]$ | $500$
$[-0.993,-0.99]$ | $1000$ | $[-0.1,-0.05]$ | $500$
$[-0.99,-0.95]$ | $2000$ | $[-0.05,-0.025]$ | $500$
$[-0.95,-0.935]$ | $5000$ | $[-0.025,-0.0125]$ | $1000$
$[-0.935,-0.9]$ | $8000$ | $[-0.0125,-0.00625]$ | $2000$
$[-0.9,-0.85]$ | $4000$ | $[-0.00625,-0.003125]$ | $4000$
$[-0.85,-0.8]$ | $2000$ | $[-0.003125,-0.0015625]$ | $8000$
$[-0.8,-0.7]$ | $2000$ | $[-0.0015625,-0.0012]$ | $4000$
Table 1: The interval $[-0.9999,-0.0012]$ is split into these $32$
subintervals which are then further split uniformly into the given number of
subintervals. The first $11$ use the hybrid approach corresponding to $\alpha$
near $-1$ and the last $8$ use the hybrid approach corresponding to $\alpha$
near $0$, the ones in between use the default approach.

Due to the large number of subintervals, explicit bounds for each case is not
given here, we refer to the repository [14] where the data can be found. The
bounds are visualized in Figures 5(a), 5(b) and 5(c). However, these figures
only give a rough picture since the number of subintervals is larger than the
number of pixels in the picture and the subintervals are concentrated near
$\alpha=-1$ and $\alpha=0$.

For the precise values of $N_{\alpha,0}$ and $N_{\alpha,1}$ we refer to the
repository [14]. In general $N_{\alpha,0}$ takes it maximum value near
$\alpha=-1$ and minimum value near $\alpha=0$, varying from $5799$ to $2$. For
$N_{\alpha,1}$ it varies between $0$ and $16$.

The approach for bounding $n_{\alpha}$, $\delta_{\alpha}$ and $D_{\alpha}$ is
given in the following three lemmas. The total runtime for the calculations on
Dardel were around 12000 core hours (corresponding to a wall time of around 48
hours using 256 cores).

######  Lemma 12.5.

For all $\alpha\in I_{2}$ the value $N_{\alpha}$ is bounded. A sketch of the
bound is given in Figure 5(a), for the precise bound for each value of
$\alpha$ we refer to the repository [14].

###### Proof.

Similar to in Lemma 12.1 the maximum of $N_{\alpha}$ is in practice attained
at $x=\pi$, which makes it easy to compute an accurate bound. As seen in the
plot of $n_{\alpha}$ in Figure 5(a) it in general varies smoothly with
$\alpha$, the exception being the points where the choice of weight is
changed. We can also note that the bounds are well-behaved near $\alpha=-1$
and $\alpha=0$.

When computing the bounds the value of $\epsilon$ is not fixed, instead it is
determined dynamically. Starting with $\epsilon=3$ we check if the asymptotic
version of $N_{\alpha}$ evaluated at this point is less than
$N_{\alpha}(\pi)$. if this is the case we take this value of $\epsilon$,
otherwise we try with a slightly smaller $\epsilon$, stopping once we find one
that works. On the interval $[0,\epsilon]$ we then check that $N_{\alpha}$ is
bounded by $N_{\alpha}(\pi)$.

For the interval $[\epsilon,\pi]$ we compute an enclosure of the maximum.

In both cases we use Taylor expansions of degree $0$ when computing the
bounds.

For the hybrid case when $\alpha$ is close to $-1$ the approach is exactly the
same as above. For the hybrid case when $\alpha$ is close to $0$ we use the
same approach as in Lemma 12.9 for computing a bound.

The computational time varies with $\alpha$, on average the computations took
4.05 seconds for each subinterval with a minimum and maximum of 0.018 and 144
seconds respectively. ∎

######  Lemma 12.6.

For all $\alpha\in I_{2}$ the value $\delta_{\alpha}$ is bounded. A sketch of
the bound is given in Figure 5(b), for the precise bound for each value of
$\alpha$ we refer to the repository [14].

###### Proof.

The value of $\delta_{\alpha}$ is heavily dependent on the precise
approximation used. Very small changes in the approximation can give
relatively large changes for $\delta_{\alpha}$. This is clearly seen in Figure
5(b) where the bound is seen to vary a lot from subinterval to subinterval.
The patches where the bound is more or less continuous correspond to certain
values of $N_{\alpha,0}$ in the approximation, when the value changes from one
to the other it gives a large change in defect. There are also many isolated
points where the bound is very different from the nearby points, this is
mostly due to numerical instabilities when computing the coefficients for the
approximation.

Similar to in the previous lemma the value of $\epsilon$ is not fixed but
chosen dynamically. The precise choice is based on comparing the asymptotic
and the non-asymptotic versions at different values and see where the
asymptotic one starts to give tighter enclosures. The bounds on $[0,\epsilon]$
and $[\epsilon,\pi]$ are then computed separately, and their maximum is
returned.

In both cases we use Taylor expansions, the degree is tuned depending on
$\alpha$ and is between $2$ and $6$.

For the hybrid case when $\alpha$ is close to $-1$ the approach is exactly the
same as above. For the hybrid case when $\alpha$ is close to $0$ some
modifications needs to be made. For the interval $[0,\epsilon]$ we use Taylor
models in $\alpha$ to compute enclosures for the coefficients in the expansion
at $x=0$. Once we have the expansion we are able to compute Taylor expansions
in $x$ of it, where we use degree $8$. For the interval $[\epsilon,\pi]$ we
are not able to compute Taylor expansions in $x$ and have to fall back to
using direct enclosures.

The computational time varies with $\alpha$, on average the computations took
37 seconds for each subinterval with a minimum and maximum of 0.11 and 1084
seconds respectively. ∎

######  Lemma 12.7.

For all $\alpha\in I_{2}$ the value $D_{\alpha}$ is bounded by a value smaller
than $1$. A sketch of the bound is given in Figure 5(c), for the precise bound
for each value of $\alpha$ we refer to the repository [14].

###### Proof.

The computation of the bound of $D_{\alpha}$ is the most time-consuming part.
For that reason we do not attempt to compute a very accurate upper bound but
only as much as we need for Lemma 12.8 to go through. We need $D_{\alpha}$ to
satisfy the inequality

$D_{\alpha}<1-2\sqrt{n_{\alpha}\delta_{\alpha}}.$

Taking the upper bounds of $n_{\alpha}$ and $\delta_{\alpha}$ from the
previous two lemmas we compute the value that $D_{\alpha}$ needs to be bounded
by. To give a little headroom we subtract $2^{-26}$ to get the bound that we
use. This bound is seen in Figure 5(c), and we prove that $D_{\alpha}$ is
bounded by this value for all the subintervals.

To get an understanding of how this bound compares to the actual value of
$D_{\alpha}$ we also compute a non-rigorous approximation of it. This
approximation is computed by evaluating $\mathcal{T}_{\alpha}$ on a few points
in the interval $[0,\pi]$ and taking the maximum, but with no control of what
happens in between these points. This estimate can also be seen in Figure
5(c). From this estimate we can also compute an estimate of
$\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}$, which is the value we want
$\delta_{\alpha}$ to be less than, this estimate is seen in Figure 5(b).

Similar to in the previous two lemmas the value of $\epsilon$ is not fixed but
chosen dynamically. It is determined by starting at $\epsilon=1$ and then
taking smaller and smaller values until the asymptotic version of
$\mathcal{T}_{\alpha}$ evaluated at $\epsilon$ satisfies the prescribed bound.
We then prove that the bound holds on both $[0,\epsilon]$ and
$[\epsilon,\pi]$.

For the interval $[0,\epsilon]$ we do not make use of Taylor expansions but
only use direct evaluation. For the interval $[\epsilon,\pi]$ we are not able
to compute Taylor expansion of $U_{\alpha}$, and hence not of
$\mathcal{T}_{\alpha}$. We do however make one optimization to get better
enclosures. Recall that

$\mathcal{T}_{\alpha}(x)=\frac{U_{\alpha}(x)}{\pi|w_{\alpha}(x)||u_{\alpha}(x)|}.$

For a given interval x both $U_{\alpha}(\textbf{{x}})$ and
$w_{\alpha}(\textbf{{x}})$ are evaluated directly. To enclose
$u_{\alpha}(\textbf{{x}})$ we first check if it is monotone by computing the
derivative and checking that it is non-zero. If it is monotone we compute an
enclosure by evaluating it at the endpoints of x.

Both the hybrid cases use exactly the same approach.

The computational time varies with $\alpha$, on average the computations took
72 seconds for each subinterval with a minimum and maximum of 0.740.68 and
1075 seconds respectively. ∎

######  Lemma 12.8.

For all $\alpha\in I_{2}$ the following inequality is satisfied

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}.$

###### Proof.

Using the computed upper bounds from Lemma 12.5, 12.6 and 12.7 it is straight
forward to check that the inequality

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}$

holds for each of the $72028$ subintervals. Since the union of these
subintervals cover the entire of $I_{2}$ it follows that the inequality holds
for all $\alpha\in I_{2}$. ∎

(a)

(b)

(c)

Figure 5: Plot of bounds of $n_{\alpha}$, $\delta_{\alpha}$ and $D_{\alpha}$
for $\alpha\in I_{2}$. For $\delta_{\alpha}$ the estimated goal is given by
$\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}$ computed with the upper bound of
$n_{\alpha}$ and the estimate of $D_{\alpha}$. For $D_{\alpha}$ the non-
rigorous estimate is also included.

###  12.3 Bounds for $I_{3}$

For this interval the bounds are given in a slightly different form. As
$\alpha$ gets close to $0$ the value of $\delta_{\alpha}$ tends to zero and
$D_{\alpha}$ tends to one. To be able to say something about the inequality

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}$

in Proposition 2.2 in the limit we therefore need information about the rate
that they go to zero and one respectively. This is however not the case for
$n_{\alpha}$, for which we have the following lemma:

######  Lemma 12.9.

The constant $n_{\alpha}$ satisfies the inequality
$n_{\alpha}\leq\bar{n}_{3}=1.58$ for all $\alpha\in I_{3}$.

###### Proof.

A plot of $N_{\alpha}(x)$ on the interval $[0,\pi]$ is given in Figure 6(a),
clearly hinting at the maximum being attained at $x=\pi$. We use the
asymptotic expansion on the entire interval, i.e. $\epsilon=\pi$. The
computation of the enclosure takes less than a second and gives us

$n_{\alpha}\in[1.57\pm 2.45\cdot 10^{-3}],$

which is upper bounded by $\bar{n}_{3}$. ∎

For $D_{\alpha}$ and $\delta_{\alpha}$ we give bounds which are functions of
$\alpha$.

######  Lemma 12.10.

The constant $\delta_{\alpha}$ satisfies the inequality
$\delta_{\alpha}\leq\bar{\Delta}_{\delta}\cdot\alpha^{2}$, where
$\bar{\Delta}_{\delta}=0.0077$, for all $\alpha\in I_{3}$.

###### Proof.

For each $x\in[0,\pi]$ we can compute a Taylor model $M_{F}(x)$ of degree $1$
in $\alpha$ of $F_{\alpha}(x)$ centered at $\alpha=0$ and valid for $\alpha\in
I_{3}$. This means that $M_{F}(x)$ satisfies

$F_{\alpha}(x)\in p_{M_{F}(x)}(\alpha)+\Delta_{M_{F}(x)}\alpha^{2}$

for all $\alpha\in I_{3}$, where $p_{M_{F}(x)}$ is the degree $1$ polynomial
associated with the Taylor model and $\Delta_{M_{F}(x)}$ is the remainder.
From Lemma 10.1 we get that $p_{M_{F}(x)}=0$ for all $x\in[0,\pi]$, hence

$|F_{\alpha}(x)|\in|\Delta_{M_{F}(x)}|\alpha^{2}.$

This gives us

$\delta_{\alpha}=\sup_{x\in[0,\pi]}|F_{\alpha}(x)|\in\left(\sup_{x\in[0,\pi]}|\Delta_{M_{F}(x)}|\right)\alpha^{2}=:\Delta_{\delta}\alpha^{2},$

where we with the supremum of intervals we mean the interval whose upper
endpoint is the supremum of all upper endpoints and the lower endpoint is the
supremum of all lower endpoints. A plot of $\Delta_{M_{F}(x)}$ on the interval
$[0,\pi]$ is given in Figure 6(b), we are interested in enclosing
$\Delta_{\delta}$.

We use the asymptotic expansion on the entire interval, i.e. $\epsilon=\pi$.
Enclosing the supremum we get the interval

$\Delta_{\delta}\subseteq[0,7.65\cdot 10^{-3}],$

for which an upper bound is given by $\bar{\Delta}_{\delta}$. The runtime is
around 3 seconds. ∎

######  Lemma 12.11.

The constant $D_{\alpha}$ satisfies the inequality $D_{\alpha}\leq
1+\bar{\Delta}_{D}\cdot\alpha$, where $\bar{\Delta}_{D}=0.226$, for all
$\alpha\in I_{3}$.

###### Proof.

The proof is similar to the previous lemma. For each $x\in[0,\pi]$ we can
compute a Taylor model $M_{\mathcal{T}}(x)$ of degree $0$ in $\alpha$ of
$\mathcal{T}_{\alpha}(x)$ centered at $\alpha=0$ and valid for $\alpha\in
I_{3}$. This means that $M_{\mathcal{T}}(x)$ satisfies

$\mathcal{T}_{\alpha}(x)\in
p_{M_{\mathcal{T}}(x)}(\alpha)+\Delta_{M_{\mathcal{T}}(x)}\alpha$

for all $\alpha\in I_{3}$, where $p_{M_{\mathcal{T}}(x)}$ is the degree $0$
polynomial associated with the Taylor model and $\Delta_{M_{\mathcal{T}}(x)}$
is the remainder. From Lemma 11.8 we get that $p_{M_{\mathcal{T}}(x)}=1$ for
all $x\in[0,\pi]$, hence

$\mathcal{T}_{\alpha}(x)\in 1+\Delta_{M_{\mathcal{T}}(x)}\alpha.$

This gives us

$D_{\alpha}=\sup_{x\in[0,\pi]}|\mathcal{T}_{\alpha}(x)|\in
1+\left(\inf_{x\in[0,\pi]}\Delta_{M_{\mathcal{T}}(x)}\right)\alpha=:1+\Delta_{D}\alpha,$

where we take the infimum since $\alpha$ is negative. A plot of
$\Delta_{M_{\mathcal{T}}(x)}$ is given in Figure 6(c), we are interested in
enclosing $\Delta_{D}$.

We take $\epsilon=1$. From Figure 6(c) it seems like the infimum is attained
at $x=0$, we therefore compute the infimum on the interval $[0,\epsilon]$
first and then only prove that the value on $[\epsilon,\pi]$ is lower bounded
by this. Enclosing the infimum on $[0,\epsilon]$ we get the enclosure

$\Delta_{D}\subset[0.2413\pm 0.0151],$

which is then checked to be a lower bound for $[\epsilon,\pi]$. This computed
enclosure is then lower bounded by $\bar{\Delta}_{D}$. The runtime is around 7
seconds. ∎

Combining the above we get the following result.

######  Lemma 12.12.

For all $\alpha\in I_{3}$ the following inequality is satisfied

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}.$

###### Proof.

From lemma 12.9 and 12.11 we get

$\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}\geq\frac{\bar{\Delta}_{D}^{2}}{4\bar{n}_{3}}\cdot\alpha^{2}$

for all $\alpha\in I_{3}$. Combining this with lemma 12.10 means we only have
to check the inequality

$\bar{\Delta}_{\delta}<\frac{\bar{\Delta}_{D}^{2}}{4\bar{n}_{3}},$

which indeed holds. ∎

(a)

(b)

(c)

Figure 6: Plots for $\alpha\in I_{3}$. The leftmost figure shows a plot of
$N_{\alpha}(x)$. The two other figures show enclosures of $\Delta_{M_{F}(x)}$
and $\Delta_{M_{\mathcal{T}}(x)}$, the remainder terms of the Taylor models
for $F_{\alpha}(x)$ and $\mathcal{T}_{\alpha}(x)$ respectively. The dashed
green lines shows the upper/lower bounds given in Lemmas 12.9, 12.10 and
12.11. The dotted red line in the middle figure shows
$\frac{\bar{\Delta}_{D}^{2}}{4\bar{n}_{3}}$, the value that $\Delta_{\delta}$
needs to be smaller than.

##  13 Proof of Theorem 1.1

We are now ready to give the proof of Theorem 1.1.

###### Proof of Theorem 1.1.

Consider the operator $G_{\alpha}$ from (11) given by

$G_{\alpha}[v]=(I-T_{\alpha})^{-1}(-F_{\alpha}-N_{\alpha}v^{2}).$

Combining Lemmas 12.3, 12.7 and 12.11 we have $\|T_{\alpha}\|<1$ for all
$\alpha\in(-1,0)$, so the inverse of the operator $I-T_{\alpha}$ is well-
defined. Combining Lemmas 12.4, 12.8 and 12.12 gives us that the inequality

$\delta_{\alpha}<\frac{(1-D_{\alpha})^{2}}{4n_{\alpha}}$

holds for all $\alpha\in(-1,0)$. This, together with Proposition 2.2 and
Banach fixed-point theorem, proves that for

$\epsilon_{\alpha}=\frac{1-D_{\alpha}-\sqrt{(1-D_{\alpha})^{2}-4\delta_{\alpha}n_{\alpha}}}{2n_{\alpha}}$

the operator $G_{\alpha}$ has a unique fixed-point $v_{\alpha}$ in
$X_{\epsilon_{\alpha}}\subseteq L^{\infty}(\mathbb{T})$.

By the construction of the operator $G_{\alpha}$ this means that the function

$u(x)=u_{\alpha}(x)+w_{\alpha}(x)v_{\alpha}(x)$

solves (5), given by

$\frac{1}{2}u^{2}=-\mathcal{H}^{\alpha}[u].$

For any wave speed $c\in\mathbb{R}$ we then have that the function

$\varphi(x)=c-u(x)$

is a traveling wave solution to (1). This proves the existence of a
$2\pi$-periodic highest cusped traveling wave.

The asymptotic behavior of $u_{\alpha}$ close to $x=0$ is given by

$u_{\alpha}(x)=\nu_{\alpha}|x|^{-\alpha}+\mathcal{O}(|x|^{-\alpha+p_{\alpha}}).$

with $\nu_{\alpha}$ as in Lemma 4.1. The precise value for $p_{\alpha}$ varies
depending on the type of construction used, but in all cases it satisfies
$-\alpha+p_{\alpha}>1$. Furthermore the weight satisfies

$w_{\alpha}(x)=\mathcal{O}(|x|^{p})$

for some $p$ with $-\alpha<p\leq 1$. Hence

$u(x)=u_{\alpha}(x)+w_{\alpha}(x)v_{\alpha}(x)=\nu_{\alpha}|x|^{-\alpha}+\mathcal{O}(|x|^{p})$

and

$\varphi(x)=c-\nu_{\alpha}|x|^{-\alpha}+\mathcal{O}(|x|^{p}),$

as we wanted to show. ∎

##  Appendix A Removable singularities

In several cases we have to compute enclosures of functions with removable
singularities. For example the function

$\Gamma(1-s)\cos(\pi(1-s)/2)$

comes up when computing $C_{s}$ through equation (39) and has a removable
singularity whenever $s$ is a positive even integer. For this we follow the
same approach as in [15, Appendix A], where more details are given. The main
tool is the following lemma:

######  Lemma A.1.

Let $m\in\mathbb{Z}_{\geq 0}$ and let $I$ be an interval containing zero.
Consider a function $f(x)$ with a zero of order $n$ at $x=0$ and such that
$f^{(m+n)}(x)$ is absolutely continuous on $I$. Then for all $x\in I$ we have

$\frac{f(x)}{x^{n}}=\sum_{k=0}^{m}f_{k+n}(0)x^{k}+f_{m+n+1}(\xi)x^{m+1}$

for some $\xi$ between $0$ and $x$. Furthermore, if $f^{m+n+p}(x)$ is
absolutely continuous for $p\in\mathbb{Z}_{\geq 0}$ we have

$\frac{d^{p}}{dx^{p}}\frac{f(x)}{x^{n}}=\sum_{k=0}^{m}\frac{(k+p)!}{k!}f_{k+n+p}(0)x^{k}+\frac{(m+p+1)!}{(m+1)!}f_{m+n+p+1}(\xi)x^{m+1}$

for some $\xi$ between $0$ and $x$.

We also make use of the first statement of the lemma for $x\in\mathbb{C}$, the
proof is the same as for $x\in\mathbb{R}$.

##  Appendix B Taylor models

An important tool for handling the limit $\alpha\to 0$ is the use of Taylor
models for enclosing $\delta_{\alpha}$ and $D_{\alpha}$. We here give a brief
introduction to Taylor models, for a more thorough introduction we refer to
[44].

With a Taylor model we mean what in [44] is refereed to as a _Taylor model
with relative error_. We use the following definition, compare with [44,
Definition 2.3.2].

######  Definition B.1.

A Taylor model $M=(p,\Delta)$ of degree $n$ for a function $f$ on an interval
$I$ centered at a point $x_{0}\in I$ is a polynomial $p$ of degree $n$
together with an interval $\Delta$, satisfying that for all $x\in I$ there is
$\delta\in\Delta$ such that

$f(x)-p(x-x_{0})=\delta(x-x_{0})^{n+1}.$

For an $n+1$ times differentiable function $f$ the polynomial $p$ is the
Taylor polynomial of degree $n$ of $f$ centered at $x_{0}$ [44, Lemma 2.3.3].
A Taylor model is thus given by a truncated Taylor expansion plus a bound on
the remainder term valid on some interval $I$. From Taylor’s theorem we see
that we can take $\Delta$ to be an enclosure of $\frac{f^{n+1}(x)}{(n+1)!}$ on
the interval $I$.

We can perform arithmetic on Taylor models. Given two functions $f$ and $g$
with corresponding Taylor models $M_{f}=(p_{f},\Delta_{f})$ and
$M_{g}=(p_{g}+\Delta_{g})$ we can compute a Taylor model of $f+g$ as
$M_{f+g}=(p_{f}+p_{g},\Delta_{f}+\Delta_{g})$, similarly for $f-g$. With
slightly more work we can compute a Taylor model of $f\cdot g$, see [44,
Algorithm 2.3.6], as well as of $f/g$, see [44, Algorithm 2.3.12]. The
division can directly handle removable singularities for $f/g$, if there is a
removable singularity of order $k$ then we need Taylor models of degree $n+k$
for $f$ and $g$ to get a Taylor model of degree $n$ for $f/g$. We can also
compose Taylor models with arbitrary functions, given a Taylor model $M_{f}$
of $f$ and a function $g$ we can compute a Taylor model $M_{g\circ f}$ of
$g\circ f$, see [44, Algorithm 2.3.8].

###  B.1 Taylor models in $\alpha$ of expansions in $x$

For both $F_{\alpha}(x)$ and $\mathcal{T}_{\alpha}(x)$ we need to compute
expansions in $x$ to evaluate them near $x=0$. For $\alpha\in I_{3}$ we then
need to compute Taylor models in $\alpha$ from these expansions, see Section
10.3 and 11.2.3. This is done by computing Taylor models of the coefficients
in the expansion in $x$.

For example, consider the problem of computing an expansion in $x$ of
$u_{\alpha}(x)$, used e.g. when computing $\frac{x^{-\alpha}}{u_{\alpha}(x)}$.
For $\alpha\in I_{3}$ we have from Lemma 4.2, and using that $N_{\alpha,1}=0$
in this case,

$u_{\alpha}(x)=\sum_{j=0}^{N_{\alpha,0}}a_{\alpha,j}^{0}|x|^{-\alpha+jp_{\alpha}}+\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=1}^{N_{\alpha,0}}a_{\alpha,j}\zeta(1-\alpha+jp_{\alpha}-2m)\right)x^{2m}.$

It is straight forward to compute Taylor models of the coefficients
$a_{\alpha,j}^{0}$ and

$\frac{(-1)^{m}}{(2m)!}\left(\sum_{j=1}^{N_{\alpha,0}}a_{\alpha,j}\zeta(1-\alpha+jp_{\alpha}-2m)\right).$

for $m<M$. What remains is computing a Taylor model bounding the tail. To
compute a Taylor model enclosing a sum of the form

$\sum_{m=M}^{\infty}c_{\alpha,m}x^{2m}.$

it is enough to have a method for enclosing

$\frac{d^{k}}{d\alpha^{k}}\sum_{m=M}^{\infty}c_{\alpha,m}x^{2m}.$

for $k\geq 0$. For the Clausen functions this is given in Lemma C.6.

###  B.2 Taylor model of $p_{\alpha}$

The above is enough to compute Taylor models of functions in the paper for
which we have explicit expressions. For $p_{\alpha}$ from (17) we only have an
implicit equation defining it, and we have to take a slightly different
approach.

Consider the equation

$f(x,y)=0.$

We want to compute a Taylor model of $y=y(x)$. Consider a Taylor model of
degree $n$, centered at $x_{0}$ and which should be valid on the interval $I$.
As mentioned above it is enough to compute the Taylor polynomial of $y(x)$ at
$x=x_{0}$ of degree $n$ and enclose $\frac{y^{n+1}(x)}{(n+1)!}$ on $I$.

Let $y_{0}$ be such that $f(x_{0},y_{0})=0$. We first consider the case when
$f_{y}(x_{0},y_{0})\not=0$. If $f$ is sufficiently smooth we can use the
implicit function theorem to compute a Taylor polynomial of $y$ at $x_{0}$ to
degree $n$. Expanding $f(x,y(x))$ at $x_{0}$ we get

$f(x_{0},y_{0})+(f_{x}(x_{0},y_{0})+y^{\prime}(x_{0})f_{y}(x_{0},y_{0}))(x-x_{0})\\\
+\frac{1}{2}(f_{xx}(x_{0},y_{0})+2y^{\prime}(x_{0})f_{xy}(x_{0},y_{0})+y^{\prime}(x_{0})f_{yy}(x_{0},y_{0})+y^{\prime\prime}(x_{0})f_{y}(x_{0},y_{0}))(x-x_{0})^{2}\cdots=0.$

Solving for $y^{\prime}(x_{0})$, $y^{\prime\prime}(x_{0})$, etc., we get

$\displaystyle y^{\prime}(x_{0})$
$\displaystyle=\frac{f_{x}(x_{0},y(x_{0}))}{f_{y}(x_{0},y(x_{0}))},$
$\displaystyle y^{\prime\prime}(x_{0})$
$\displaystyle=\frac{f_{xx}(x_{0},y_{0})+2y^{\prime}(x_{0})f_{xy}(x_{0},y_{0})+y^{\prime}(x_{0})f_{yy}(x_{0},y_{0})}{2f_{y}(x_{0},y(x_{0}))}$

and similarly for higher orders.

While it is possible to use the above formulas for the derivatives to enclose
$\frac{y^{n+1}(x)}{(n+1)!}$ on $I$ this gets very complicated for $p_{\alpha}$
at $\alpha=0$ due to the large number of removable singularities to handle.
Instead, we take a slightly different, more direct, approach for computing
$\Delta$. We want to find $\Delta$ such that for all $x\in I$ there is
$\delta\in\Delta$ such that

$f(x,p(x-x_{0})+\delta(x-x_{0})^{n+1})=0.$

If we let

$g(x,\delta)=f(x,p(x-x_{0})+\delta(x-x_{0})^{n+1})$

we want $\Delta$ to be an enclosure of a root in $\delta$ of $g$ for all $x\in
I$. To make the derivative in $\delta$ non-zero at $x=x_{0}$ we normalize $g$
as

$h(x,\delta)=\frac{g(x,\delta)}{(x-x_{0})^{n+1}}.$

Given a guess $\Delta=[\underline{\Delta},\overline{\Delta}]$, to prove that
this is an enclosure of a root we only need to verify that
$h(x,\underline{\Delta})$ and $h(x,\overline{\Delta})$ both have constant
signs for all $x\in I$ and that the two signs are different. The guess for
$\Delta$ can be given using non-rigorous means, for example by looking at the
error at the endpoints of $I$. Only the verification of the signs at the
endpoints of $\Delta$ needs to be done rigorously.

For $p_{\alpha}$ we get from (17) the function

$f(\alpha,p)=\frac{\Gamma(2\alpha-p)\cos\left(\frac{\pi}{2}(2\alpha-p)\right)}{\Gamma(\alpha-p)\cos\left(\frac{\pi}{2}(\alpha-p)\right)}-\frac{2\Gamma(2\alpha)\cos(\pi\alpha)}{\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)}.$

In this case $\alpha=0$ we have $f_{p}(0,p_{0})=0$, and we cannot apply the
above approach directly. For the coefficients of polynomial of the Taylor
model we expand

$f(\alpha,p_{0}+p_{1}\alpha+p_{2}\alpha^{2})$

at $\alpha=0$. The constant coefficient is exactly zero, and we then solve for
$p_{0}$, $p_{1}$ and $p_{2}$ making the three coming coefficients zero. For
the remainder term of the Taylor model we instead we consider the function
$\frac{f(\alpha,p)}{\alpha}$, for which the above approach does work.

##  Appendix C Computing Clausen functions

To be able to compute bounds of $D_{\alpha}$, $\delta_{\alpha}$ and
$n_{\alpha}$ it is critical that we can compute accurate enclosures of
$C_{s}(x)$ and $S_{s}(x)$, including expansions in the argument and
derivatives in the parameter. For this we follow the same approach as in [15,
Appendix B] with a few additions and improvements. We here give a shortened
version of the approach while highlighting the additions and improvements.

We start by going through how to compute $C_{s}(x)$ and $S_{s}(x)$ for
$s,x\in\mathbb{R}$. Since both $C_{s}(x)$ and $S_{s}(x)$ are $2\pi$-periodic
we can reduce it to $x=0$ or $0<x<2\pi$.

For $x=0$ and $s>1$ have $C_{s}(0)=\zeta(s)$ and $S_{s}(0)=0$. For $s\leq 1$
both functions typically diverge at $x=0$.

For $0<x<2\pi$ we can compute the Clausen functions by going through the
polylog function,

$C_{s}(x)=\operatorname{Re}\left(\operatorname{Li}_{s}(e^{ix})\right),\quad
S_{s}(x)=\operatorname{Im}\left(\operatorname{Li}_{s}(e^{ix})\right).$

However it is computationally beneficial to instead go through the periodic
zeta function [18, Sec. 25.13],

$F(x,s):=\operatorname{Li}_{s}(e^{2\pi ix})=\sum_{n=1}^{\infty}\frac{e^{2\pi
inx}}{n^{s}},$

for which we have

$C_{s}(x)=\operatorname{Re}F\left(\frac{x}{2\pi},s\right),\quad
S_{s}(x)=\operatorname{Im}F\left(\frac{x}{2\pi},s\right).$

Using [18, Eq. 25.13.2] we get, for $0<x<2\pi$,

$\displaystyle C_{s}(x)$
$\displaystyle=\frac{\Gamma(1-s)}{(2\pi)^{1-s}}\cos(\pi(1-s)/2)\left(\zeta\left(1-s,\frac{x}{2\pi}\right)+\zeta\left(1-s,1-\frac{x}{2\pi}\right)\right),$
(39) $\displaystyle S_{s}(x)$
$\displaystyle=\frac{\Gamma(1-s)}{(2\pi)^{1-s}}\sin(\pi(1-s)/2)\left(\zeta\left(1-s,\frac{x}{2\pi}\right)-\zeta\left(1-s,1-\frac{x}{2\pi}\right)\right).$
(40)

This formulation works well as long as $s$ is not a non-negative integer. For
non-negative integers we have to handle some removable singularities, for
details see [15, Appendix B].

Compared to [15] we also need to compute $C_{s}(x)$ for real $s$ but complex
$x$ during the rigorous integration discussed in Appendix D In those cases $s$
is never an integer, and we use Equation (39) which holds for
$0<\operatorname{Re}x<2\pi$.

###  C.1 Interval arguments

Let $\textbf{{x}}=[\underline{x},\overline{x}]$ and
$\textbf{{s}}=[\underline{s},\overline{s}]$ be two finite intervals, we are
interested in computing an enclosure of $C_{\textbf{{s}}}(\textbf{{x}})$ and
$S_{\textbf{{s}}}(\textbf{{x}})$. Due to the periodicity we can reduce it to
three different cases for x

  1. 1.

x doesn’t contain a multiple of $2\pi$, by adding or subtracting a suitable
multiple of $2\pi$ we can assume that $0<\textbf{{x}}<2\pi$;

  2. 2.

x has a diameter of at least $2\pi$, it then covers a full period and can
without loss of generality be taken as $\textbf{{x}}=[0,2\pi]$;

  3. 3.

x contains a multiple of $2\pi$ but has a diameter less than $2\pi$, by adding
or subtracting a suitable multiple of $2\pi$ we can take x such that
$-2\pi<\underline{x}\leq 0\leq\overline{x}<2\pi$.

The second and third case are handled exactly as in [15], for the first case
we make some minor improvements.

For $C_{s}$ we make use of the following lemma which is a slightly upgraded
version of [15, Lemma B.1], see also [22, Lemma B.1].

######  Lemma C.1.

For all $s\in\mathbb{R}$ the Clausen function $C_{s}(x)$ is monotone in $x$ on
the interval $(0,\pi)$. For $s>0$ it is decreasing. For $s\leq 0$ the sign of
the derivative is the same as that of $-\sin\left(\frac{\pi}{2}s\right)$.

###### Proof.

The derivative in $x$ is given by $-S_{s-1}(x)$. For $s>1$ the proof is the
same as in [15, Lemma B.1].

For $s<1$ we use equation (40) together with [18, Eq. 25.11.25] to get

$S_{s-1}(x)=\frac{\sin\left(\frac{\pi}{2}(2-s)\right)}{(2\pi)^{2-s}}\int_{0}^{\infty}t^{1-s}e^{-\frac{xt}{2\pi}}\frac{1-e^{(x/\pi-1)t}}{1-e^{-t}}\
dt,$

which sign depends only on the value $s$ and is the same as that of
$\sin\left(\frac{\pi}{2}s\right)$. In particular, for $0<s<1$ we have
$\sin\left(\frac{\pi}{2}s\right)>0$ so $C_{s}(x)$ is decreasing for $0<s<1$.

For $s=1$ the result follows directly from that $C_{1}(x)=-\log(2\sin(x/2))$
on the interval. ∎

For $S_{s}$ we have the following, slightly weaker, result

######  Lemma C.2.

For $s\leq 1$ the Clausen function $S_{s}(x)$ is monotone in $x$ on the
interval $(0,2\pi)$. The sign of the derivative is the same as that of
$-\cos\left(\frac{\pi}{2}s\right)$.

###### Proof.

The derivative in $x$ is given by $C_{s-1}(x)$. For $s<1$ we use equation (39)
together with [18, Eq. 25.11.25] to get

$C_{s-1}(x)=\frac{\cos\left(\frac{\pi}{2}(2-s)\right)}{(2\pi)^{2-s}}\int_{0}^{\infty}t^{1-s}e^{-\frac{xt}{2\pi}}\frac{1+e^{(x/\pi-1)t}}{1-e^{-t}}\
dt,$

which sign depends only on the value $s$ and is the same as that of
$-\cos\left(\frac{\pi}{2}s\right)$.

For $s=1$ the result follows directly from that
$S_{1}(x)=\frac{\pi}{2}-\frac{x}{2}$ on the interval. ∎

For $s\leq 1$ the extrema of $S_{s}(\textbf{{x}})$ are thus always attained at
$x=\underline{x}$ and $\overline{x}$. We handle $s>1$ by computing and
enclosure of the derivative
$S_{\textbf{{s}}}^{\prime}(\textbf{{x}})=C_{\textbf{{s}}-1}(\textbf{{x}})$, if
the enclosure of the derivative doesn’t contain zero then the function is
monotone, and we evaluate it at the endpoints, if the derivative contains zero
we instead use the midpoint approximation
$S_{\textbf{{s}}}(\textbf{{x}})=S_{\textbf{{s}}}(x_{0})+(\textbf{{x}}-x_{0})C_{\textbf{{s}}-1}(\textbf{{x}})$
where $x_{0}$ is the midpoint of x.

For s we follow the same approach as in [15]. The only difference is that we
also have to handle s overlapping $1$, in which case the implementation of the
deflated zeta function,

$\underline{\zeta}(s,x)=\zeta(s,x)+\frac{1}{1-s}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\gamma_{n}(x)(s-1)^{n},$
(41)

in Arb does not work directly. It only supports $s$ exactly equal to $1$, and
not intervals containing $1$. Looking at the implementation it is however
fairly easy to adapt it to also work for intervals containing $1$, which we
have done.

###  C.2 Expansion in $x$

We now go through how to compute expansions of the Clausen functions in the
argument $x$. In general the procedure is the same as in [15], except that we
also have to handle the case when $s$ overlaps positive integers for the
expansion at $x=0$.

At $x=0$ we have the following asymptotic expansions [22]

$\displaystyle C_{s}(x)$
$\displaystyle=\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)|x|^{s-1}+\sum_{m=0}^{\infty}(-1)^{m}\zeta(s-2m)\frac{x^{2m}}{(2m)!};$
(42) $\displaystyle S_{s}(x)$
$\displaystyle=\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\operatorname{sgn}(x)|x|^{s-1}+\sum_{m=0}^{\infty}(-1)^{m}\zeta(s-2m-1)\frac{x^{2m+1}}{(2m+1)!}.$
(43)

For positive integers we have to handle the poles of $\Gamma(s)$ at non-
positive integers and the pole of $\zeta(s)$ at $s=1$.

For $C_{s}(x)$ with positive even integers $s$ the only problematic term is

$\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)|x|^{s-1}$

which has a removable singularity. Similarly, for $S_{s}(x)$ with positive odd
integers $s$. For $C_{s}(x)$ with positive odd integers $s$ and $S_{s}$ with
positive even integers $s$ the singularities are not removable. The only case
we encounter in the paper is for $C_{s}$, for which we make use of the
following lemma to bound the sum of the two singular terms.

######  Lemma C.3.

Let $m\geq 1$ and $s\in\left[2m+\frac{1}{2},2m+\frac{3}{2}\right]$, then

$\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)|x|^{s-1}+(-1)^{m}\zeta(s-2m)\frac{x^{2m}}{(2m)!}=K_{1}|x|^{s-1}+K_{2}x^{2m}+K_{3}x^{2m}\frac{|x|^{s-(2m+1)}-1}{s-(2m+1)}$

with

$\displaystyle K_{1}$
$\displaystyle=\frac{\frac{\Gamma(2m+2-s)}{(1-s)_{2m}}\sin\left(\frac{\pi}{2}s\right)-\frac{(-1)^{m}}{(2m!)}}{2m+1-s},$
$\displaystyle K_{2}$
$\displaystyle=-\frac{(-1)^{m}\underline{\zeta}(s-2m)}{(2m)!},$ $\displaystyle
K_{3}$ $\displaystyle=-\frac{(-1)^{m}}{(2m)!}.$

###### Proof.

The result follows directly from adding and subtracting

$(-1)^{m}\frac{|x|^{s-1}}{(2m+1-s)(2m)!}$

and collecting the terms. ∎

Computing $K_{2}$ and $K_{3}$ is straight forward, for $K_{1}$ there is a
removable singularity to handle. The function

$x^{2m}\frac{|x|^{s-(2m+1)}-1}{s-(2m+1)}$

also has a removable singularity, to compute an accurate enclosure we use the
following lemma.

######  Lemma C.4.

For $x\not=0$ the function

$\frac{|x|^{t}-1}{t}$

is non-decreasing in $t$. In the limit $t\to 0$ it is equal to $\log|x|$.

###### Proof.

The derivative in $t$ is given by

$\frac{1+(t\log|x|-1)|x|^{t}}{t^{2}}.$ (44)

The sign depends on the numerator, which we can write as

$1+(t\log|x|-1)e^{t\log|x|}.$

Letting $v=t\log|x|$ we can write this as $1+(v-1)e^{v}$, which has the unique
root $v=0$ and is positive for other values of $v$. It follows that (44) is
non-negative and hence the function non-decreasing.

For the limit $t\to 0$ we directly get

$\lim_{t\to 0}\frac{|x|^{t}-1}{t}=\lim_{t\to 0}\frac{\log|x|\
|x|^{t}}{1}=\log|x|.$

∎

These two lemmas are only strictly required for interval s overlapping
positive odd integers. However, it gives better enclosures also when s is
close to such integers, even if it doesn’t overlap them. This is used to
compute better enclosures in some cases.

The tails for the asymptotic expansions are bounded using the following lemma,
see also [22, Lemma 2.1]. We omit the proof since it is very similar to that
in [22].

######  Lemma C.5.

Let $s\geq 0$, $2M\geq s+1$ and $|x|<2\pi$, we then have the following bounds
for the tails in equations (42) and (43)

$\displaystyle\left|\sum_{m=M}^{\infty}(-1)^{m}\zeta(s-2m)\frac{x^{2m}}{(2m)!}\right|$
$\displaystyle\leq
2(2\pi)^{1+s-2M}\left|\sin\left(\frac{\pi}{2}s\right)\right|\zeta(2M+1-s)\frac{x^{2M}}{4\pi^{2}-x^{2}},$
$\displaystyle\left|\sum_{m=M}^{\infty}(-1)^{m}\zeta(s-2m-1)\frac{x^{2m+1}}{(2m+1)!}\right|$
$\displaystyle\leq
2(2\pi)^{s-2M}\left|\cos\left(\frac{\pi}{2}s\right)\right|\zeta(2M+2-s)\frac{x^{2M+1}}{4\pi^{2}-x^{2}}.$

###  C.3 Derivatives in $s$

For $C_{s}^{(\beta)}(x)$ and $S_{s}^{(\beta)}(x)$ we use (39) and (40) and
differentiate directly in $s$. When $s$ is not an integer this is handled
directly using Taylor arithmetic, for integers we use the approach in Appendix
A to handle the removable singularities.

To get asymptotic expansions at $x=0$ we take the expansions (42) and (43) and
differentiate them with respect to $s$. Giving us

$\displaystyle C_{s}^{(\beta)}(x)$
$\displaystyle=\frac{d}{ds^{\beta}}\left(\Gamma(1-s)\sin\left(\frac{\pi}{2}s\right)|x|^{s-1}\right)+\sum_{m=0}^{\infty}(-1)^{m}\zeta^{(\beta)}(s-2m)\frac{x^{2m}}{(2m)!};$
(45) $\displaystyle S_{s}^{(\beta)}(x)$
$\displaystyle=\frac{d}{ds^{\beta}}\left(\Gamma(1-s)\cos\left(\frac{\pi}{2}s\right)\operatorname{sgn}(x)|x|^{s-1}\right)+\sum_{m=0}^{\infty}(-1)^{m}\zeta^{(\beta)}(s-2m-1)\frac{x^{2m+1}}{(2m+1)!}.$
(46)

These formulas work well when $s$ is not a positive odd integer for
$C_{s}^{(\beta)}(x)$ or a positive even integer for $S_{s}^{(\beta)}(x)$, and
the derivatives can be computed using Taylor expansions. We mostly make use of
the functions $C_{2}^{(1)}(x)$ and $C_{3}^{(1)}(x)$, in which case the
expansions can be computed explicitly using [3, Eq. 16], for $|x|<2\pi$ we
have

$\displaystyle C_{2}^{(1)}(x)=$
$\displaystyle\zeta^{(1)}(2)-\frac{\pi}{2}|x|\log|x|-(\gamma-1)\frac{\pi}{2}|x|+\sum_{m=1}^{\infty}(-1)^{m}\zeta^{(1)}(2-2m)\frac{x^{2m}}{(2m)!}$
$\displaystyle C_{3}^{(1)}(x)=$
$\displaystyle\zeta^{(1)}(3)-\frac{1}{4}x^{2}\log^{2}|x|+\frac{3-2\gamma}{4}x^{2}\log|x|-\frac{36\gamma-12\gamma^{2}-24\gamma_{1}-42+\pi^{2}}{48}x^{2}$
$\displaystyle+\sum_{m=2}^{\infty}(-1)^{m}\zeta^{(1)}(3-2m)\frac{x^{2m}}{(2m)!}.$

Where $\gamma_{n}$ is the Stieltjes constant and $\gamma=\gamma_{0}$. To bound
the tails we have the following lemma from [15].

######  Lemma C.6.

Let $\beta\geq 1$, $s\geq 0$, $2M\geq s+1$ and $|x|<2\pi$, we then have the
following bounds:

$\left|\sum_{m=M}^{\infty}(-1)^{m}\zeta^{(\beta)}(s-2m)\frac{x^{2m}}{(2m)!}\right|\\\
\leq\sum_{j_{1}+j_{2}+j_{3}=\beta}\binom{\beta}{j_{1},j_{2},j_{3}}2\left(\log(2\pi)+\frac{\pi}{2}\right)^{j_{1}}(2\pi)^{s-1}|\zeta^{(j_{3})}(1+2M-s)|\sum_{m=M}^{\infty}\left|p_{j_{2}}(1+2m-s)\left(\frac{x}{2\pi}\right)^{2m}\right|,$
$\left|\sum_{m=M}^{\infty}(-1)^{m}\zeta^{(\beta)}(s-2m-1)\frac{x^{2m+1}}{(2m+1)!}\right|\\\
\leq\sum_{j_{1}+j_{2}+j_{3}=\beta}\binom{\beta}{j_{1},j_{2},j_{3}}2\left(\log(2\pi)+\frac{\pi}{2}\right)^{j_{1}}(2\pi)^{s-2}|\zeta^{(j_{3})}(2+2M-s)|\sum_{m=M}^{\infty}\left|p_{j_{2}}(2+2m-s)\left(\frac{x}{2\pi}\right)^{2m+1}\right|.$

Here $p_{j_{2}}$ is given recursively by

$p_{k+1}(s)=\psi^{(0)}(s)p_{k}(s)+p^{\prime}_{k}(s),\quad p_{0}=1,$

where $\psi^{(0)}$ is the polygamma function. It is given by a linear
combination of terms of the form

$(\psi^{(0)}(s))^{q_{0}}(\psi^{(1)}(s))^{q_{1}}\cdots(\psi^{(j_{2}-1)}(s))^{q_{j_{2}-1}}.$

We have the following bounds

$\sum_{m=M}^{\infty}\left|(\psi^{(0)}(1+2m-s))^{q_{0}}\cdots(\psi^{(j_{2}-1)}(1+2m-s))^{q_{j_{2}-1}}\left(\frac{x}{2\pi}\right)^{2m}\right|\\\
\leq|(\psi^{(1)}(1+2M-s))^{q_{1}}\cdots(\psi^{(j_{2}-1)}(1+2M-s))^{q_{j_{2}-1}}|\frac{1}{2^{q_{0}/2}}(2\pi)^{-2M}\Phi\left(\frac{x^{2}}{4\pi^{2}},-\frac{q_{0}}{2},M+\frac{1}{2}\right)x^{2M}$

and

$\sum_{m=M}^{\infty}\left|(\psi^{(0)}(2+2m-s))^{q_{0}}\cdots(\psi^{(j_{2}-1)}(2+2m-s))^{q_{j_{2}-1}}\left(\frac{x}{2\pi}\right)^{2m+1}\right|\\\
\leq|(\psi^{(1)}(2+2M-s))^{q_{1}}\cdots(\psi^{(j_{2}-1)}(2+2M-s))^{q_{j_{2}-1}}|\frac{1}{2^{q_{0}/2}}(2\pi)^{-2M-1}\Phi\left(\frac{x^{2}}{4\pi^{2}},-\frac{q_{0}}{2},M+1\right)x^{2M+1}$

where

$\Phi(z,s,a)=\sum_{m=0}^{\infty}\frac{z^{m}}{(a+m)^{s}},$

is the Lerch transcendent.

For $\alpha\in I_{1}$ the expansions in $x$ of $u_{\alpha}$ and
$\mathcal{H}^{\alpha}[u_{\alpha}]$ are given in Lemma 4.6. Bounding the
remainder terms for the sums

$\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}a_{\alpha,0}\left(\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)\right)x^{2m}$

and

$\sum_{m=2}^{\infty}\frac{(-1)^{m}}{(2m)!}a_{\alpha,0}\left(\zeta(1-2\alpha-2m)-\zeta(2-\alpha+(1+\alpha)^{2}/2-2m)\right)x^{2m}$

requires a bound similar to that in Lemma C.6 above. Factoring out
$a_{\alpha,0}(1+\alpha)$, which is bounded, we are left bounding

$\sum_{m=1}^{\infty}\frac{(-1)^{m}}{(2m)!}\frac{\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)}{1+\alpha}x^{2m}$

and

$\sum_{m=2}^{\infty}\frac{(-1)^{m}}{(2m)!}\frac{\zeta(1-2\alpha-2m)-\zeta(2-\alpha+(1+\alpha)^{2}/2-2m)}{1+\alpha}x^{2m}.$

The following lemma allows us to write them in a form similar to that in Lemma
C.6.

######  Lemma C.7.

For $-1<\alpha<0$ and $M_{1}\geq 1$ we have

$\left|\sum_{m=M_{1}}^{\infty}\frac{(-1)^{m}}{(2m)!}\frac{\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)}{1+\alpha}x^{2m}\right|\\\
\leq\left|\sum_{m=M_{1}}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta^{\prime}(s_{1,m}-2m)x^{2m}\right|+\frac{1+\alpha}{2}\left|\sum_{m=M_{1}}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta^{\prime}(s_{2,m}-2m)x^{2m}\right|$

with $s_{1,m}\in[1-\alpha,2]$ and $s_{2,m}\in[2,2+(1+\alpha)^{2}/2]$.
Similarly, for $M_{2}\geq 2$,

$\left|\sum_{m=M_{2}}^{\infty}\frac{(-1)^{m}}{(2m)!}\frac{\zeta(1-2\alpha-2m)-\zeta(2-\alpha+(1+\alpha)^{2}/2-2m)}{1+\alpha}x^{2m}\right|\\\
\leq
2\left|\sum_{m=M_{2}}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta^{\prime}(s_{3,m}-2m)x^{2m}\right|+\left(1-\frac{1+\alpha}{2}\right)\left|\sum_{m=M_{2}}^{\infty}\frac{(-1)^{m}}{(2m)!}\zeta^{\prime}(s_{4,m}-2m)x^{2m}\right|$

with $s_{3,m}\in[1-2\alpha,3]$ and $s_{4,m}\in[2-\alpha+(1+\alpha)^{2}/2,3]$.

###### Proof.

The first inequality follows directly from that

$\frac{\zeta(1-\alpha-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)}{1+\alpha}=\frac{\zeta(1-\alpha-2m)-\zeta(2-2m)}{1+\alpha}+\frac{\zeta(2-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)}{1+\alpha}$

together with

$\frac{\zeta(1-\alpha-2m)-\zeta(2-2m)}{1+\alpha}=\zeta^{\prime}(s_{1,m}-2m)$

for some $s_{1,m}\in[1-\alpha,2]$ and

$\frac{\zeta(2-2m)-\zeta(2+(1+\alpha)^{2}/2-2m)}{1+\alpha}=\frac{1+\alpha}{2}\zeta^{\prime}(s_{2,m}-2m)$

for some $s_{2,m}\in[2,2+(1+\alpha)^{2}/2]$. The second one is similar. ∎

Lemma C.6 cannot be directly applied to the result of the above lemma since
the argument of the zeta function depends on $m$. This can however be fixed,
as the following lemma shows.

######  Lemma C.8.

Let $\beta\geq 1$, $\textbf{{s}}=[\underline{s},\overline{s}]$ with
$\underline{s}\geq 0$, $2M\geq\overline{s}+1$, $s_{m}\in\textbf{{s}}$ for
$m\geq M$ and $|x|<2\pi$, we then have the following bounds:

$\left|\sum_{m=M}^{\infty}(-1)^{m}\zeta^{(\beta)}(s_{m}-2m)\frac{x^{2m}}{(2m)!}\right|\\\
\leq\sum_{j_{1}+j_{2}+j_{3}=\beta}\binom{\beta}{j_{1},j_{2},j_{3}}2\left(\log(2\pi)+\frac{\pi}{2}\right)^{j_{1}}(2\pi)^{\overline{s}-1}|\zeta^{(j_{3})}(1+2M-\overline{s})|\sum_{m=M}^{\infty}\left|p_{j_{2}}(1+2m-s_{m})\left(\frac{x}{2\pi}\right)^{2m}\right|,$
$\left|\sum_{m=M}^{\infty}(-1)^{m}\zeta^{(\beta)}(s_{m}-2m-1)\frac{x^{2m+1}}{(2m+1)!}\right|\\\
\leq\sum_{j_{1}+j_{2}+j_{3}=\beta}\binom{\beta}{j_{1},j_{2},j_{3}}2\left(\log(2\pi)+\frac{\pi}{2}\right)^{j_{1}}(2\pi)^{s-2}|\zeta^{(j_{3})}(2+2M-s)|\sum_{m=M}^{\infty}\left|p_{j_{2}}(2+2m-s_{m})\left(\frac{x}{2\pi}\right)^{2m+1}\right|.$

Here $p_{j_{2}}$ is the same as in Lemma C.6 and given recursively by

$p_{k+1}(s)=\psi^{(0)}(s)p_{k}(s)+p^{\prime}_{k}(s),\quad p_{0}=1,$

where $\psi^{(0)}$ is the polygamma function. It is given by a linear
combination of terms of the form

$(\psi^{(0)}(s))^{q_{0}}(\psi^{(1)}(s))^{q_{1}}\cdots(\psi^{(j_{2}-1)}(s))^{q_{j_{2}-1}}.$

We have the following bounds

$\sum_{m=M}^{\infty}\left|(\psi^{(0)}(1+2m-s_{m}))^{q_{0}}\cdots(\psi^{(j_{2}-1)}(1+2m-s_{m}))^{q_{j_{2}-1}}\left(\frac{x}{2\pi}\right)^{2m}\right|\\\
\leq|(\psi^{(1)}(1+2M-\overline{s}))^{q_{1}}\cdots(\psi^{(j_{2}-1)}(1+2M-\overline{s}))^{q_{j_{2}-1}}|\frac{1}{2^{q_{0}/2}}(2\pi)^{-2M}\Phi\left(\frac{x^{2}}{4\pi^{2}},-\frac{q_{0}}{2},M+\frac{1}{2}\right)x^{2M}$

and

$\sum_{m=M}^{\infty}\left|(\psi^{(0)}(2+2m-s_{m}))^{q_{0}}\cdots(\psi^{(j_{2}-1)}(2+2m-s_{m}))^{q_{j_{2}-1}}\left(\frac{x}{2\pi}\right)^{2m+1}\right|\\\
\leq|(\psi^{(1)}(2+2M-\overline{s}))^{q_{1}}\cdots(\psi^{(j_{2}-1)}(2+2M-\overline{s}))^{q_{j_{2}-1}}|\frac{1}{2^{q_{0}/2}}(2\pi)^{-2M-1}\Phi\left(\frac{x^{2}}{4\pi^{2}},-\frac{q_{0}}{2},M+1\right)x^{2M}$

###### Proof.

The proof is the same as for Lemma C.6, and we therefore omit the details. The
only difference is that care has to be taken so that the termwise bounds holds
for all $s_{m}$. ∎

##  Appendix D Rigorous integration with singularities

In this section we discuss how to compute enclosures of the integrals
$U_{\alpha,1,1}(x)$, $U_{\alpha,1,2}(x)$ and $U_{\alpha,2}(x)$ in the non-
asymptotic case. We focus on the case when $w_{\alpha}(x)$ is not given by
$|x|$, since otherwise Lemma 11.5 allows us to directly compute the integral.
In particular this means that we don’t have to consider the case when
$\alpha\to 0$, though we have to handle $\alpha\to-1$.

Recall that

$\displaystyle U_{\alpha,1,1}(x)$
$\displaystyle=-x\int_{0}^{r_{\alpha,x}}\hat{I}_{\alpha}(x,t)w_{\alpha}(tx)\
dt,$ $\displaystyle U_{\alpha,1,2}(x)$
$\displaystyle=x\int_{r_{\alpha,x}}^{1}\hat{I}_{\alpha}(x,t)w_{\alpha}(tx)\
dt,$ $\displaystyle U_{\alpha,2}(x)$
$\displaystyle=\int_{x}^{\pi}I_{\alpha}(x,y)w_{\alpha}(y)\ dy.$

The integrand for $U_{1,2}$ has a (integrable) singularity at $t=1$ and the
integrand for $U_{2}$ has one at $y=x$. As a first step we split these off to
handle them separately.

Let

$\displaystyle U_{\alpha,1,2}(x)=$ $\displaystyle
x\int_{r_{\alpha,x}}^{1-\delta_{U,1}}\hat{I}_{\alpha}(x,t)w_{\alpha}(tx)\
dt+x\int_{1-\delta_{U,1}}^{1}\hat{I}_{\alpha}(x,t)w_{\alpha}(tx)\
dt=U_{\alpha,1,2,1}(x)+U_{\alpha,1,2,2}(x),$ $\displaystyle U_{\alpha,2}(x)=$
$\displaystyle\int_{x}^{x+\delta_{U,2}}I_{\alpha}(x,y)w_{\alpha}(y)\
dy+\int_{x+\delta_{U,2}}^{\pi}I_{\alpha}(x,y)w_{\alpha}(y)\
dy=U_{\alpha,2,1}(x)+U_{\alpha,2,2}(x).$

These integrals are handled by noticing that $w_{\alpha}$ is bounded on the
interval of integration. This allows us to compute an enclosure as

$\displaystyle U_{\alpha,1,2,2}(x)\in$ $\displaystyle\
xw_{\alpha}(x[1-\delta_{U,1},1])\int_{1-\delta_{U,1}}^{1}\hat{I}_{\alpha}(x,t)\
dt,$ $\displaystyle U_{\alpha,2,1}(x)\in$ $\displaystyle\
w_{\alpha}([x,x+\delta_{U,2}])\int_{x}^{x+\delta_{U,2}}I_{\alpha}(x,y)\ dy.$

The integrals can be computed explicitly, using that

$\displaystyle\int\hat{I}_{\alpha}(x,t)\ dt$
$\displaystyle=\frac{1}{x}\left(-S_{1-\alpha}(x(1-t))+S_{1-\alpha}(x(1+t))-2S_{1-\alpha}(xt)\right),$
$\displaystyle\int I_{\alpha}(x,y)\ dy$
$\displaystyle=-S_{1-\alpha}(x-y)+S_{1-\alpha}(x+y)-2S_{1-\alpha}(y).$

From which we get

$\int_{1-\delta_{U,1}}^{1}\hat{I}_{\alpha}(x,t)\
dt=\frac{1}{x}\Big{(}S_{1-\alpha}(2x)-2S_{1-\alpha}(x)+S_{1-\alpha}(x\delta_{U,1})-S_{1-\alpha}(x(2-\delta_{U,1}))+2S_{1-\alpha}(x(1-\delta_{U,1}))\Big{)}$

and

$\int_{x}^{x+\delta_{U,2}}I_{\alpha}(x,y)\
dy=-S_{1-\alpha}(-\delta_{U,2})+S_{1-\alpha}(2x+\delta_{U,2})-2S_{1-\alpha}(x+\delta_{U,2})-S_{1-\alpha}(2x)+2S_{1-\alpha}(x).$

For $\alpha\in I_{1}$ we use a slightly modified approach which gives better
bounds. Using that $w_{\alpha}(x)=|x|^{(1-\alpha)/2}\log(2e+1/|x|)$ in this
case we have

$U_{\alpha,1,2,2}(x)=x^{1+(1-\alpha)/2}\int_{1-\delta_{U,1}}^{1}\hat{I}_{\alpha}(x,t)t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt\\\ \in
x^{1+(1-\alpha)/2}[1-\delta_{U,1},1]^{-(1+\alpha)/2}\log(2e+1/(x[1-\delta_{U,1},1]))\int_{1-\delta_{U,1}}^{1}\hat{I}_{\alpha}(x,t)t\
dt$

and can use the same approach as in Lemma 11.5 for computing the integral.

What remains is computing $U_{\alpha,1,1}$, $U_{\alpha,1,2,1}$ and
$U_{\alpha,2,2}$. In this case the integrands are bounded everywhere on the
intervals of integration. To enclose the integrals we make use of two
different rigorous numerical integrators, depending on the value of $\alpha$.
For $\alpha\in I_{2}$ we use the rigorous numerical integrator implemented by
Arb [38]. For functions that are analytic on the interval the integrator uses
Gaussian quadratures with error bounds computed through complex magnitudes, we
therefore need to evaluate the integrands on complex intervals. When the
function is not analytic it falls back to naive enclosures using interval
arithmetic. For $\alpha\in I_{1}$ it is harder to enclose the integrand for
complex values, we therefore use a quadrature rule given by

$\int_{a}^{b}f(y)\
dy\in\frac{b-a}{2}\left(f(y_{1})+f(y_{2})\right)+\frac{1}{4320}(b-a)^{5}f^{(4)}([a,b]),$

where

$y_{1}=\frac{b+a}{2}+\frac{b-a}{2\sqrt{3}},\quad
y_{2}=\frac{b+a}{2}-\frac{b-a}{2\sqrt{3}},$

which doesn’t require us to evaluate on complex balls, see e.g. [46]. On
intervals where the function is not two times differentiable it falls back to
naive enclosures using interval arithmetic.

For $U_{\alpha,1,1}$ the integrand is bounded but not analytic at $t=0$.
Slightly more work is required to compute an enclosure of the integrand for
intervals containing $t=0$. The only non-trivial part of the integrand is the
term $C_{-\alpha}(xt)w_{\alpha}(xt)$ where the first factor diverges at $t=0$
and the second factor tends to zero. For $\alpha\in I_{2}$ this is handled by
expanding the Clausen function at zero, which allows us to simplify the terms
and bound them individually. For $\alpha\in I_{1}$ this doesn’t work directly
because the individual terms are not bounded as $\alpha\to-1$, instead we use
the following lemma.

######  Lemma D.1.

For $\alpha\in\left(-1,-\frac{1}{2}\right)$ and $x\in(0,\pi)$ the function

$C_{-\alpha}(xt)|xt|^{(1-\alpha)/2}\log(2e+1/|xt|)$

is increasing in $t$ on the interval $(0,t_{0})$ for

$t_{0}=\left(-\frac{2\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1}\left(-\alpha-\frac{1}{2}\right)}{\zeta(-\alpha)}\right)^{\frac{1}{\alpha+1}}.$

###### Proof.

For $t\geq 0$ can remove the absolute values and rewrite the function as

$C_{-\alpha}(xt)(xt)^{1/2}\cdot(xt)^{-\alpha/2}\log(2e+1/(xt)).$

It is enough to prove that the two factors both are increasing in $t$.

For the first factor, $C_{-\alpha}(xt)(xt)^{1/2}$, we expand the Clausen
function, giving us

$C_{-\alpha}(xt)(xt)^{1/2}=\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)(xt)^{-\alpha-1/2}+\sum_{m=0}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(xt)^{2m+1/2}}{(2m)!}.$

All terms in the sum with $m\geq 1$ are positive, due to the location of the
zeros of the zeta function on the negative real axis, and hence increasing in
$t$. We are left with proving that

$\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)(xt)^{-\alpha-1/2}+\zeta(-\alpha)(xt)^{1/2}$

is increasing in $t$. Differentiating with respect to $t$ gives us

$\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1/2}\left(-\alpha-\frac{1}{2}\right)t^{-\alpha-3/2}+\frac{1}{2}\zeta(-\alpha)x^{1/2}t^{-1/2}.$

Since we are only interested in the sign we can multiply by $t^{\alpha+3/2}$,
giving us

$\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1/2}\left(-\alpha-\frac{1}{2}\right)+\frac{1}{2}\zeta(-\alpha)x^{1/2}t^{\alpha+1}.$

The positiveness of the first term together with $1+\alpha>0$ means that it is
positive at $t=0$. The unique root for $t>0$ is given by

$\left(-\frac{2\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha-1}\left(-\alpha-\frac{1}{2}\right)}{\zeta(-\alpha)}\right)^{\frac{1}{\alpha+1}},$

which is exactly the value for $t_{0}$.

For the second factor, $(xt)^{-\alpha/2}\log(2e+1/(xt))$, differentiation with
respect to $t$ gives us

$-xt^{-\frac{\alpha}{2}-1}\frac{1+\alpha(1+2ext)\log(2e+1/(xt))}{2(1+2ext)}.$

Since $xt^{-\frac{\alpha}{2}-1}$ and $1+2ext$ are positive the sign is the
same as that for

$-(1+\alpha(1+2ext)\log(2e+1/(xt))).$

For $x,t>0$ a lower bound is given by $-1-2\alpha$, which is positive for
$\alpha<-\frac{1}{2}$ and hence for $\alpha\in I_{1}$. The derivative is hence
positive for all $t>0$ and the factor is always increasing. ∎

For the endpoint $r_{\alpha,x}$ of $U_{\alpha,1,1}$ and $U_{\alpha,1,2,1}$ we
don’t have an exact value but only an enclosure. The width of this enclosure
depends on how large the intervals for $\alpha$ and $x$ are. The integrators
we use don’t handle wide endpoints very well, it is therefore beneficial to
slightly modify the endpoints. If we have the enclosure
$r_{\alpha,x}\in[\underline{r}_{\alpha,x},\overline{r}_{\alpha,x}]$ then for
$U_{\alpha,1,1}$ we only integrate up to $\underline{r}_{\alpha,x}$ and for
$U_{\alpha,1,2,1}$ we integrate from $\overline{r}_{\alpha,x}$. The part we
have missed is given by

$\int_{\underline{r}_{\alpha,x}}^{\overline{r}_{\alpha,x}}|\hat{I}_{\alpha}(x,t)|w_{\alpha}(xt)\
dt.$

Since the interval centered at the root of $\hat{I}_{\alpha}(x,t)$ the
integrand will be very small. It is therefore enough to just compute a naive
enclosure given by the diameter of the interval times an enclosure of the
integrand on the interval.

##  Appendix E Details for evaluating $F_{\alpha}(x)$ with $\alpha\in I_{1}$
and $x$ near zero

We here explain how to bound (28), given by

$\frac{\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}}{\Gamma(1+\alpha)\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})x^{1-\alpha}}$

for $\alpha\in I_{1}$ and $x$ close to zero. This is used when bounding
$F_{\alpha}(x)$ for $x\in[0,\epsilon]$. To simplify the notation we in this
section let $p_{\alpha}=1+\alpha+(1+\alpha)^{2}/2$, note that
$p_{\hat{\alpha}}$ still denotes a numerical value, we also assume that
$0<x<1$.

From Lemma 4.6 we can get the expansion of
$\mathcal{H}^{\alpha}[u_{\alpha}](x)+\frac{1}{2}u_{\alpha}(x)^{2}$ at $x=0$.
To handle the cancellations between some of the terms in the expansion we
extract them and handle them separately.

To begin with we take out the leading terms in the expansions of $u_{\alpha}$
and $\mathcal{H}^{\alpha}[u_{\alpha}]$, namely

$\displaystyle P$
$\displaystyle=a_{\alpha,0}\left(\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)-\Gamma(-1-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1+(1+\alpha)^{2}/2)\right)x^{p_{\alpha}}\right)x^{-\alpha},$
$\displaystyle Q$
$\displaystyle=-a_{\alpha,0}\Big{(}\Gamma(2\alpha)\cos\left(\pi\alpha\right)$
$\displaystyle\quad\quad-\Gamma(-1+\alpha-(1+\alpha)^{2}/2)\cos\left(\frac{\pi}{2}(1-\alpha+(1+\alpha)^{2}/2)\right)x^{p_{\alpha}}$
$\displaystyle\quad\quad+\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-\alpha+(1+\alpha)^{2}/2))x^{2(1+\alpha)}\Big{)}x^{-2\alpha},$

and treat them together. For small values of $j$ the terms
$-a_{j}\tilde{C}_{1\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}$ in
$\mathcal{H}^{\alpha}[u_{\alpha}]$ have large cancellations between the
$x^{-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}$ and $x^{2}$ terms in their
expansion, we therefore also treat these separately. More precisely we
consider the terms

$\sum_{j=1}^{M}-\hat{A}_{\alpha,j}^{0}x^{-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}+\frac{a_{\hat{\alpha},j}\zeta(-1-\alpha-\hat{\alpha}+jp_{\hat{\alpha}})}{2}x^{2}$

Where $M\leq N_{0}$ is some fixed limit for which terms we treat separately.

###  E.1 First part

For the first part we are interested in the term

$\frac{Q+P^{2}/2}{\Gamma(1+\alpha)\log(1/x)(1-x^{p_{\alpha}})x^{1-\alpha}}$

with $P$ and $Q$ as above. If we let
$c(\alpha)=\Gamma(\alpha)\cos\left(\frac{\pi}{2}\alpha\right)$ we can write
$P$ and $Q$ as

$\displaystyle P$ $\displaystyle=a_{\alpha,0}(c(\alpha)-c(\alpha-
p_{\alpha})x^{p_{\alpha}})x^{-\alpha};$ $\displaystyle Q$
$\displaystyle=-a_{\alpha,0}(c(2\alpha)-c(2\alpha-
p_{\alpha})x^{p_{\alpha}})x^{-2\alpha}+\frac{a_{\alpha,0}}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha}))x^{2}.$

This gives us

$Q+P^{2}/2=a_{\alpha,0}\Big{(}(a_{\alpha,0}c(\alpha)^{2}/2-c(2\alpha))x^{-2\alpha}+(c(2\alpha-
p_{\alpha})-a_{\alpha,0}c(\alpha)c(\alpha-
p_{\alpha}))x^{-2\alpha+p_{\alpha}}\\\
+\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha}))x^{2}+\frac{a_{\alpha,0}}{2}c(\alpha-
p_{\alpha})^{2}x^{-2\alpha+2p_{\alpha}}\Big{)},$

where we can note that $2<-2\alpha+2p_{\alpha}$. The value of $a_{\alpha,0}$
is taken such that the first term is identically equal to zero, we are thus
left with

$Q+P^{2}/2=a_{\alpha,0}\Big{(}(c(2\alpha-
p_{\alpha})-a_{\alpha,0}c(\alpha)c(\alpha-
p_{\alpha}))x^{-2\alpha+p_{\alpha}}\\\
+\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha}))x^{2}+\frac{a_{\alpha,0}}{2}c(\alpha-
p_{\alpha})^{2}x^{-2\alpha+2p_{\alpha}}\Big{)}.$

Now, cancelling the $x^{1-\alpha}$ and reordering the terms a bit gives us

$\frac{Q+P^{2}/2}{\Gamma(1+\alpha)\log(1/x)(1-x^{p_{\alpha}})x^{1-\alpha}}=\frac{a_{\alpha,0}}{\Gamma(1+\alpha)}\Big{(}\frac{(c(2\alpha-
p_{\alpha})-a_{\alpha,0}c(\alpha)c(\alpha-
p_{\alpha}))x^{-1-\alpha+p_{\alpha}}}{\log(1/x)(1-x^{p_{\alpha}})}\\\
+\frac{\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha}))x^{1+\alpha}+\frac{a_{\alpha,0}}{2}c(\alpha-
p_{\alpha})^{2}x^{-1-\alpha+2p_{\alpha}}}{\log(1/x)(1-x^{p_{\alpha}})}\Big{)}.$

The factor $\frac{a_{\alpha,0}}{\Gamma(1+\alpha)}$ can be handled the same way
as in Lemma 8.2. We consider the two terms separately.

The first term we split as

$\frac{x^{-1-\alpha+p_{\alpha}}}{\log(1/x)}\frac{1+\alpha}{1-x^{p_{\alpha}}}\frac{c(2\alpha-
p_{\alpha})-a_{\alpha,0}c(\alpha)c(\alpha-p_{\alpha})}{1+\alpha}$

For the first factor we have $-1-\alpha+p_{\alpha}=(1+\alpha)^{2}/2>0$, so the
factor it is zero for $x=0$ and increasing in $x$, allowing us to compute an
enclosure. The second factor is also increasing in $x$, for $x=0$ it is
$1+\alpha$ and for non-zero $x$ we can handle the removable singularity in
$\alpha$. For the third factor we note that
$a_{\alpha,0}=2c(2\alpha)/c(\alpha)^{2}$, giving us

$\frac{c(2\alpha-p_{\alpha})-a_{\alpha,0}c(\alpha)c(\alpha-
p_{\alpha})}{1+\alpha}=\frac{c(2\alpha-p_{\alpha})-2c(2\alpha)c(\alpha-
p_{\alpha})/c(\alpha)}{1+\alpha}.$

This can also be computed by handling the removable singularities.

The second term we split in a similar way

$\frac{\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha}))x^{1+\alpha}+\frac{a_{\alpha,0}}{2}c(\alpha-
p_{\alpha})^{2}x^{-1-\alpha+2p_{\alpha}}}{\log(1/x)(1-x^{p_{\alpha}})}\\\
=\frac{x^{1+\alpha}}{2}\frac{1+\alpha}{1-x^{p_{\alpha}}}\frac{\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha}))+\frac{a_{\alpha,0}}{2}c(\alpha-
p_{\alpha})^{2}x^{(1+\alpha)^{2}}}{(1+\alpha)\log(1/x)},$

where in the last exponent we have used
$-2-\alpha+2p_{\alpha}=(1+\alpha)^{2}$. The first factor is easily handled,
and the second factor is the same as in the previous term. For the third
factor we let

$\displaystyle v_{\alpha}$
$\displaystyle=\frac{1}{2}(\zeta(-1-2\alpha)-\zeta(-1-2\alpha+p_{\alpha})),$
$\displaystyle w_{\alpha}$ $\displaystyle=\frac{a_{\alpha,0}}{2}c(\alpha-
p_{\alpha})^{2}.$

Allowing us to write it as

$\frac{v_{\alpha}+w_{\alpha}x^{(1+\alpha)^{2}}}{(1+\alpha)\log(1/x)}=\frac{v_{\alpha}-w_{\alpha}}{1+\alpha}\frac{1}{\log(1/x)}+(1+\alpha)w_{\alpha}\frac{1-x^{(1+\alpha)^{2}}}{(1+\alpha)^{2}\log(1/x)}$

The first factor in the first term has a removable singularity we can handle,
and its second factor is easily enclosed. For the second term we can enclose
$(1+\alpha)w_{\alpha}$ which has a removable singularity at $\alpha=-1$. For
the remaining factor we let $t=(1+\alpha)^{2}\log(x)$ and rewrite it as

$\frac{1-x^{(1+\alpha)^{2}}}{(1+\alpha)^{2}\log(1/x)}=\frac{x^{(1+\alpha)^{2}}-1}{(1+\alpha)^{2}\log(x)}=\frac{e^{t}-1}{t}.$
(47)

This function is increasing in $t$, so it is enough to get an enclosure of $t$
and evaluate at the endpoints of $t$.

###  E.2 Second part

For the second part we are interested in the term

$\sum_{j=1}^{M}\frac{-\hat{A}_{\alpha,j}^{0}x^{-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}+\frac{a_{\hat{\alpha},j}\zeta(-1-\alpha-\hat{\alpha}+jp_{\hat{\alpha}})}{2}x^{2}}{\Gamma(1+\alpha)\log(1/x)(1-x^{p_{\alpha}})x^{1-\alpha}}=\frac{1}{\Gamma(1+\alpha)(1-x^{p_{\alpha}})}\sum_{j=1}^{M}\frac{-\hat{A}_{\alpha,j}^{0}x^{-\alpha-\hat{\alpha}+jp_{\hat{\alpha}}}+\frac{a_{\hat{\alpha},j}\zeta(-1-\alpha-\hat{\alpha}+jp_{\hat{\alpha}})}{2}x^{2}}{\log(1/x)x^{1-\alpha}}.$

We begin by noting that

$\frac{1}{\Gamma(1+\alpha)(1-x^{p_{\alpha}})}$

is increasing in $x$. For $x=0$ it is lower bounded by $0$ and increasing in
$\alpha$. For $x>0$ it has a removable singularity at $\alpha=-1$ which we can
handle. What remains is to handle the sum, this we do term by term.

Let $1\leq j\leq M$ and
$r_{j}=-1-\hat{\alpha}+jp_{\hat{\alpha}}=(j-1)(1+\hat{\alpha})+(1+\hat{\alpha})^{2}/2$,
note that $r_{j}>0$, this gives us

$\frac{-\hat{A}_{\alpha,j}^{0}x^{1-\alpha+r_{j}}+\frac{a_{j}\zeta(-\alpha+r_{j})}{2}x^{2}}{\log(1/x)x^{1-\alpha}}=\frac{-\hat{A}_{\alpha,j}^{0}x^{r_{j}}+\frac{a_{\hat{\alpha},j}\zeta(-\alpha+r_{j})}{2}x^{1+\alpha}}{\log(1/x)}.$

Note that which exponent is the largest depends on the precise choice of both
$j$ and $\alpha$ and is in general not the same for all $\alpha\in I_{1}$.
Using that

$\hat{A}_{\alpha,j}^{0}=\Gamma(\alpha+\hat{\alpha}-jp_{\hat{\alpha}})\cos\left((\alpha+\hat{\alpha}-jp_{\hat{\alpha}})\frac{\pi}{2}\right)a_{\hat{\alpha},j}=\Gamma(\alpha-1-r_{j})\cos\left((\alpha-1-r_{j})\frac{\pi}{2}\right)a_{\hat{\alpha},j}$

we get

$a_{\hat{\alpha},j}\frac{-\Gamma(\alpha-1-r_{j})\cos\left((\alpha-1-r_{j})\frac{\pi}{2}\right)x^{r_{j}}+\frac{\zeta(-\alpha+r_{j})}{2}x^{1+\alpha}}{\log(1/x)}.$

The $a_{\hat{\alpha},j}$ is a constant, we therefore focus on the rest. Adding
and subtracting $\frac{\zeta(-\alpha+r_{j})}{2}x^{r_{j}}$ we can write this as

$\left(-\Gamma(\alpha-1-r_{j})\cos\left((\alpha-1-r_{j})\frac{\pi}{2}\right)+\frac{\zeta(-\alpha+r_{j})}{2}\right)\frac{x^{r_{j}}}{\log(1/x)}+\frac{\zeta(-\alpha+r_{j})}{2}\frac{x^{1+\alpha}-x^{r_{j}}}{\log(1/x)}.$

For the first term we can easily enclose the factor depending on $x$ using
that it is monotone. The factor depending on $\alpha$ we either enclose
directly, if $-\alpha-1-r_{j}\not=0$, or handle the removable singularity
otherwise.

For the second term we use the deflated zeta function (41) to split it further
into two terms

$\frac{\underline{\zeta}(-\alpha+r_{j})}{2}\frac{x^{1+\alpha}-x^{r_{j}}}{\log(1/x)}-\frac{1}{2(1+\alpha-
r_{j})}\frac{x^{1+\alpha}-x^{r_{j}}}{\log(1/x)}.$

The first term can be enclosed directly. The second term we write as

$\frac{x^{r_{j}}-x^{1+\alpha}}{2(r_{j}-(1+\alpha))\log(1/x)}.$

We first consider the case when $r\geq 1+\alpha$, we then factor out
$x^{1+\alpha}$, giving us

$\frac{x^{1+\alpha}}{2}\frac{x^{r_{j}-(1+\alpha)}-1}{(r_{j}-(1+\alpha))\log(1/x)}=\frac{x^{1+\alpha}}{2}\frac{1-x^{r_{j}-(1+\alpha)}}{(r_{j}-(1+\alpha))\log(x)}$

If we let $t=(r_{j}-(1+\alpha))\log(x)$ the second factor can be written as

$\frac{1-e^{t}}{t}$

which we handle similarly to (47), note that since $r_{j}\geq 1+\alpha$ and
$x<1$ we have $t\leq 0$.

Next we consider the case when $r_{j}<1+\alpha$, depending on the value of
$r_{j}$ this case might or might not occur. We then factor out $x^{r_{j}}$,
giving us

$\frac{x^{r_{j}}}{2}\frac{1-x^{1+\alpha-
r_{j}}}{(r_{j}-(1+\alpha))\log(1/x)}=\frac{x^{r_{j}}}{2}\frac{1-x^{1+\alpha-
r_{j}}}{(1+\alpha-r_{j})\log(x)}.$

We here let $t=1+\alpha-r_{j}$ and the procedure is then the same as the
previous case.

###  E.3 Remaining part

Similarly to in the above section we factor out
$\frac{1}{\Gamma(1+\alpha)(1-x^{p_{\alpha}})}$ and bound it separately. For
the terms in the expansion we note that they all have an exponent greater than
$1-\alpha$, we can thus cancel the $x^{1-\alpha}$ in the denominator directly.
It is then possible to enclose the terms in the numerator and the factor
$\frac{1}{\log(1/x)}$ separately. For some terms we can however improve the
computed enclosures by incorporating the division by $\log(1/x)$. These are
the terms coming from $u_{\alpha}(x)^{2}$ containing a factor $P$. In that
case we want to enclose the factor

$\frac{a_{\alpha,0}(c(\alpha)-c(\alpha-
p_{\alpha})x^{p_{\alpha}})}{\log(1/x)}.$

This can be done using the following lemma and combining it with Lemma C.4 to
enclose $\frac{x^{p_{\alpha}}-1}{p_{\alpha}\log(1/x)}$.

######  Lemma E.1.

The factor

$a_{\alpha,0}(c(\alpha)-c(\alpha-p_{\alpha})x^{p_{\alpha}}).$

can be written as

$C_{1}x^{p_{\alpha}}-C_{2}\frac{x^{p_{\alpha}}-1}{p_{\alpha}}$

with

$C_{1}=a_{\alpha,0}(c(\alpha)-c(\alpha-p_{\alpha}))\quad\text{ and }\quad
C_{2}=a_{\alpha,0}c(\alpha)p_{\alpha}.$

Both $C_{1}$ and $C_{2}$ have removable singularities at $\alpha=-1$ and are
finite there.

###### Proof.

The expression follows directly by adding and subtracting
$a_{\alpha,0}c(\alpha)x^{p_{\alpha}}$ and collecting the terms. The removable
singularities of $C_{1}$ and $C_{2}$ can be handled by writing them as

$\displaystyle C_{1}$
$\displaystyle=a_{\alpha,0}(1+\alpha)\cdot\frac{c(\alpha)-c(\alpha-
p_{\alpha})}{1+\alpha},$ $\displaystyle C_{2}$
$\displaystyle=a_{\alpha,0}(1+\alpha)\cdot c(\alpha)(1+(1+\alpha)/2)$

and enclosing all the factors separately. ∎

##  Appendix F Details for evaluating $\mathcal{T}_{\alpha}(x)$ with
$\alpha\in I_{1}$ and $x$ near zero

In this section we give bounds for $G_{\alpha,1}$, $G_{\alpha,2}$ and
$R_{\alpha}$ occurring in Lemma 11.7.

###  F.1 $G_{\alpha,1}$

Recall that

$G_{\alpha,1}(x)=\frac{\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})}.$

To bound it we use the following lemma.

######  Lemma F.1.

For $\alpha\in(-1,0)$ and $x<1$, $G_{\alpha,1}$ satisfies

$G_{\alpha,1}(x)\leq\frac{1+\alpha}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\left(\frac{\log(2e+1/x)}{\log(1/x)}J_{\alpha,1}-\frac{1}{\log(1/x)}J_{\alpha,2}\right),$

where $J_{\alpha,1}$ and $J_{\alpha,2}$ are independent of $x$ and given by

$\displaystyle J_{\alpha,1}$
$\displaystyle=\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\
dt,$ $\displaystyle J_{\alpha,2}$
$\displaystyle=\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\log(t)\
dt.$

###### Proof.

Factoring out $1+\alpha$ we have

$G_{\alpha,1}(x)=\frac{1+\alpha}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}\\\
\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt.$

Let $J_{\alpha}(x)$ denote

$J_{\alpha}(x)=\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt$

Splitting the logarithm as

$\log(2e+1/(xt))=\log(1+2ext)-\log(x)-\log(t)$

we can split $J_{\alpha}(x)$ as

$J_{\alpha}(x)=\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\log(1+2ext)\
dt\\\
-\log(x)\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\
dt\\\
-\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\log(t)\
dt.$

Using that $\log(1+2ext)\leq\log(1+2ex)$ on the interval of integration we
have

$J_{\alpha}(x)\leq(\log(1+2ex)-\log(x))\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\
dt\\\
-\int_{0}^{1}\left|\frac{(1-t)^{-(1+\alpha)}+(1+t)^{-(1+\alpha)}-2t^{-(1+\alpha)}}{1+\alpha}\right|t^{(1-\alpha)/2}\log(t)\
dt=\log(2e+1/x)J_{\alpha,1}-J_{\alpha,2}$

and the result follows. ∎

The computation of $J_{\alpha,1}$ and $J_{\alpha,2}$ follow the same approach
as in Appendix D, rigorous numerical integration on most of the interval and
explicitly handling the singularity at $t=1$ by factoring out a bound for
$t^{(1-\alpha)/2}$ and $t^{(1-\alpha)/2}\log(t)$ respectively.

###  F.2 $G_{\alpha,2}$

Recall that

$G_{\alpha,2}(x)=\frac{\int_{1}^{\pi/x}\left((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right)t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt}{\log(1/x)(1-x^{1+\alpha+(1+\alpha)^{2}/2})}.$

Unfortunately it is not enough in this case to just factor out $1+\alpha$ and
bound the two factors independently, as for $G_{\alpha,1}$, the second factor
would not be bounded. Instead, we have to more carefully track the dependence
on $x$. We split the work from getting an upper bound into several lemmas. In
Lemma F.2 we split $G_{\alpha,2}$ into one main part, $G_{\alpha,2,M}$, and
one remainder part, $G_{\alpha,2,R}$. In Lemma F.3 we give a bound for
$G_{\alpha,2,M}$. The bound for the $G_{\alpha,2,R}$ is handled in Lemma F.4,
F.5 and F.6. These bounds work well for very small values of $x$, less than
around $10^{-10}$, but work poorly for larger values of $x$. The methods based
on direct integration, as discussed in Appendix D, work well for $x$ larger
than around $0.1$. In between these two intervals, $10^{-10}<x<0.$, we use a
slightly different approach for bounding $G_{\alpha,2}$. It is described at
the end of this section, with most of the details in Lemma F.7.

######  Lemma F.2.

For $\alpha\in(-1,0)$ and $x<1$, $G_{\alpha,2}(x)$ satisfies

$G_{\alpha,2}(x)=G_{\alpha,2,M}(x)+G_{\alpha,2,R}(x)$

with

$\displaystyle G_{\alpha,2,M}(x)=$
$\displaystyle\frac{1}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}\frac{4+2\alpha}{3}$
$\displaystyle\left(\left(D_{x}-\log
x-\frac{2}{3}\frac{1}{1+\alpha}\right)\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)+\log\left(\frac{\pi}{x}\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right),$
$\displaystyle G_{\alpha,2,R}(x)=$
$\displaystyle\frac{1}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}$
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}h_{n-k}(t)t\log(2e+1/(xt))\
dt,$

for some $D_{x}\in\left[-\log(1+2e\pi),\log(1+2e\pi)\right]$ and

$h_{k}(t)=\log(t-1)^{k}+\log(t+1)^{k}-2\log(t)^{k}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}.$

###### Proof.

Let us start by studying the integral in of $G_{\alpha,2}$, for which we use
the notation

$K_{\alpha}(x)=\int_{1}^{\pi/x}\left((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right)t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt.$

We have

$\left((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right)t^{(1-\alpha)/2}\\\
=t\left(e^{-(1+\alpha)(\log(t-1)+\log(t)/2)}+e^{-(1+\alpha)(\log(t+1)+\log(t)/2)}-2e^{-(1+\alpha)3\log(t)/2}\right)\\\
=t\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}g_{n}(t)$

with

$g_{n}(t)=\left(\log(t-1)+\frac{\log
t}{2}\right)^{n}+\left(\log(t+1)+\frac{\log
t}{2}\right)^{n}-2\left(\frac{3\log t}{2}\right)^{n}.$

Inserting this into the integral we can write it as

$K_{\alpha}(x)=\int_{1}^{\pi/x}t\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}g_{n}(t)\log(2e+1/(xt))\
dt.$

Switching the integral and the sum we get

$K_{\alpha}(x)=\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\int_{1}^{\pi/x}g_{n}(t)t\log(2e+1/(xt))\
dt.$

Next we want to look closer at

$I_{n}(x)=\int_{1}^{\pi/x}g_{n}(t)t\log(2e+1/(xt))\ dt.$

Using the binomial theorem and splitting $\frac{3\log t}{2}=\log t+\frac{\log
t}{2}$ we can write $g_{n}(t)$ as

$g_{n}(t)=\sum_{k=0}^{n-1}\binom{n}{k}\left(\frac{\log
t}{2}\right)^{k}\left(\log(t-1)^{n-k}+\log(t+1)^{n-k}-2\log(t)^{n-k}\right).$

This gives us

$I_{n}(x)=\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}\left(\log(t-1)^{n-k}+\log(t+1)^{n-k}-2\log(t)^{n-k}\right)t\log(2e+1/(xt))\
dt.$

Asymptotically as $t$ goes to infinity we have that

$\log(t-1)^{l}+\log(t+1)^{l}-2\log(t)^{l}$

behaves like

$\frac{l(l-1-\log t)\log(t)^{l-2}}{t^{2}}.$

That this indeed is the case is seen in the proof of Lemma F.6, and this will
be what gives the main part of the integral for small $x$. If we let

$h_{k}(t)=\log(t-1)^{k}+\log(t+1)^{k}-2\log(t)^{k}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}$

and

$\displaystyle I_{n,M}(x)=$
$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}\frac{(n-k)(n-k-1-\log
t)\log(t)^{n-k-2}}{t^{2}}t\log(2e+1/(xt))\ dt,$ $\displaystyle I_{n,R}(x)=$
$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}h_{n-k}(t)t\log(2e+1/(xt))\
dt,$

then we have

$I_{n}(x)=I_{n,M}(x)+I_{n,R}(x)$

and

$K_{\alpha}(x)=\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}I_{n,M}(x)+\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}I_{n,R}(x).$

Multiplying the second term with the factor

$\frac{1}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}$

gives us $G_{\alpha,2,R}$.

For $I_{n,M}(x)$ we note that

$\log(t)^{k}(n-k)(n-k-1-\log
t)\log(t)^{n-k-2}=(n-k)\left((n-k-1)\log(t)^{n-2}-\log(t)^{n-1}\right).$

Giving us

$\displaystyle I_{n,M}(x)=$
$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}(n-k)$
$\displaystyle\left((n-k-1)\int_{1}^{\pi/x}\frac{\log(t)^{n-2}}{t}\log(2e+1/(xt))\
dt-\int_{1}^{\pi/x}\frac{\log(t)^{n-1}}{t}\log(2e+1/(xt))\ dt\right)$
$\displaystyle=$
$\displaystyle\int_{1}^{\pi/x}\frac{\log(t)^{n-2}}{t}\log(2e+1/(xt))\
dt\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}(n-k)(n-k-1)$
$\displaystyle-\int_{1}^{\pi/x}\frac{\log(t)^{n-1}}{t}\log(2e+1/(xt))\
dt\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}(n-k).$

The sums can be computed explicitly,

$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}(n-k)(n-k-1)$
$\displaystyle=\left(\frac{3}{2}\right)^{n-2}n(n-1),$
$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}(n-k)$
$\displaystyle=\left(\frac{3}{2}\right)^{n-1}n.$

Inserting this back we have

$I_{n,M}(x)=\left(\frac{3}{2}\right)^{n-2}n\left((n-1)\int_{1}^{\pi/x}\frac{\log(t)^{n-2}}{t}\log(2e+1/(xt))\
dt-\frac{3}{2}\int_{1}^{\pi/x}\frac{\log(t)^{n-1}}{t}\log(2e+1/(xt))\
dt\right).$

Next we are interested in computing integrals of the form

$\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\log(2e+1/(xt))\ dt.$

Since $n\geq 1$ we need to consider $l\geq-1$. However, the only case for
which we have $l=-1$ is for $n=1$, in which case the factor $n-1$ in front is
zero. Hence, we only need to consider $l\geq 0$. Using that
${\log(2e+1/(xt))=\log(1+2ext)-\log(x)-\log(t)}$ we get

$\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\log(2e+1/(xt))\
dt=\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\log(1+2ext)\
dt-\log(x)\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\
dt-\int_{1}^{\pi/x}\frac{\log(t)^{l+1}}{t}\ dt.$

For $1\leq t\leq\pi/x$ we have $0\leq\log(1+2ext)\leq\log(1+2e\pi)$, hence

$\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\log(1+2ext)\
dt=D_{l,x}\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\ dt$

for some $D_{l,x}\in[0,\log(1+2e\pi)]$. This gives us

$\displaystyle\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\log(2e+1/(xt))\ dt$
$\displaystyle=(D_{l,x}-\log(x))\int_{1}^{\pi/x}\frac{\log(t)^{l}}{t}\
dt-\int_{1}^{\pi/x}\frac{\log(t)^{l+1}}{t}\ dt$
$\displaystyle=(D_{l,x}-\log(x))\frac{\log(\pi/x)^{l+1}}{l+1}-\frac{\log(\pi/x)^{l+2}}{l+2},$

valid for $l\geq 0$. Inserting this back into $I_{n,M}$ we get for $n\geq 2$

$\displaystyle I_{n,M}(x)=$
$\displaystyle\left(\frac{3}{2}\right)^{n-2}n\Bigg{(}(n-1)\left((D_{n-2,x}-\log(x))\frac{\log(\pi/x)^{n-1}}{n-1}-\frac{\log(\pi/x)^{n}}{n}\right)$
$\displaystyle-\frac{3}{2}\left((D_{n-1,x}-\log(x))\frac{\log(\pi/x)^{n}}{n}-\frac{\log(\pi/x)^{n+1}}{n+1}\right)\Bigg{)}$
$\displaystyle=$
$\displaystyle\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n-1}\Bigg{(}n(D_{n-2,x}-\log(x))-(n-1)\log(\pi/x)$
$\displaystyle-\frac{3}{2}(D_{n-1,x}-\log(x))\log(\pi/x)+\frac{3n}{2(n+1)}\log(\pi/x)^{2}\Bigg{)}.$

For $n=1$ we instead get

$\displaystyle I_{1,M}(x)=$
$\displaystyle\left(\frac{3}{2}\right)^{-1}\log(\pi/x)^{-1}\left(-\frac{3}{2}(D_{-1,x}-\log(x))\log(\pi/x)+\frac{3}{4}\log(\pi/x)^{2}\right).$

Inserting this into the sum for $n$ and splitting it in parts we get

$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}I_{n,M}(x)=$
$\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n-1}nD_{n-2,x}$
$\displaystyle-\log(x)\sum_{n=2}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n-1}n$
$\displaystyle-\sum_{n=2}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n}(n-1)$
$\displaystyle-\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}D_{n-1,x}$
$\displaystyle+\log(x)\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}$
$\displaystyle+\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n+1}\frac{n}{n+1}.$

We begin by noting that

$\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n-1}nD_{n-2,x}$
$\displaystyle=-(1+\alpha)\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}D_{n-1,x},$
$\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n-1}n$
$\displaystyle=-(1+\alpha)\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n},$
$\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-2}\log(\pi/x)^{n}(n-1)$
$\displaystyle=-(1+\alpha)\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n+1}\frac{n}{n+1},$

and that since $D_{l,x}\in[0,\log(1+2e\pi)]$ for all values of $l$ and $x$ we
have that

$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}D_{n-1,x}$
$\displaystyle=D_{x}\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}$

for some $D_{x}\in[-\log(1+2e\pi),\log(1+2e\pi)]$. With this we get

$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}I_{n,M}(x)=$
$\displaystyle-(D_{x}-\log(x))(2+\alpha)\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}$
$\displaystyle+(2+\alpha)\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n+1}\frac{n}{n+1}.$

Explicitly computing the two sums we have

$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n}$
$\displaystyle=\frac{2}{3}\sum_{n=1}^{\infty}(-1)^{n}\frac{\left(\frac{3}{2}(1+\alpha)\log(\pi/x)\right)^{n}}{n!}$
$\displaystyle=\frac{2}{3}\left(e^{-\frac{3}{2}(1+\alpha)\log(\pi/x)}-1\right)$
$\displaystyle=\frac{2}{3}\left(\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}-1\right),$
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}\left(\frac{3}{2}\right)^{n-1}\log(\pi/x)^{n+1}\frac{n}{n+1}$
$\displaystyle=-\left(\frac{2}{3}\right)^{2}\frac{1}{1+\alpha}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\left(\frac{3}{2}(1+\alpha)\log(\pi/x)\right)^{n+1}}{(n+1)!}n$
$\displaystyle=-\left(\frac{2}{3}\right)^{2}\frac{1}{1+\alpha}\left(1-\left(1+\frac{3}{2}(1+\alpha)\log(\pi/x)\right)e^{-\frac{3}{2}(1+\alpha)\log(\pi/x)}\right)$
$\displaystyle=-\left(\frac{2}{3}\right)^{2}\frac{1}{1+\alpha}\left(1-\left(1+\frac{3}{2}(1+\alpha)\log(\pi/x)\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right).$

This gives us

$\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}I_{n,M}(x)=-(D_{x}-\log
x)(2+\alpha)\frac{2}{3}\left(\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}-1\right)\\\
-(2+\alpha)\left(\frac{2}{3}\right)^{2}\frac{1}{1+\alpha}\left(1-\left(1+\frac{3}{2}(1+\alpha)\log(\pi/x)\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right).$

Simplifying we get

$\sum_{n=1}^{\infty}(-1)^{n}\frac{(1+\alpha)^{n}}{n!}I_{n,M}(x)=\frac{4+2\alpha}{3}\left(\left(D_{x}-\log
x-\frac{2}{3}\frac{1}{1+\alpha}\right)\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)+\log\left(\frac{\pi}{x}\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right).$

Multiplying this with the factor

$\frac{1}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}$

gives us $G_{\alpha,2,M}$. ∎

A bound for $G_{\alpha,2,M}(x)$ is given in the following lemma.

######  Lemma F.3.

For $\alpha\in(-1,0)$ and $x<\frac{1}{\pi^{3}}$, $G_{\alpha,2,M}$ satisfies
the following bound

$G_{\alpha,2,M}(x)\leq\frac{4+2\alpha}{3}\left(\frac{2\log(1+2e\pi)}{\log(1/x)}+1\right).$

###### Proof.

As a first step we split $G_{\alpha,2,M}(x)$ into two terms

$G_{\alpha,2,M}(x)=\frac{4+2\alpha}{3}\frac{1}{\log(1/x)}\frac{D_{x}}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)\\\
+\frac{4+2\alpha}{3}\frac{1}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}\left(\log\left(\frac{\pi}{x}\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}-\left(\log
x+\frac{2}{3}\frac{1}{1+\alpha}\right)\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)\right)$

For the first term we focus on the part

$\frac{D_{x}}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)=D_{x}\frac{1-e^{\frac{3}{2}(1+\alpha)(\log
x-\log\pi)}}{1-e^{(1+\alpha)(1+(1+\alpha)/2)\log x}}.$

Adding and subtracting $e^{(1+\alpha)(1+(1+\alpha)/2)\log x}$ to the numerator
we can write this as

$D_{x}\left(1+e^{(1+\alpha)(1+(1+\alpha)/2)\log
x}\frac{1-e^{\frac{3}{2}(1+\alpha)(\log
x-\log\pi)-(1+\alpha)(1+(1+\alpha)/2)\log
x}}{1-e^{(1+\alpha)(1+(1+\alpha)/2)\log x}}\right)\\\
=D_{x}\left(1+e^{(1+\alpha)(1+(1+\alpha)/2)\log
x}\frac{1-e^{(1+\alpha)\left(\frac{3}{2}(\log x-\log\pi)-(1+(1+\alpha)/2)\log
x\right)}}{1-e^{(1+\alpha)(1+(1+\alpha)/2)\log x}}\right)$

Since $(1+\alpha)(1+(1+\alpha)/2)\log x$ is negative this is bounded by

$D_{x}\left(1+\frac{1-e^{(1+\alpha)\left(\frac{3}{2}(\log
x-\log\pi)-(1+(1+\alpha)/2)\log x\right)}}{1-e^{(1+\alpha)(1+(1+\alpha)/2)\log
x}}\right)$

Since $x<\frac{1}{\pi^{3}}$ we have

$(1+\alpha)\left(\frac{3}{2}(\log x-\log\pi)-(1+(1+\alpha)/2)\log
x\right)>(1+\alpha)\left(\frac{3}{2}\left(\log x+\frac{1}{3}\log
x\right)-(1+(1+\alpha)/2)\log x\right)\\\ =(1+\alpha)(1-(1+\alpha)/2)\log x.$

From which we get

$e^{(1+\alpha)\left(\frac{3}{2}(\log x-\log\pi)-(1+(1+\alpha)/2)\log
x\right)}>e^{(1+\alpha)(1+(1+\alpha)/2)\log x}$

and hence

$\frac{1-e^{(1+\alpha)\left(\frac{3}{2}(\log x-\log\pi)-(1+(1+\alpha)/2)\log
x\right)}}{1-e^{(1+\alpha)(1+(1+\alpha)/2)\log x}}<1.$

From this we get the upper bound

$\frac{D_{x}}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)<2D_{x}.$

Together with the bound $D_{x}<\log(1+2e\pi)$ this gives us the first part of
the bound for $G_{\alpha,2,M}(x)$.

For the second term we want to prove that

$\frac{1}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}\left(\log\left(\frac{\pi}{x}\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}-\left(\log
x+\frac{2}{3}\frac{1}{1+\alpha}\right)\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)\right)$
(48)

is bounded by $1$. To begin with we note that

$\log\left(\frac{\pi}{x}\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}-\left(\log
x+\frac{2}{3}\frac{1}{1+\alpha}\right)\left(1-\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right)\\\
=\frac{1}{1+\alpha}\left(-\frac{2}{3}-(1+\alpha)\log
x+\left(\frac{2}{3}+(1+\alpha)\log(\pi)\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right).$

Allowing us to write (48) as

$\frac{1}{\left(1-x^{1+\alpha+(1+\alpha)^{2}/2}\right)(1+\alpha)\log
x}\left(\frac{2}{3}+(1+\alpha)\log
x-\left(\frac{2}{3}+(1+\alpha)\log(\pi)\right)\left(\frac{x}{\pi}\right)^{\frac{3}{2}(1+\alpha)}\right).$

If we let $s=(1+\alpha)\log x$ we can write it as

$\frac{1}{\left(1-e^{(1+(1+\alpha)/2)s}\right)s}\left(\frac{2}{3}+s-\left(\frac{2}{3}+(1+\alpha)\log\pi\right)e^{\frac{3}{2}s-\frac{3}{2}(1+\alpha)\log\pi}\right).$

Since the denominator is negative it is enough to verify that

$\frac{2}{3}+s-\left(\frac{2}{3}+(1+\alpha)\log\pi\right)e^{\frac{3}{2}s-\frac{3}{2}(1+\alpha)\log\pi}>\left(1-e^{(1+(1+\alpha)/2)s}\right)s$

to prove that the quotient is bounded by $1$. If we let $v=(1+\alpha)\log\pi$
and simplify the inequality we get

$\frac{2}{3}-\left(\frac{2}{3}-v\right)e^{\frac{3}{2}(s-v)}+se^{\frac{3+\alpha}{2}s}>0.$
(49)

The left-hand side is increasing in $v$, to see this we differentiate it with
respect to $v$ to get

$\frac{4+3v}{2}e^{\frac{3}{2}(s-v)}.$

Since $0<v$ this is positive and the value for the left-hand side of (49) is
hence lower bounded by the value for $v=0$, from which we get the inequality

$\frac{2}{3}-\frac{2}{3}e^{\frac{3}{2}s}+se^{\frac{3+\alpha}{2}s}>0.$ (50)

Note that $s=(1+\alpha)\log x<0$. We want to prove that the left-hand side is
decreasing in $s$ so that it is lower bounded by the value for $s=0$, which is
$0$. To see that the left-hand side of (50) is decreasing in $s$ we
differentiate it with respect to $s$, giving us

$-e^{\frac{3}{2}s}+e^{\frac{3+\alpha}{2}s}+s\frac{3+\alpha}{2}e^{\frac{3+\alpha}{2}s}=e^{\frac{3}{2}s}\left(e^{\frac{\alpha}{2}s}\left(1+s\frac{3+\alpha}{2}\right)-1\right).$

We want to check that

$e^{\frac{\alpha}{2}s}\left(1+s\frac{3+\alpha}{2}\right)-1<0.$ (51)

If we again differentiate the left-hand side with respect to $s$ we get

$\frac{1}{2}e^{\frac{\alpha}{2}s}\left(3+2\alpha+\frac{(3+\alpha)\alpha}{2}s\right).$

We have $3+2\alpha>0$ and $\frac{(3+\alpha)\alpha}{2}s>0$, so this is positive
and hence (51) is upper bounded by the value at $s=0$, where it is $0$. It
follows that the left-hand side of (50) is decreasing in $s$. With all of this
we have shown that (48) is bounded by $1$ and the result follows. ∎

To bound $G_{\alpha,2,R}(x)$ we factor out $1+\alpha$ from the sum and take
the absolute value termwise and move it inside the integral to get

$G_{\alpha,2,R}(x)\leq\frac{1+\alpha}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{1}{\log(1/x)}\\\
\sum_{n=1}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}|h_{n-k}(t)|t\log(2e+1/(xt))\
dt.$

Furthermore we use the bound $\log(2e+1/(xt))<\log(2e+1/x)$ to get

$G_{\alpha,2,R}(x)\leq\frac{1+\alpha}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}\frac{\log(2e+1/x)}{\log(1/x)}\sum_{n=1}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}|h_{n-k}(t)|t\
dt.$ (52)

We can bound

$\frac{1+\alpha}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}$

by handling the removable singularity at $\alpha=-1$ and the second factor by
writing it as

$\frac{\log(2e+1/x)}{\log(1/x)}=1+\frac{\log(1+2ex)}{\log(1/x)}.$

For the remaining part the first term in the sum, with $n=1$, is treated
explicitly in Lemma F.4. For the remaining terms we split the interval of
integration into two parts and bound them separately. More precisely we look
at the two terms

$\displaystyle G_{\alpha,2,R,1}(x)=$
$\displaystyle\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{2}\log(t)^{k}|h_{n-k}(t)|t\
dt,$ $\displaystyle G_{\alpha,2,R,2}(x)=$
$\displaystyle\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{2}^{\pi/x}\log(t)^{k}|h_{n-k}(t)|t\
dt$

which are handled in Lemma F.5 and F.6.

######  Lemma F.4.

For $\alpha\in(-1,0)$ and $x<1$ the first term for the sum in (52) is bounded
by $\frac{1}{2}$.

###### Proof.

The term is given by

$\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}\int_{1}^{\pi/x}\log(t)^{k}|h_{n-k}(t)|t\
dt.$

Inserting $n=1$ gives us

$\int_{1}^{\pi/x}|h_{1}(t)|t\ dt.$

We have

$h_{1}(t)=\log(t-1)+\log(t+1)-2\log(t)+\frac{1}{t^{2}},$

giving us

$\int_{1}^{\pi/x}|h_{1}(t)|t\
dt=\int_{1}^{\pi/x}\left|\log(t-1)+\log(t+1)-2\log(t)+\frac{1}{t^{2}}\right|t\
dt.$

Note that for $t>1$ we have

$h_{1}(t)=\log(1-1/t)+\log(1+1/t)+\frac{1}{t^{2}}=-\sum_{n=2}^{\infty}\frac{1}{nt^{2n}}<0,$

so the absolute value can be replaced with a minus sign. If we also integrate
to infinity we get the upper bound

$\int_{1}^{\pi/x}|h_{1}(t)|t\
dt\leq\int_{1}^{\infty}-\left(\log(t-1)+\log(t+1)-2\log(t)+\frac{1}{t^{2}}\right)t\
dt.$

The integral can be computed explicitly to be $\frac{1}{2}$, giving us the
required upper bound. ∎

######  Lemma F.5.

For $\alpha\in(-1,0)$ and $x<1$, $G_{\alpha,2,R,1}$ satisfies the following
bound

$G_{\alpha,2,R,1}(x)\leq
2\left(\sqrt{e}\frac{1+\alpha}{-\alpha}+9\frac{e^{3(1+\alpha)}-3(1+\alpha)-1}{3(1+\alpha)}+(2+\alpha)e^{3(1+\alpha)/2}-1\right).$

###### Proof.

As a first step our goal is to compute an upper bound of the integral

$H_{k}=\int_{1}^{2}\log(t)^{k}|h_{n-k}(t)|t\ dt.$

On the interval of integration we have

$\log(t)^{k}t\leq 2\log(2)^{k}$

and hence

$H_{k}\leq 2\log(2)^{k}\int_{1}^{2}\left|h_{n-k}(t)\right|\ dt.$

For $1<t<2$ we have

$\displaystyle\left|h_{k}(t)\right|$
$\displaystyle=\left|\log(t-1)^{k}+\log(t+1)^{k}-2\log(t)^{k}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}\right|$
$\displaystyle\leq(-1)^{k}\log(t-1)^{k}+\log(t+1)^{k}+2\log(t)^{k}+\frac{k(k-1)}{t^{2}}\log(t)^{k-2}+\frac{k}{t^{2}}\log(t)^{k-1}$
$\displaystyle\leq(-1)^{k}\log(t-1)^{k}+\log(3)^{k}+2\log(2)^{k}+k(k-1)\log(2)^{k-2}+k\log(2)^{k-1}.$

This, together with $\int_{1}^{2}(-1)^{k}\log(t-1)^{k}\ dt=k!$, gives us

$\int_{1}^{2}\left|h_{n-k}(t)\right|\
dt\leq(n-k)!+\log(3)^{n-k}+2\log(2)^{n-k}+(n-k)(n-k-1)\log(2)^{n-k-2}+(n-k)\log(2)^{n-k-1}.$

Combining this with $\log(2)<1<\log(3)<2$ and $n\geq 2$ we get

$H_{k}\leq 2\left((n-k)!+3\cdot 2^{n}+(n-k)^{2}\right).$

This gives us the upper bound

$G_{\alpha,2,R,1}(x)\leq
2\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\Bigg{(}\sum_{k=0}^{n-1}\binom{n}{k}\frac{(n-k)!}{2^{k}}+3\cdot
2^{n}\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}+\sum_{k=0}^{n-1}\binom{n}{k}\frac{(n-k)^{2}}{2^{k}}\Bigg{)}.$

Explicitly computing the inner sums we have

$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{(n-k)!}{2^{k}}$
$\displaystyle=\sqrt{e}n\Gamma\left(n,\frac{1}{2}\right)<\sqrt{e}n!,$
$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{1}{2^{k}}$
$\displaystyle=\left(\frac{3}{2}\right)^{n}-\frac{1}{2^{n}}<\left(\frac{3}{2}\right)^{n},$
$\displaystyle\sum_{k=0}^{n-1}\binom{n}{k}\frac{(n-k)^{2}}{2^{k}}$
$\displaystyle=\frac{n(2n+1)}{3}\left(\frac{3}{2}\right)^{n-1},$

here $\Gamma(n,z)$ is the upper incomplete gamma function, which satisfies
$\Gamma(n,z)<\Gamma(n)=(n-1)!$ for $z>0$. Inserting this back gives us

$G_{\alpha,2,R,1}(x)\leq
2\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\left(\sqrt{e}n!+3\cdot
2^{n}\left(\frac{3}{2}\right)^{n}+\frac{n(2n+1)}{3}\left(\frac{3}{2}\right)^{n-1}\right).$

Splitting it into the three sums we have

$G_{\alpha,2,R,1}(x)\leq
2\left(\sqrt{e}\sum_{n=2}^{\infty}(1+\alpha)^{n-1}+9\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}3^{n-1}+\frac{1}{3}\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}n(2n+1)\left(\frac{3}{2}\right)^{n-1}\right).$

For the individual sums we get

$\displaystyle\sum_{n=2}^{\infty}(1+\alpha)^{n-1}$
$\displaystyle=\frac{1+\alpha}{-\alpha},$
$\displaystyle\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}3^{n-1}$
$\displaystyle=\frac{e^{3(1+\alpha)}-3(1+\alpha)-1}{3(1+\alpha)},$
$\displaystyle\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}n(2n+1)\left(\frac{3}{2}\right)^{n-1}$
$\displaystyle=3(2+\alpha)e^{3(1+\alpha)/2}-3.$

Combining all of this we get the bound

$G_{\alpha,2,R,1}(x)\leq
2\left(\sqrt{e}\frac{1+\alpha}{-\alpha}+9\frac{e^{3(1+\alpha)}-3(1+\alpha)-1}{3(1+\alpha)}+(2+\alpha)e^{3(1+\alpha)/2}-1\right),$

which is what we wanted to prove. ∎

######  Lemma F.6.

For $\alpha\in(-1,0)$ and $x<1$, $G_{\alpha,2,R,2}$ satisfies the following
bound

$G_{\alpha,2,R,2}(x)<192\log(2)\frac{1+\alpha}{-\alpha}.$

###### Proof.

To bound the integral we need to better understand the behavior of $h_{k}(t)$.
Recall that we have

$h_{k}(t)=\log(t-1)^{k}+\log(t+1)^{k}-2\log(t)^{k}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}.$

As a first step we use that

$\displaystyle\log(t-1)^{k}+\log(t+1)^{k}-2\log(t)^{k}$
$\displaystyle=\left(\log(t)+\log(1-1/t)\right)^{k}+\left(\log(t)+\log(1+1/t)\right)^{k}-2\log(t)^{k}$
$\displaystyle=\sum_{j=0}^{k-1}\binom{k}{j}\log(t)^{j}\left(\log(1-1/t)^{k-j}+\log(1+1/t)^{k-j}\right)$

to get

$h_{k}(t)=\sum_{j=0}^{k-1}\binom{k}{j}\log(t)^{j}\left(\log(1-1/t)^{k-j}+\log(1+1/t)^{k-j}\right)-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}.$

Next, using that

$\log(1-1/t)=-t\sum_{i=0}^{\infty}\frac{1}{i+1}\frac{1}{t^{i}},\quad\log(1+1/t)=t\sum_{i=0}^{\infty}(-1)^{i}\frac{1}{i+1}\frac{1}{t^{i}},$

we can write it as

$h_{k}(t)=\sum_{j=0}^{k-1}\binom{k}{j}\log(t)^{j}\frac{1}{t^{k-j}}\left((-1)^{k-j}\left(\sum_{i=0}^{\infty}\frac{1}{i+1}\frac{1}{t^{i}}\right)^{k-j}+\left(\sum_{i=0}^{\infty}(-1)^{i}\frac{1}{i+1}\frac{1}{t^{i}}\right)^{k-j}\right)\\\
-\frac{k(k-1-\log t)\log(t)^{k-2}}{t^{2}}.$

Using formulas for powers of power series (see e.g. [28]) we have

$\displaystyle\left(\sum_{i=0}^{\infty}\frac{1}{i+1}\frac{1}{t^{i}}\right)^{j}$
$\displaystyle=\sum_{i=0}^{\infty}c_{i,j}\frac{1}{t^{i}},$
$\displaystyle\left(\sum_{i=0}^{\infty}(-1)^{i}\frac{1}{i+1}\frac{1}{t^{i}}\right)^{j}$
$\displaystyle=\sum_{i=0}^{\infty}d_{i,j}\frac{1}{t^{i}}.$

with

$c_{0,j}=1,\quad
c_{i,j}=\frac{1}{i}\sum_{l=1}^{i}\frac{lj-i+l}{l+1}c_{i-l,j},\ i\geq 1$

and

$d_{0,j}=1,\quad
d_{i,j}=\frac{1}{i}\sum_{l=1}^{i}(-1)^{l}\frac{lj-i+l}{l+1}d_{i-l,j},\ i\geq
1.$

By induction it can be checked that $d_{i,j}=(-1)^{i}c_{i,j}$, hence

$\left(\sum_{i=0}^{\infty}(-1)^{i}\frac{1}{i+1}\frac{1}{t^{i}}\right)^{j}=\sum_{i=0}^{\infty}(-1)^{i}c_{i,j}\frac{1}{t^{i}}.$

Inserting this back into $h_{k}(t)$ we get

$h_{k}(t)=\sum_{j=0}^{k-1}\binom{k}{j}\log(t)^{j}\frac{1}{t^{k-j}}\sum_{i=0}^{\infty}\left((-1)^{k-j}+(-1)^{i}\right)c_{i,k-j}\frac{1}{t^{i}}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}},$

which we can rewrite as

$h_{k}(t)=\sum_{j=0}^{k-1}\sum_{i=0}^{\infty}\binom{k}{j}\left((-1)^{k-j}+(-1)^{i}\right)c_{i,k-j}\log(t)^{j}\frac{1}{t^{k-j+i}}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}.$

Reordering the sums we can further write it as

$\displaystyle h_{k}(t)$
$\displaystyle=\sum_{m=0}^{\infty}\sum_{j=\max(k-m,0)}^{k-1}\binom{k}{j}\left((-1)^{k-j}+(-1)^{m+j-k}\right)c_{m+j-k,k-j}\log(t)^{j}\frac{1}{t^{m}}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}$
$\displaystyle=\sum_{m=1}^{\infty}\sum_{j=\max(k-m,0)}^{k-1}\binom{k}{j}\left((-1)^{k-j}+(-1)^{m+j-k}\right)c_{m+j-k,k-j}\log(t)^{j}\frac{1}{t^{m}}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}$
$\displaystyle=\sum_{m=1}^{\infty}\sum_{j=\max(k-m,0)}^{k-1}\binom{k}{j}(-1)^{k-j}\left(1+(-1)^{m}\right)c_{m+j-k,k-j}\log(t)^{j}\frac{1}{t^{m}}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}$
$\displaystyle=2\sum_{m=1}^{\infty}\sum_{j=\max(k-2m,0)}^{k-1}\binom{k}{j}(-1)^{k-j}c_{2m+j-k,k-j}\log(t)^{j}\frac{1}{t^{2m}}-\frac{k(k-1-\log
t)\log(t)^{k-2}}{t^{2}}.$

The $m=1$ term exactly cancels the subtracted term, giving us

$h_{k}(t)=2\sum_{m=2}^{\infty}\sum_{j=\max(k-2m,0)}^{k-1}\binom{k}{j}(-1)^{k-j}c_{2m+j-k,k-j}\log(t)^{j}\frac{1}{t^{2m}}.$

Taking the absolute value we have

$\left|h_{k}(t)\right|\leq
2\sum_{m=2}^{\infty}\sum_{j=\max(k-2m,0)}^{k-1}\binom{k}{j}c_{2m+j-k,k-j}\log(t)^{j}\frac{1}{t^{2m}}.$

Using that $\log(t)^{j}\leq 1+\log(t)^{k-1}$ and $t^{-2m}\leq
16t^{-4}2^{-2m}$for $t\geq 2$ we get the upper bound

$\left|h_{k}(t)\right|\leq
32\frac{1+\log(t)^{k-1}}{t^{4}}\sum_{m=2}^{\infty}\sum_{j=\max(k-2m,0)}^{k-1}\binom{k}{j}\frac{c_{2m+j-k,k-j}}{2^{2m}}.$

Note that the sum no longer depends on $t$, we denote it by $S_{k}$.

Inserting this back into $G_{\alpha,2,R,2}$ we get the upper bound

$G_{\alpha,2,R,2}(x)\leq
32\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{S_{n-k}}{2^{k}}\int_{2}^{\pi/x}\frac{\log(t)^{k}+\log(t)^{n-1}}{t^{3}}\
dt.$

The integral can be explicitly computed to be

$\int_{2}^{\pi/x}\frac{\log(t)^{k}+\log(t)^{n-1}}{t^{3}}\
dt=\frac{\Gamma(k+1,2\log(2))-\Gamma\left(k+1,2\log\left(\frac{\pi}{x}\right)\right)}{2^{k+1}}+\frac{\Gamma(n,2\log(2))-\left(n,2\log\left(\frac{\pi}{x}\right)\right)}{2^{n}}.$

The differences between the incomplete gamma functions are bounded by
$\Gamma(k+1)=k!$ and $\Gamma(n)=(n-1)!$ respectively, giving us

$\int_{2}^{\pi/x}\frac{\log(t)^{k}+\log(t)^{n-1}}{t^{3}}\
dt\leq\frac{k!}{2^{k+1}}+\frac{(n-1)!}{2^{n}}.$

For $0\leq k\leq n-1$ we have $\frac{k!}{2^{k+1}}<2\frac{(n-1)!}{2^{n}}$ and
hence

$\int_{2}^{\pi/x}\frac{\log(t)^{k}+\log(t)^{n-1}}{t^{3}}\ dt\leq
3\frac{(n-1)!}{2^{n}}.$

Inserting this back into $G_{\alpha,2,R,2}$ gives

$\displaystyle G_{\alpha,2,R,2}(x)$ $\displaystyle\leq
96\sum_{n=2}^{\infty}\frac{(1+\alpha)^{n-1}}{n!}\sum_{k=0}^{n-1}\binom{n}{k}\frac{S_{n-k}}{2^{k}}\frac{(n-1)!}{2^{n}}$
$\displaystyle=48\sum_{n=2}^{\infty}\left(\frac{1+\alpha}{2}\right)^{n-1}\frac{1}{n}\sum_{k=0}^{n-1}\binom{n}{k}\frac{S_{n-k}}{2^{k}}.$

The sum no longer depends on $x$ and to bound it we need to analyze

$S_{k}=\sum_{m=2}^{\infty}\sum_{j=\max(k-2m,0)}^{k-1}\binom{k}{j}\frac{c_{2m+j-k,k-j}}{2^{2m}}.$

The bound $\binom{k}{j}\leq 2^{k}$ gives us

$S_{k}\leq
2^{k}\sum_{m=2}^{\infty}\sum_{j=\max(k-2m,0)}^{k-1}\frac{c_{2m+j-k,k-j}}{2^{2m}}.$

Starting from $m=0$ and adding back the odd powers of $2$ gives us

$S_{k}\leq
2^{k}\sum_{m=0}^{\infty}\sum_{j=\max(k-m,0)}^{k-1}\frac{c_{m+j-k,k-j}}{2^{m}}.$

Reordering the sums we have

$\sum_{m=0}^{\infty}\sum_{j=\max(k-m,0)}^{k-1}\frac{c_{m+j-k,k-j}}{2^{m}}=\sum_{j=0}^{k-1}\frac{1}{2^{k-j}}\sum_{i=0}^{\infty}c_{i,k-j}\frac{1}{2^{i}}.$

From the definition of $c_{i,j}$ we get

$\sum_{i=0}^{\infty}c_{i,k-j}\frac{1}{2^{i}}=\left(\sum_{i=0}^{\infty}\frac{1}{i+1}\frac{1}{2^{i}}\right)^{k-j}=\left(-2\log(1-1/2)\right)^{k-j}=\log(4)^{k-j}.$

Inserting this back into $S_{k}$ we have

$S_{k}\leq
2^{k}\sum_{j=0}^{k-1}\left(\frac{\log(4)}{2}\right)^{k-j}=2^{k}\sum_{j=0}^{k-1}\log(2)^{k-j}=2^{k}\log(2)\frac{1-\log(2)^{k}}{1-\log(2)}.$

We have $\frac{1-\log(2)^{k}}{1-\log(2)}\leq 4$ and hence $S_{k}\leq
2^{k+2}\log(2)$. Inserting this back into $G_{\alpha,2,R,2}$ gives

$\displaystyle G_{\alpha,2,R,2}(x)$
$\displaystyle=48\sum_{n=2}^{\infty}\left(\frac{1+\alpha}{2}\right)^{n-1}\frac{1}{n}\sum_{k=0}^{n-1}\binom{n}{k}\frac{2^{k+2}\log(2)}{2^{k}}$
$\displaystyle=192\log(2)\sum_{n=2}^{\infty}\left(\frac{1+\alpha}{2}\right)^{n-1}\frac{1}{n}\sum_{k=0}^{n-1}\binom{n}{k}$
$\displaystyle=192\log(2)\sum_{n=2}^{\infty}\left(\frac{1+\alpha}{2}\right)^{n-1}\frac{1}{n}(2^{n}-1)$
$\displaystyle\leq 192\log(2)\sum_{n=2}^{\infty}(1+\alpha)^{n-1}$
$\displaystyle=192\log(2)\frac{1+\alpha}{-\alpha},$

which is what we wanted to prove. ∎

As mentioned in the beginning of the section the above methods work well for
very small values of $x$. For small, but not that small, values we have to
take another approach. We let

$G_{\alpha,2,1}(x)=\frac{1}{\log(1/x)}\int_{1}^{2}\frac{(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}}{1+\alpha}t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt$

and

$G_{\alpha,2,2}(x)=\frac{1}{\log(1/x)}\int_{2}^{\pi/x}\frac{(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}}{1+\alpha}t^{(1-\alpha)/2}\log(2e+1/(xt))\
dt,$

giving

$G_{\alpha,2}(x)=\frac{1+\alpha}{1-x^{1+\alpha+(1+\alpha)^{2}/2}}(G_{\alpha,2,1}(x)+G_{\alpha,2,2}(x)).$

The integral for $G_{\alpha,2,2}$ is bounded on the interval of integration
and can be computed using numerical integration, as discussed in Appendix D.
The only problematic part is evaluating

$\frac{(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}}{1+\alpha},$

which has a removable singularity at $\alpha=-1$, this can be handled as
described in Appendix A. The integral for $G_{\alpha,2,1}$ has a singularity
at $t=1$ and to compute it we use the following lemma.

######  Lemma F.7.

We have

$G_{\alpha,2,1}(x)=\frac{C}{\log(1/x)}\frac{1+2^{-\alpha}-3^{-\alpha}+\alpha(-4+5\cdot
2^{-\alpha}-2\cdot 3^{-\alpha})}{(\alpha-1)\alpha(\alpha+1)}$

for some

$C\in[2^{-(1+\alpha)/2}\log(2e+1/(2x)),\log(2e+1/x)].$

###### Proof.

For $t\in[1,2]$ we have

$\log(2e+1/(xt))\in[\log(2e+1/(2x)),\log(2e+1/x)].$

Since the integrand is positive it follows that

$G_{\alpha,2,2}(x)=\frac{C_{1}}{\log(1/x)}\int_{1}^{2}\frac{(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}}{1+\alpha}t^{(1-\alpha)/2}\
dt,$

for some $C_{1}\in[\log(2e+1/(2x)),\log(2e+1/x)]$. Furthermore we have
$t^{(1-\alpha)/2}=tt^{-(1+\alpha)/2}$ and

$t^{-(1+\alpha)/2}\in[2^{-(1+\alpha)/2},1].$

This gives us

$G_{\alpha,2,2}(x)=\frac{C_{1}C_{2}}{\log(1/x)}\int_{1}^{2}\frac{(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}}{1+\alpha}t\
dt,$

for some $C_{2}\in[2^{-(1+\alpha)/2},1]$. The integral can now be computed
explicitly to be

$\int_{1}^{2}\frac{(t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}}{1+\alpha}t\
dt=\frac{1+2^{-\alpha}-3^{-\alpha}+\alpha(-4+5\cdot 2^{-\alpha}-2\cdot
3^{-\alpha})}{(\alpha-1)\alpha(\alpha+1)}.$

Finally we note that if we let $C=C_{1}C_{2}$ then

$C\in[2^{-(1+\alpha)/2}\log(2e+1/(2x)),\log(2e+1/x)].$

This gives us the result. ∎

###  F.3 $R_{\alpha}$

Recall that

$R_{\alpha}(x)=2\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt.$

To begin with we can note that since $0<-\alpha<1$ we have
$(-1)^{m}\zeta(-\alpha-2m)>0$ for all $m=1,2,\dots$ due to the zeros of the
zeta function on the real line being exactly the even negative integers. As a
consequence all terms in the sum are positive. The following lemma gives a
bound

######  Lemma F.8.

For $\alpha\in(-1,0)$, $x<1$ and $M\geq 1$, $R_{\alpha}(x)$ satisfies the
following bound

$R_{\alpha}(x)<2\log(2e+1/\pi)\Bigg{(}\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
+\pi^{2}2(2\pi)^{1-\alpha-2M}\left|\sin\left(\frac{\pi}{2}\alpha\right)\right|\zeta(2M+1+\alpha)\frac{(\pi(1+1/\pi))^{2M}}{4\pi^{2}-(\pi(1+1/\pi))^{2}}\Bigg{)}.$

###### Proof.

Looking at the integral in the inner sum we write

$\log(2e+1/(xt))=\log(1+2ext)-\log(xt),$

giving us

$\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt=\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(1+2ext)\
dt-\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(xt)\ dt.$

For the first integral we compute an upper bound using that $\log(1+2ext)$ is
bounded by $\log(1+2e\pi)$, for the second integral we integrate directly.

$\displaystyle\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(1+2ext)\ dt$
$\displaystyle\leq\log(1+2e\pi)\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\ dt$
$\displaystyle=\log(1+2e\pi)\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{\pi}{x}\right)^{2k+1+(1-\alpha)/2}$
$\displaystyle\leq\frac{\log(1+2e\pi)}{2k+1+(1-\alpha)/2}\left(\frac{\pi}{x}\right)^{2(k+1)}.$

The second integral can be computed explicitly to be

$\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(xt)\
dt=\frac{(2k+1+(1-\alpha)/2)\log\pi-1}{(2k+1+(1-\alpha)/2)^{2}}\left(\frac{\pi}{x}\right)^{2k+1+(1-\alpha)/2},$

and we note that this is positive for $k\geq 0$. An upper bound for the full
integral is hence given by

$\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt\leq\frac{\log(2e+1/\pi)}{2k+1+(1-\alpha)/2}\left(\frac{\pi}{x}\right)^{2(k+1)}$

For the inner sum we thus get

$\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{\pi/x}t^{2k+(1-\alpha)/2}\log(2e+1/(xt))\
dt\leq\log(2e+1/\pi)\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{\pi}{x}\right)^{2(k+1)}.$

Factoring out $\left(\frac{\pi}{x}\right)^{2m}$ and inserting into
$R_{\alpha}$ we have

$R_{\alpha}(x)\leq
2\log(2e+1/\pi)\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}.$

Taking $M\geq 1$ we split this into one finite sum and one tail as

$R_{\alpha}(x)\leq
2\log(2e+1/\pi)\Bigg{(}\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
+\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\Bigg{)}.$

The finite sum can be enclosed directly, for the second sum we note that for
$x<1$ we have

$\displaystyle\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}$
$\displaystyle<\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1}\left(\frac{1}{\pi}\right)^{2(m-1-k)}$
$\displaystyle<\sum_{k=0}^{m}\binom{2m}{2k}\frac{1}{2k+1}\left(\frac{1}{\pi}\right)^{2(m-1-k)}$
$\displaystyle=\pi^{1-2m}\frac{(\pi+1)^{2m}-(\pi-1)^{2m}+\pi(\pi+1)^{2m}+\pi(\pi-1)^{2m}}{2(1+2m)}$
$\displaystyle\leq\pi^{1-2m}(\pi+1)^{2m}\frac{1+2\pi}{6}$
$\displaystyle=\pi^{2}\left(1+\frac{1}{\pi}\right)^{2m}.$

and hence

$\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
\leq\pi^{2}\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(\pi(1+1/\pi))^{2m}}{(2m)!}.$

The last sum is the same as for the tail of the Clausen function occurring in
Lemma C.5, since $\pi(1+1/\pi)<2\pi$ this gives us

$\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(\pi(1+1/\pi))^{2m}}{(2m)!}\leq
2(2\pi)^{1-\alpha-2M}\left|\sin\left(\frac{\pi}{2}\alpha\right)\right|\zeta(2M+1+\alpha)\frac{(\pi(1+1/\pi))^{2M}}{4\pi^{2}-(\pi(1+1/\pi))^{2}}.$

Finally we then get

$R_{\alpha}(x)<2\log(2e+1/\pi)\Bigg{(}\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+(1-\alpha)/2}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
+\pi^{2}2(2\pi)^{1-\alpha-2M}\left|\sin\left(\frac{\pi}{2}\alpha\right)\right|\zeta(2M+1+\alpha)\frac{(\pi(1+1/\pi))^{2M}}{4\pi^{2}-(\pi(1+1/\pi))^{2}}\Bigg{)},$

which is what we wanted to prove. ∎

##  Appendix G Details for evaluating $\mathcal{T}_{\alpha}(x)$ with $\alpha$
near $-1$ and $x$ near zero

In this section we discuss how to bound

$\frac{U_{\alpha}(x)}{\log(1/x)x^{-\alpha+p}}$

for small $x$ in the hybrid case when $\alpha$ is near $-1$, see Section
11.2.4.

In this case the weight is given by $w_{\alpha}(x)=|x|^{p}\log(2e+1/|x|)$,
giving us

$U_{\alpha}(x)=x^{1+p}\int_{0}^{\pi/x}|\hat{I}_{\alpha}(x,t)|t^{p}\log(2e+1/(xt))\
dt.$

Using that

$\log(2e+1/(xt))=\log(1+2ext)+log(1/x)+log(1/t)$

we split $U_{\alpha}$ into three parts

$\displaystyle U_{\alpha}^{1}(x)$
$\displaystyle=x^{1+p}\int_{0}^{\pi/x}|\hat{I}_{\alpha}(x,t)|t^{p}\log(1+2ext)\
dt,$ $\displaystyle U_{\alpha}^{2}(x)$
$\displaystyle=x^{1+p}\int_{0}^{\pi/x}|\hat{I}_{\alpha}(x,t)|t^{p}\ dt,$
$\displaystyle U_{\alpha}^{3}(x)$
$\displaystyle=x^{1+p}\int_{0}^{\pi/x}|\hat{I}_{\alpha}(x,t)|t^{p}\log(1/t)\
dt,$

satisfying

$U_{\alpha}(x)=U_{\alpha}^{1}(x)+\log(1/x)U_{\alpha}^{2}(x)+U_{\alpha}^{3}(x).$

The integral $U_{\alpha}^{2}$ is the same as what we would get with the weight
$w_{\alpha}(x)=|x|^{p}$, and we can therefore use Lemma 11.6 to bound it.

For $U_{\alpha}^{1}$ we can use that $\log(1+2ext)$ is bounded on the interval
of integration to factor it out. We then recover the same integral as for
$U_{\alpha}^{2}$, and we can bound it in the same way. If $x$ is very small
this work well, but for $x$ larger than around $10^{-5}$ this gives fairly
poor bounds. In that case we compute part of the integral using a rigorous
integrator instead.

The computation of $U_{\alpha}^{3}$ is the most complicate part. Similar to in
Lemma 11.6 we expand the integrand and split it into two main parts and one
remainder part as

$\displaystyle U_{\alpha,M,1}^{3}(x)$
$\displaystyle=\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha+p}\int_{0}^{1}\left|(1-t)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1}\right|t^{p}\log(1/t)\
dt$ $\displaystyle U_{\alpha,M,2}^{3}(x)$
$\displaystyle=\Gamma(1+\alpha)\sin\left(-\frac{\pi}{2}\alpha\right)x^{-\alpha+p}\int_{1}^{\pi/x}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\log(1/t)\
dt$ $\displaystyle U_{\alpha,R}^{3}(x)$
$\displaystyle=2x^{1+p}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\int_{0}^{\pi/x}t^{2k+p}\log(1/t)\
dt,$

with

$U_{\alpha}^{3}(x)\leq
U_{\alpha,M,1}^{3}(x)+U_{\alpha,M,2}^{3}(x)+U_{\alpha,R}^{3}(x).$

For $U_{\alpha,M,1}^{3}$ the integral doesn’t depend on $x$. We compute it by
using a rigorous integrator from $0$ to $0.999$ and from $0.999$ to $1$ we use
that $\log(1/t)$ is bounded to factor it out and integrate explicitly.

For $U_{\alpha,M,2}^{3}$ we note that $\log(1/t)$ is negative and the integral
is hence decreasing in $x$. To get an upper bound it is thus enough to work
with an upper bound of $x$. We split the interval of integration into several
parts that are all treated slightly different. We let

$\displaystyle U_{\alpha,M,2,1}^{3}(x)$
$\displaystyle=\int_{1}^{1.001}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\log(1/t)\
dt,$ $\displaystyle U_{\alpha,M,2,2}^{3}(x)$
$\displaystyle=\int_{1.001}^{10^{10}}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\log(1/t)\
dt,$ $\displaystyle U_{\alpha,M,2,3}^{3}(x)$
$\displaystyle=\int_{10^{10}}^{10^{50}}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\log(1/t)\
dt,$ $\displaystyle U_{\alpha,M,2,4}^{3}(x)$
$\displaystyle=\int_{10^{50}}^{\pi/x}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\log(1/t)\
dt.$

If $\pi/x\geq 10^{50}$ we then have

$U_{\alpha,M,2}^{3}=U_{\alpha,M,2,1}^{3}+U_{\alpha,M,2,2}^{3}+U_{\alpha,M,2,3}^{3}+U_{\alpha,M,2,4}^{3}.$

If $\pi/x$ is less than $10^{50}$ we simply skip the integrals above $\pi/x$
and cut the last one off at $\pi/x$. The integral $U_{\alpha,M,2,2}^{3}$ is
computed using a rigorous integrator. For the remaining ones we use that

$\int_{a}^{b}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\log(1/t)\
dt\geq\log(1/a)\int_{a}^{b}((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\
dt$

and integrate explicitly. The primitive function is given by

$\int((t-1)^{-\alpha-1}+(1+t)^{-\alpha-1}-2t^{-\alpha-1})t^{p}\ dt\\\
=-\frac{t^{p-\alpha}}{\alpha-p}\left({}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;\frac{1}{t}\right)+{}_{2}F_{1}\left(1+\alpha,\alpha-p;1+\alpha-p;-\frac{1}{t}\right)-2\right).$

The ${}_{2}F_{1}$ functions are evaluated using the hypergeometric series,
valid when $1+\alpha-p$ is not an integer and $t>1$, with bounds from [42].

For $U_{\alpha,R}^{3}(x)$ we use the following lemma to compute a bound.

######  Lemma G.1.

Let $0<\epsilon<\frac{\pi}{2}$, for $\alpha\in(-1,0)$, $x\in[0,\epsilon]$ and
$-\alpha<p<1$ with $1+\alpha\not=p$ we have

$\frac{U_{\alpha,R}^{3}(x)}{x^{-\alpha+p}\log(1/x)}\leq\frac{2x^{2+\alpha-p}\pi^{p-1}}{\log(1/x)}\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{(2k+1+p)^{2}}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
-\frac{2x^{2+\alpha-p}\pi^{p-1}\log(\pi/x)}{\log(1/x)}\sum_{m=1}^{M-1}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
+\frac{6x^{2+\alpha-p}\pi^{p-1}}{\log(1/x)}\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(\frac{3\pi}{2})^{2m}}{(2m)!}.$

###### Proof.

We have

$\int_{0}^{\pi/x}t^{2k+p}\log(1/t)\
dt=\frac{1-(2k+1+p)\log(\pi/x)}{(2k+1+p)^{2}}\left(\frac{\pi}{x}\right)^{2k+1+p},$

giving us

$\frac{U_{\alpha,R}^{3}(x)}{x^{-\alpha+p}\log(1/x)}=\frac{2x^{1+\alpha}}{\log(1/x)}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{x^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1-(2k+1+p)\log(\pi/x)}{(2k+1+p)^{2}}\left(\frac{\pi}{x}\right)^{2k+1+p}.$

Using a similar approach as in Lemma 11.6 we can rewrite this as

$\frac{U_{\alpha,R}^{3}(x)}{x^{-\alpha+p}\log(1/x)}\\\
=\frac{2x^{2+\alpha-p}\pi^{p-1}}{\log(1/x)}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1-(2k+1+p)\log(\pi/x)}{(2k+1+p)^{2}}\left(\frac{x}{\pi}\right)^{2(m-1-k)}.$

Splitting it into two sums we can write it as

$\frac{U_{\alpha,R}^{3}(x)}{x^{-\alpha+p}\log(1/x)}=\frac{2x^{2+\alpha-p}\pi^{p-1}}{\log(1/x)}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{(2k+1+p)^{2}}\left(\frac{x}{\pi}\right)^{2(m-1-k)}\\\
-\frac{2x^{2+\alpha-p}\pi^{p-1}\log(\pi/x)}{\log(1/x)}\sum_{m=1}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}\left(\frac{x}{\pi}\right)^{2(m-1-k)}.$

We can treat the first few terms in the sums explicitly, what remains is
bounding the tails

$\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{(2k+1+p)^{2}}\left(\frac{x}{\pi}\right)^{2(m-1-k)}$
(53)

and

$\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}\left(\frac{x}{\pi}\right)^{2(m-1-k)}.$
(54)

For (54) the factor in front is negative, we hence only need to compute a
lower bound. Since all terms in the sum are positive it is trivially lower
bounded by zero.

For (53) we need an upper bound. We can notice that

$\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{\pi^{2m}}{(2m)!}\sum_{k=0}^{m-1}\binom{2m}{2k}\frac{1}{2k+1+p}\left(\frac{x}{\pi}\right)^{2(m-1-k)}$

gives an upper bound and that this sum also occurs in the proof of Lemma 11.6.
Following the same approach we therefore get that an upper bound is given by

$3\sum_{m=M}^{\infty}(-1)^{m}\zeta(-\alpha-2m)\frac{(\frac{3\pi}{2})^{2m}}{(2m)!}.$

∎

## Acknowledgments

The author was partially supported by the ERC Starting Grant ERC-StG-
CAPA-852741 as well as MICINN (Spain) research grant number PID2021-
125021NA-I00. This material is based upon work supported by the National
Science Foundation under Grant No. DMS-1929284 while the author was in
residence at the Institute for Computational and Experimental Research in
Mathematics in Providence, RI, during the program “Hamiltonian Methods in
Dispersive and Wave Evolution Equations”. The author was partially supported
by the Swedish-American foundation for the visit. We are also thankful for the
hospitality of the Princeton Department of Mathematics and the Brown
University Department of Mathematics where parts of this paper were done. The
computations were enabled by resources provided by the National Academic
Infrastructure for Supercomputing in Sweden (NAISS) and the Swedish National
Infrastructure for Computing (SNIC) at the PDC Center for High Performance
Computing, KTH Royal Institute of Technology, partially funded by the Swedish
Research Council through grant agreements no. 2022-06725 and no. 2018-05973.
The author would like to thank Javier Gómez-Serrano for his guidance and Erik
Wahlén and Gabriele Brüll for fruitful discussions about highest waves for
related equations.

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