# Counting functions for sums of rational powers of integers

Trevor Wine

###### Abstract.

Counting functions are constructed for sums of integers raised to a fixed
positive rational power. That is, given values formed by
$u_{1}^{j/k}+u_{2}^{j/k}+...+u_{l}^{j/k}$, $u_{i}\in\mathbb{Z}^{+}$, the
number of values less than or equal to a given $w>0$ is determined. The
counting functions developed are framed in terms of convolution exponentials,
and are closely related to the Riemann zeta function. At the conclusion,
several estimates are derived, with special emphasis on the case of sums of
square roots, i.e. $j=1$, $k=2$.

###### Key words and phrases:

sums of integers raised to rational powers, sums of square roots of integers,
convolution exponential, Riemann zeta function, counting function

###### 2010 Mathematics Subject Classification:

11M06, 11N45

##  1\. Introduction

This paper develops counting functions for sums of positive rational powers,
$j/k$ ($k>1$, $j/k$ in lowest terms), of positive integers. In other words,
given the set
$\mathbb{M}_{j,k}\equiv\\{u_{1}^{j/k}+u_{2}^{j/k}+...+u_{l}^{j/k}|u\in\mathbb{Z}^{+}\\}$,
the number of elements at or below a given $w$ is determined (it should be
noted, as the set notation implies, the values in $\mathbb{M}$ are unique–for
example, $\sqrt{12}+\sqrt{48}$ and $\sqrt{3}+\sqrt{75}$ are counted only
once). More formally, we study the class of functions in $j$ and $k$ defined
by,

(1.1) $S_{j,k}(w)=\\#\\{v\in\mathbb{M}_{j,k}|v\leq w\\}.$

The main result of the paper is the following theorem for $S_{j,k}$:

######  Theorem 1.

$S_{j,k}(w)=\int_{t=0^{+}}^{w}\exp^{*}(dI_{j,k})dt,$

where $\exp^{*}$ denotes the convolution exponential, and where the
exponential power function, $I_{j,k}$, is given in terms of the Riemann zeta
function and its zeros,

$\begin{split}I_{j,k}(w)=&\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j}+\lim_{m\to\infty}\stackrel{{\scriptstyle*}}{{\sum_{|\mbox{Im
}\rho|<Y_{m}/(jk)}}}\frac{\zeta(\rho/j+1)\zeta(\rho/k)}{\zeta^{\prime}(\rho)}\frac{{w}^{\rho/j}}{\rho}+\ln
w\\\ &+\gamma-j\left(1-\frac{1}{k}\right)\ln
2\pi+\mathcal{O}\left(w^{-3/(2j)+\epsilon}\right),\end{split}$

($\rho$ are the non-trivial zeros of the Riemann zeta function, $\zeta$; *
indicates this form for the summands is valid for simple zeta zeros only; the
$Y_{m}$ are specifically chosen to ensure convergence of the sum, as will be
shown; $\gamma$ is the Euler-Mascheroni constant).

The sum over the zeta zeros and the lower order terms may be bounded as
follows,

$I_{j,k}(w)=\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j}+\mathcal{O}(w^{1/j}),$

(where this bound may be improved, conditional on the Riemann hypothesis).

The theorem will be proved in sections 3.7 and 3.8, after the necessary
groundwork.

The main approach of the paper is to borrow the Euler product form for the
Riemann zeta function, and substitute a set of relevant ‘additive primes’, for
the usual multiplicative primes. Specifically, the counting functions,
$S_{j,k}(w)$, are built based on an analog to the familiar Riemann zeta
formulation as an Euler product,

(1.2) $\zeta(s)=\sum_{n}n^{-s}=\prod_{p}\left(1-p^{-s}\right)^{-1}\mbox{ Re
$s>1$},$

namely,

(1.3)
$Z_{j,k}(s)=1+\sum_{v\in\mathbb{M}_{j,k}}e^{-sv}=\prod_{i}\left(1-e^{-sn_{i}^{j/k}}\right)^{-1},$

where the $n_{i}$ are k-free integers, and region of convergence will be
determined below. The analog to the product form follows since just as any
integer can be written uniquely as $p_{1}^{j_{1}}...p_{t}^{j_{t}}$, the
$p_{i}$ multiplicative primes, so any value in $\mathbb{M}_{j,k}$ can be
written uniquely as $m_{1}n_{1}^{j/k}+...+m_{l}n_{l}^{j/k}$,
$m_{i}\in\mathbb{Z}^{+}$, where each $n_{i}$ is a _unique_ k-free number; the
set $\\{n_{i}^{j/k}\\}_{i}$ constitutes the desired ‘additive primes’. To help
see the legitimacy of this unique ‘prime factorization’, consider that for any
$u_{1}^{j/k}+u_{2}^{j/k}+...+u_{l}^{j/k}$, each $u_{i}$ can be uniquely
factored into a k-free, and a ‘k-full’ portion. Repeated addition among like
k-free terms (certainly at least when $u_{i}$ itself is k-free) can then yield
any integer multiple of a given $n_{i}^{j/k}$.

##  2\. $\mathbf{S_{j,k}(w)}$ as a convolution exponential

###  2.1. $\mathbf{Z_{j,k}(s)}$ product and sum forms.

To develop $Z_{j,k}$, both the product and sum forms in $(\ref{Z_jk})$ will be
shown to converge to the same analytic function in the region
$H\equiv\\{s|\mbox{Re }s=a>0\\}$.

To examine the product form first, a basic theorem on infinite products (see
[9], vol. 1) will be useful:

######  Theorem.

If every term of the sequence of functions $\\{v_{i}(s)\\}$ is analytic on
domain G, and if $v_{i}(s)\neq 1$ for all $s$ in G, and if there exists a
convergent series $\sum_{n}M_{n}$ such that $|v_{i}(s)|\leq M_{n}$ for all $s$
in G, then the product,

$\prod_{i}\left(1-v_{i}(s)\right)^{-1}$

converges uniformly on G to a nonvanishing analytic function $f(s)$.

Application of the theorem to the product form for $Z_{j,k}(s)$,

(2.1) $\prod_{i}\left(1-e^{-sn_{i}^{j/k}}\right)^{-1},$

follows easily when the bounding $M_{n}$ are applied over the domain
$H_{\delta}\equiv\\{s|\mbox{Re }s=a\geq\delta>0\\}$, as $M_{n}=e^{-\delta
n^{j/k}}$, producing

(2.2) $\sum_{n}M_{n}<\int_{x=0}^{\infty}e^{-\delta
x^{j/k}}dx=\frac{k}{j}\frac{1}{\delta^{k/j}}\int_{y=0}^{\infty}y^{k/j-1}e^{-y}dy=\frac{k}{j}\Gamma\left(\frac{k}{j}\right)\frac{1}{\delta^{k/j}},$

and completing the requirements of the theorem. The product form therefore
converges to an analytic function in the $H$ half-plane and does so uniformly
in any $H_{\delta}$.

To prove convergence of the sum form of $Z_{j,k}(s)$, consider
$s\in\mathbb{R}\bigcap H=x$, and notice

$1+\sum_{v\in\mathbb{M}_{j,k},v<N}e^{-xv}<\prod_{i=1}^{M}\left(1-e^{-xn_{i}^{j/k}}\right)^{-1}<\prod_{i=1}^{\infty}\left(1-e^{-xn_{i}^{j/k}}\right)^{-1}<B,$

where the leftmost inequality follows by choosing $M$ large enough in the
product to cover all $v<N$ in the sum, and the rightmost inequality follows
from convergence of the product. The infinite sum, being monotone and bounded,
converges. Then from a well known result for Dirichlet series, if the series
converges at $\mbox{Re}(s)=x$ then it converges in the half-plane
$\\{s|\mbox{Re}(s)>x\\}$ (see [9], vol. 2). This proves convergence of the sum
in $(\ref{Z_jk})$ in half-plane $H$, and by analytic continuation equals the
same analytic function as product form $(\ref{Z_product_form})$.

###  2.2. Initial construction of $\mathbf{S_{j,k}}$; deriving a form for
$\mathbf{\ln Z(s)}$.

Applying Perron’s formula to the Dirichlet series, $Z_{j,k}(s)$ from above,
the main formula for the counting functions $(\ref{Sjk})$ will be,

(2.3) $S_{j,k}(w)=\frac{1}{2\pi
i}\int_{s=a-i\infty}^{a+i\infty}e^{sw}\frac{e^{\ln Z(s)}}{s}ds-1,$

where $\mbox{Re }s>0$, and with the important restrictions that integer powers
are not allowed, and that the fraction $j/k$ must be in lowest form. The $-1$
correction factor prevents counting the extraneous $1$ in the sum form of
$(\ref{Z_jk})$, the value $v=0$ not being an element of $\mathbb{M}_{j,k}$.
(For details on applying Perron’s formula to Dirichlet series to extract
counting functions, see for example [6].) Note also that $e^{\ln Z(s)}$ has
been used instead of $Z(s)$ in the integrand of $(\ref{S_integral_1})$. This
is because, as we will see, a convenient expression for $\ln Z(s)$ is much
easier to derive than one for $Z(s)$.

Before developing an expression for $\ln Z_{j,k}(s)$, two of its sum forms
will be shown. The first follows from the Euler product form for $Z_{j,k}(s)$
of equation $(\ref{Z_jk})$. We get, on applying the logarithm,

$\ln
Z_{j,k}(s)=\sum_{i}-\ln(1-e^{-sn_{i}^{j/k}})=\sum_{i}\left(e^{-sn_{i}^{j/k}}+\frac{e^{-2sn_{i}^{j/k}}}{2}+\frac{e^{-3sn_{i}^{j/k}}}{3}+...\right)$
(2.4) $=\sum_{m,i}\frac{1}{m}e^{-smn_{i}^{j/k}}$

(where the condition to assure convergence of the Taylor expansion of the
logarithm, $|e^{-sn_{i}^{j/k}}|<1$, holds everywhere in the $\mbox{Re}^{+}$
half-plane $H$).

Another sum form for $(\ref{ln_Z_jk_sum_form})$ will be useful:

(2.5) $=\sum_{l=1}^{\infty}a_{l}e^{-s\sqrt[k]{l}},$

where

$a_{l}=\left\\{\begin{array}[]{ll}0&\mbox{ if $l\neq m^{k}n_{i}^{j}$ },\\\
\frac{1}{m}&\mbox{ if $l=m^{k}n_{i}^{j}$},\end{array}\right.$

In both sums, $m\in\mathbb{Z}^{+}$, and, as above, the $n_{i}$ index the
k-free integers.

To develop an analytic expression for $\ln Z(s)$, denoting $Q_{k}(t)$ as the
k-free counting function, and $dQ_{k}$ as having point masses of weight $1$ at
each k-free integer, we have the following Stieltjes integral,

(2.6) $\ln Z_{j,k}(s)=-\int_{t=0}^{\infty}\ln(1-e^{-st^{j/k}})dQ_{k}(t).$

Next, in a well-known result the k-free counting function may be written in
terms of the Riemann zeta function,

$Q_{k}(t)=\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(z)t^{z}}{\zeta(kz)z}dz$

when $\mbox{Re }z=x>1$. (This follows from an application of Perron’s formula
to,

(2.7) $\sum_{n}\frac{\mu_{k}(n)}{n^{s}}=\frac{\zeta(s)}{\zeta(ks)},$

where $\mu_{k}(n)$ denotes the characteristic function for the k-free integers
(not to be confused with the (signed) Mobius function), as may be readily
derived by writing each zeta function in its product form.) Equation
$(\ref{ln_Z_dQ_integral})$ may then be integrated by parts (noting
$Q_{k}(t)=0$ when $t<1$ and $\ln(1-e^{-st^{j/k}})Q_{k}(t)\rightarrow 0$ as
$t\rightarrow\infty$). Substituting $u=t^{j/k}$ we then have for have for
$(\ref{ln_Z_dQ_integral})$,

(2.8) $=\int_{u=0}^{\infty}\frac{s}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{u^{(k/j)z}}{e^{su}-1}\frac{\zeta(z)}{\zeta(kz)z}dzdu.$

To obtain an expression for $\ln Z(s)$, we’ll swap the order of integration.
The swap will be proved by first assuming it is true, then confirming the
result equals one of the sum forms for $\ln Z_{j,k}(s)$ just above. This
approach has the advantage of better illustrating the role of the resulting
tri-zeta term in the integrand, which is central to both $\ln Z_{j,k}(s)$ and
the eventual integral form for $I_{j,k}(w)$.

Swapping produces,

(2.9) $\frac{s}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(z)}{\zeta(kz)z}\int_{u=0}^{\infty}\frac{u^{(k/j)z}}{e^{su}-1}dudz=\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(z)}{\zeta(kz)z}\frac{1}{s^{(k/j)z}}\int_{v=0}^{s\infty}\frac{v^{(k/j)z}}{e^{v}-1}dvdz$

after the substitution $v=su$. Now by common results in Mellin transforms
(see, for example, [4]),

(2.10)
$\int_{0}^{\infty}\frac{y^{\beta-1}}{e^{y}-1}dy=\Gamma(\beta)\sum_{n=1}^{\infty}\frac{1}{n^{\beta}}=\Gamma(\beta)\zeta(\beta)$

where $\mbox{Re }\beta$ must be greater than $1$. Noting the integrand of
$(\ref{ln_Z_jk_post_swap})$ is bounded as

$\left|\frac{v^{(k/j)z}}{e^{v}-1}\right|<c_{0}\frac{\left(|s|R\right)^{(k/j)x}e^{(k/j)(\pi/2)|y|}}{e^{aR}}$

along the arc from $Rs$ to $R|s|$ where $a>0$, $R>>0$ and $v=re^{i\theta}$,
the contour integral along this arc goes to zero as $R\rightarrow\infty$,
producing by Cauchy’s theorem,

$\int_{v=0}^{s\infty}\frac{v^{(k/j)z}}{e^{v}-1}dv=\Gamma((k/j)z+1)\zeta((k/j)z+1).$

Putting it all together, with the change of variables $z/j\rightarrow z$, we
have,

(2.11) $\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{k\Gamma(kz)}{s^{kz}}dz$

where $\mbox{Re }z>1/j$ and $\mbox{Re }s>0$.

To now show the swap was valid and that $(\ref{ln_Z_jk_tri_zeta})$ in fact
equals $\ln Z_{j,k}(s)$, notice the tri-zeta term can be expanded as a double
series,

$\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}=\left(\sum_{n=1}^{\infty}\frac{\mu_{k}(n)}{n^{jz}}\right)\left(\sum_{m=1}^{\infty}\frac{1/m}{m^{kz}}\right).$

Since both series are absolutely convergent for $\mbox{Re}(z)>1/j$,
distributivity produces a convergent double series:

(2.12)
$\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}=\sum_{m,i}\frac{1}{m}\left(m^{k}n_{i}^{j}\right)^{-z}=\sum_{l}a_{l}l^{-z},$

where as usual, the $n_{i}$ index the k-frees, and $a_{l}$ is defined as in
equation $(\ref{ln_Z_sum_form})$. This produces, for
$(\ref{ln_Z_jk_tri_zeta})$,

$=\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\left(\sum_{l=1}^{\infty}a_{l}l^{-z}\right)\frac{k\Gamma(kz)}{s^{kz}}dz.$

This can readily be evaluated by swapping the order of summation and
integration, the swap justified by the dominated convergence theorem. (Note
that

$\sum_{l=1}^{\infty}\int_{z=x-i\infty}^{x+i\infty}\left|a_{l}l^{-z}\frac{k\Gamma(kz)}{s^{kz}}\right|dz=k\sum_{l}a_{l}l^{-x}\int_{z}\left|\frac{\Gamma(kz)}{s^{kz}}\right|dz<\infty,$

where the final inequality follows from the convergence of the gamma function
integral, noting that if $s=re^{i\theta}$ and $z=qe^{i\phi}$, then
$|s^{-(k/j)z}|<c_{1}e^{(k/j)y(\pi/2-\epsilon)}$ and
$\Gamma((k/j)z)<c_{2}e^{-(k/j)y\pi/2}$, where we’ve assumed
$|\arg\\{z,s\\}|<\pi$ for the estimates, and since $\mbox{Re }s>0$ means
$|\theta|<\pi/2$, while $|\phi|\rightarrow\pi/2$ in the limit of large $|y|$.
The dominated convergence theorem now may be applied to the partial sums in
$l$, using $g(z)=\sum_{l}ka_{l}l^{-x}|\Gamma(kz)s^{-kz}|$ as the dominating
function.)

The post-swapped form then is,

$\sum_{l=1}^{\infty}a_{l}\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}k\Gamma(kz)(s\sqrt[k]{l})^{-kz}dz.$

By common results on Mellin transforms (see for example [5]), provided
$\mbox{Re }s>0$,

$e^{-s}=\frac{1}{2\pi i}\int_{z=x-i\infty}^{x+i\infty}\Gamma(z)s^{-z}dz,$

producing for each summand integral,

$a_{l}\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}k\Gamma(kz)(s\sqrt[k]{l})^{-kz}dz=a_{l}e^{-s\sqrt[k]{l}}$

producing the series, $\sum_{l}a_{l}e^{-s\sqrt[k]{l}}$. Compare with
$(\ref{ln_Z_sum_form})$. This proves $(\ref{ln_Z_jk_tri_zeta})$ is a valid
expression for $\ln Z_{j,k}(s)$.

###  2.3. $\mathbf{I_{j,k}(w)}$ introduction and its formulation in terms of
the Riemann zeta function.

Define the function $I_{j,k}$ as the step function formed by Perron’s formula
applied to $\ln Z_{j,k}(s)$,

(2.13) $I_{j,k}(w)=\frac{1}{2\pi
i}\int_{s=a-i\infty}^{a+i\infty}\frac{e^{sw}}{s}\ln Z_{j,k}(s)ds,$

valid for $\mbox{Re }s>0$, $w>0$.

Combining this with the first sum form for $\ln Z_{j,k}(s)$,
$(\ref{ln_Z_jk_sum_form})$, $I_{j,k}(w)$ is a step function with value $0$ at
$w=0$, with jumps of height $1$ at every $n_{i}^{j/k}$, $n_{i}$ a k-free, and
in general of height $\frac{1}{m}$ at every $mn_{i}^{j/k}$, and where its
value at each jump is found by averaging the height of the step just above and
the step just below–that is in general,

(2.14)
$I_{j,k}(w)=\frac{1}{2}\left(\sum_{mn_{i}^{j/k}<w}\frac{1}{m}+\sum_{mn_{i}^{j/k}\leq
w}\frac{1}{m}\right)$

Now to develop the expression for $I_{j,k}$, combine $(\ref{I_function_ln_Z})$
with the integral form for $\ln Z_{j,k}(s)$ in (2.11) to give,

(2.15) $I_{j,k}(w)=\frac{1}{2\pi i}\int_{s=a-i\infty}^{a+i\infty}\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{e^{sw}}{s}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{k\Gamma(kz)}{s^{kz}}dzds,$

where $a>0$ and $x>1/j$. To evaluate this double integral, we’ll swap the
order of integration, using the same approach as the double integral for $\ln
Z_{j,k}(s)$. That is, we’ll assume the swap is true and then show the result
is equal to the pre-swapped form. This approach again helps to highlight the
function of the tri-zeta term.

Assuming the swap, we can apply the inverse Laplace transform to the function
$s^{-kz-1}$ in the inner integral, noting the following relation (see, for
example, [5], Table 17.13),

$\frac{\Gamma(kz+1)}{z2\pi
i}\int_{s=a-i\infty}^{a+i\infty}\frac{e^{sw}}{s}s^{-kz}ds=\frac{w^{kz}}{z}$

to give for the post-swapped form of $(\ref{I_pre_swap})$,

(2.16) $I_{j,k}^{\verb"~"}=\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}dz.$

Applying the single-index sum form of the tri-zeta term from
$(\ref{tri_zeta_positive})$ produces

$I_{j,k}^{\verb"~"}=\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\sum_{l=1}^{\infty}a_{l}l^{-z}\frac{{w}^{kz}}{z}dz.$

It is straightfoward to swap this integral and sum, as for all $l>w^{k}$, the
integral vanishes (see for example [4], section 3.5, for a very similar case).
We have

$I_{j,k}^{\verb"~"}=\sum_{l=1}^{\infty}a_{l}\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\left(\frac{w^{k}}{l}\right)^{z}\frac{1}{z}dz$

which, by noticing each integral is just a step function with step of height
$1/m$ at $w=mn_{i}^{j/k}$, shows by $(\ref{I_steps})$ that
$I_{j,k}^{\verb"~"}=I_{j,k}$, making the fomula in $(\ref{I_jk_post_swap})$
valid for $I_{j,k}$.

###  2.4. The convolution exponential, and an overall form for
$\mathbf{S_{j,k}(w)}$.

Differentiating the $S_{j,k}$ step function of $(\ref{S_integral_1})$ with
respect to $w$, we get the distributional derivative as a sum of delta
distributions at points $v\in\mathbb{M}_{j,k}$, as well as at $v=0$. Taking
the Laplace transform in the sense of generalized functions, and integrating
over the series of delta distributions produces,

(2.17) $S_{j,k}(w)=\int_{0^{-}}^{w}\mathcal{L}^{-1}\left(e^{\ln
Z(s)}\right)[t]dt-1.$

Here we note that throughout the paper the $S_{j,k}$ integral’s lower limit
will be chosen as either $t=0^{-}$ or $t=0^{+}$, depending on the context. The
difference amounts to whether the point mass at $w=0$ ($\delta_{0}(w)$) is
captured. If it is, as in $(\ref{S_as_Laplace})$, since $0$ is not in
$\mathbb{M}_{j,k}$, a $-1$ corrective term is required.

Next note that by $(\ref{I_function_ln_Z})$, $I_{j,k}$ can be written as,

(2.18) $I_{j,k}(w)=\mathcal{L}^{-1}\left(\frac{\ln Z_{j,k}(s)}{s}\right),$

and let $dI$ be its distributional derivative,

$dI_{j,k}=\mathcal{L}^{-1}\left(\ln Z_{j,k}(s)\right).$

Applying the convolution theorem for Laplace transforms,
$\mathcal{L}^{-1}(\mathcal{L}f\mathcal{L}g)=f*g$, we have
$\mathcal{L}^{-1}((\ln Z)^{m})=dI^{*m}$, and we may now write the integrand of
$(\ref{S_as_Laplace})$ as a convolution exponential with argument $dI$:

(2.19) $\mathcal{L}^{-1}\left(\mbox{exp}{(\ln Z})\right)=\exp^{*}(dI),$

where by the definition of the convolution exponential,

$\exp^{*}(dI)=\delta_{0}+dI+\frac{dI*dI}{2!}+...=\mathcal{L}^{-1}\left(1+\ln
Z+\frac{(\ln Z)^{2}}{2!}+...\right),$

where we’ve used the shorthand $\delta_{c}(w)=\delta(w-c)$. Note that $dI$ may
be regarded as a discrete measure with support only on $[1,\infty)$. The
support of the $k$th convolution power of $dI$, $dI^{*k}$, will then be
limited to $[k,\infty)$. This means that for any $w\in\mathbb{R}^{+}$, the
exponential series of $(\ref{convo_exp_dI})$ as a function of $w$ is composed
only of finitely many point masses up to and including $\lfloor w\rfloor$.
This ensures $(\ref{convo_exp_dI})$ converges and is well-defined, as a
counting measure.

The final form for $S_{j,k}$ is,

(2.20)
$S_{j,k}(w)=\int_{t=0^{+}}^{w}\exp^{*}(dI_{j,k})dt=\int_{t=0^{-}}^{w}\exp^{*}(dI_{j,k})dt-1.$

######  Remark 1.

Note that in parallel with $(\ref{S_integral_1})$ and $(\ref{S_as_conv_exp})$,
with $\zeta(s)$ denoting the Riemann zeta function, and $N(w)$ the natural
number counting function, a similar though less convenient formula exists
relating the two,

$N(w)=\frac{1}{2\pi
i}\int_{s=a-i\infty}^{a+i\infty}w^{s}\frac{e^{\ln\zeta(s)}}{s}ds=\int_{t=1^{-}}^{w}\exp^{*}(e^{v}dJ(e^{v}))[\ln
t]\frac{dt}{t},$

where $J$ is defined as the step function with jumps of $1/n$ at each $p^{n}$,
$p$ prime (see [4], section 1.11), and $e^{v}dJ(e^{v})$ produces point masses
of weight $1/n$ at each $v=\ln(p^{n})$.

The convolution exponential formula, $(\ref{S_as_conv_exp})$, combined with
the formulation $(\ref{I_pre_swap})$ for $I_{j,k}$, now produces an expression
for $S_{j,k}$ in terms of the Riemann zeta function. The remainder of the
paper involves estimates on $I_{j,k}$.

##  3\. Developing $\mathbf{I_{j,k}(w)}$

###  3.1. Overview of methods.

To develop an estimate on $I_{j,k}(w)$, we have two formulas available. One is
as a finite sum of k-free counting functions ($Q_{k}$’s), and the other is the
contour integral of $(\ref{I_jk_post_swap})$. Because the contour integral
form offers a more comprehensive approach, it will be used in the main
estimates for $I_{j,k}(w)$, though the sum-of-$Q_{k}$’s version will also be
used for a general order bound.

#### The function $\mathbf{I_{j,k}(w)}$ as a sum of $\mathbf{Q_{k}}$’s

A quick estimate of $I_{j,k}(w)$ can be derived by continuing the parallel
between the multiplicative primes’ step function $J(w)$ mentioned in Remark 1,
and the parallel step function, $I(w)$. That is, just as $J(x)$ is related to
$\pi(x)$, the prime counting function, by,

$J(x)=\pi(x)+\frac{1}{2}\pi(x^{1/2})+\frac{1}{3}\pi(x^{1/3})+...,$

the following relation between $I_{j,k}(w)$ and $Q_{k}(w)$ (the k-free
counting function) is easy to verify,

(3.1)
$I_{j,k}(w)=Q_{k}\left(\left(w\right)^{k/j}\right)+\frac{1}{2}Q_{k}\left(\left(\frac{w}{2}\right)^{k/j}\right)+...=\sum_{m}\frac{1}{m}Q_{k}\left(\left(\frac{w}{m}\right)^{k/j}\right),$

where the series terminates when $m>w$.

######  Remark 2.

A reader only interested in first order approximations to $S_{j,k}(w)$ may at
this point skip ahead to section 3.8, where a first order approximation to
$I_{j,k}(w)$ is derived using just the $Q_{k}$ series in
$(\ref{I_jk_as_Q_sums})$. Any of the subsequent first order approximations
derived for $S_{j,k}(w)$ at that point will be valid, based on the
developments to this point.

#### The function $\mathbf{I_{j,k}(w)}$ as a sum of residues

Examining the integrand of $(\ref{I_jk_post_swap})$,

$\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z},$

let $\rho$ as usual represent the non-trivial zeros of the zeta function. The
integrand has poles at $z=1/j$, $z=\rho/jk$, $z=0$, and
$z=-2/(jk),-4/(jk),...$ except for at $z=-2m/(jk)$ when $m=(2q+1)j/2$ or
$m=qk$, $m$ and $q$ in $\mathbb{Z}^{+}$ (where the former equality can only
apply when $j$ is even). The pole at the least negative position then is at
$z=-2/(jk)$ and the strategy will be to displace the vertical contour of
$(\ref{I_jk_post_swap})$ to just skirt the line $x=-2/(jk)$. The Cauchy
residue theorem can then be applied to the region between the old and new
contours, while an order bound estimate will be found for the new, displaced
contour itself (the ‘remainder integral’). Two fundamentally different cases
arise in bounding the remainder integral. When $j=1$ and $k=2$, the remainder
integral does not converge absolutely and the most work is needed to find an
order bound. For $j=1$, $k\geq 3$, and for $j\geq 2$, $k\geq 2$, the remainder
integral converges absolutely and an order bound follows easily.

The next sections address expressing $I_{j,k}(w)$ as a sum of residues, with
remainder.

###  3.2. Residue form for $\mathbf{I_{j,k}(w)}$

The first step will be to displace the vertical contour of
$(\ref{I_jk_post_swap})$ to just skirt the first negative pole, $x=-2/(jk)$,
in the $+\mbox{Re}$ direction. Later sections will address application of the
Cauchy residue theorem.

To understand the obstacles involved in displacing the vertical contour to
$x<0$, we’ll have a look at the magnitude of the integrand of $I_{j,k}(w)$’s
contour integral, $(\ref{I_jk_post_swap})$, for large $|y|$. To accomplish
this, the following reformulation of the integrand of $(\ref{I_jk_post_swap})$
is useful. It follows from the reflection formula for the Riemann zeta
function, $\zeta(z)=2^{z}\pi^{z-1}\sin(\pi z/2)\Gamma(1-z)\zeta(1-z)$:

(3.2) $I_{j,k}(w)=\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{2}{k}(2\pi)^{(k+j-jk)z}\frac{\cos(kz\pi/2)\sin(jz\pi/2)}{\sin(jkz\pi/2)}\frac{\Gamma(-kz)\Gamma(-jz)}{\Gamma(-jkz)}\frac{\zeta(-kz)\zeta(1-jz)}{\zeta(1-jkz)}\frac{{w}^{kz}}{z}dz.$

This gives a magnitude for the integrand, after making large $|y|$
approximations to all but the tri-zeta term, of

(3.3) $\approx c_{0}|y|^{(jk-
j-k)x}|y|^{-3/2}\left|\frac{\zeta(-kz)\zeta(1-jz)}{\zeta(1-jkz)}\right|w^{kx},$

where $c_{0}$ is constant for fixed $x$.

The main obstacle to displacing the vertical contour is in bounding the new,
‘post-reflected’ tri-zeta term. The following lemmas are useful:

######  Lemma 1.

The reciprocal of the Riemann zeta function,

$\frac{1}{\zeta(z)},$

is bounded by a constant along a given vertical contour with $x>1$ and
uniformly in the half-plane $x\geq 1+\delta$, $\delta>0$.

###### Proof.

This follows immediately from the relation,

$\frac{1}{\zeta(z)}=\sum_{n}\frac{\mu(n)}{n^{z}},$

where $\mu(n)$ is the Mobius function, since,

$\left|\sum_{n}\frac{\mu(n)}{n^{z}}\right|\leq\sum_{n}\frac{|\mu(n)|}{n^{x}}<\sum_{n}\frac{1}{n^{x}}=\zeta(x)<\infty,$

whenever $x>1$. The uniformity condition follows from
$|\zeta(z)|\leq\zeta(1+\delta)$ whenever $x\geq 1+\delta$. ∎

######  Lemma 2.

(see [4], Section 9.2) The zeta function may be bounded as follows (Lindelöf),

(3.4) $|\zeta(z)|=\mathcal{O}(\ln|y|)\mbox{ whenever $x\geq 1$, for all
$|y|\geq 2$},$ (3.5) $|\zeta(z)|=\mathcal{O}(|y|^{1/2-x/2+\epsilon})\mbox{
whenever $0\leq x\leq 1$, for all $|y|\geq 1$, $\epsilon>0$},$ (3.6)
$|\zeta(z)|=\mathcal{O}(|y|^{1/2-x+\epsilon})\mbox{ whenever $x\leq 0$, for
all $|y|>y_{1}$ , $\epsilon>0$}.$

Applying the lemmas to the tri-zeta term of $(\ref{I_jk_flipped_large_y})$ we
have, when $x\leq-\delta<0$,

$|\zeta(1-jz)|=\mathcal{O}(1)$

and by Lemma 1,

$\left|\frac{1}{\zeta(1-jkz)}\right|=\mathcal{O}(1).$

Also, by Lemma 2’s equations $(\ref{zeta_bound_n1})$ and
$(\ref{zeta_bound_n2})$, we have,

$|\zeta(-kz)|=\left\\{\begin{array}[]{ll}\mathcal{O}(|y|^{1/2-k|x|/2+\epsilon})&\mbox{
when $-1/k<x\leq-\delta$},\\\ \mathcal{O}(\ln|y|)&\mbox{ when
$x\leq-1/k$}.\end{array}\right.$

Using $(\ref{I_jk_flipped_large_y})$ and the results just above, the overall
bound on the integrand of $(\ref{I_jk_post_swap})$ when $|y|\geq 2$ becomes

(3.7)
$\left|\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}\right|=\left\\{\begin{array}[]{ll}\mathcal{O}\left(|y|^{(jk-
j-k/2)x-1+\epsilon}\right)&\mbox{ when $-1/k<x\leq-\delta$},\\\
\mathcal{O}\left(|y|^{(jk-j-k)x-3/2+\epsilon}\right)&\mbox{ when
$x\leq-1/k$}.\end{array}\right.$

To examine various ranges of $j$ and $k$, recalling the remarks following
$(\ref{S_integral_1})$, that $j/k$ be in lowest form:

$\bf{j=1}$:

When $j=1$, the bound reduces to

(3.8) $\mathcal{O}\left(|y|^{(k/2-1)x-1+\epsilon}\right)\mbox{ when
$-1/k<x\leq-\delta$}$

and

(3.9) $\mathcal{O}\left(|y|^{-x-3/2+\epsilon}\right)\mbox{ when $x\leq-1/k$}.$

Notice

  1. (1):

By $(\ref{I_remainder_bel_1_k})$, vertical contours have no chance of
convergence if $x\leq-3/2$.

  2. (2):

Also by $(\ref{I_remainder_bel_1_k})$, vertical contours won’t converge
absolutely if $x\leq-1/2$. And if additionally $k=2$, then by equation
$(\ref{I_remainder_abv_1_k})$, vertical contours won’t converge absolutely if
$x\in(-1/2,-\delta]$ either, making the case $j=1$, $k=2$ unique since
vertical contours won’t converge absolutely whenever $x<0$.

  3. (3):

If $k\geq 3$, equation $(\ref{I_remainder_bel_1_k})$ shows the vertical
contour will converge absolutely whenever $-1/2+\epsilon<x\leq-1/k$.
Specifically, notice $x=-c_{c}/(jk)$, where $c_{c}=3/2-\epsilon_{0}$, for some
suitably chosen $\epsilon_{0}$, lies in this region of absolute convergence.

$\bf{j\geq 2}$:

In general, the bound to ensure absolute convergence when in $-1/k<x<-\delta$
is

(3.10) $x<\frac{-\epsilon}{jk-j-k/2},$

and notice that since $jk-j-k\geq 1$ when $j\geq 2$ ($k\geq 2$), the bound
when $x\leq-1/k$ is,

(3.11) $\mathcal{O}\left(|y|^{x-3/2+\epsilon}\right).$

So by $(\ref{I_a_remainder_abv_1_k})$, we again have absolute convergence
along the contour $x=-c_{c}/(jk)$, $c_{c}$ as above.

#### Displacing the vertical contour.

To develop the displaced contour in all cases of $j,k$, let $C_{r}$ be a
contour with $x$ fixed in $1/j<x$ and $|y|<h$, let $C_{l}$ be the contour with
$x=-c_{c}/(jk)$ and $|y|<h$, and let contours $C_{t}$ and $C_{b}$ be the
segments at $y=h$ and $y=-h$ that complete the rectangle initiated by $C_{r}$,
$C_{l}$. See Figure 1.

Figure 1. _Displacing the $I_{j,k}(w)$ vertical contour._

To justify shifting the vertical contour, we need to show that the integrand
of $(\ref{I_jk_post_swap})$ along contours $C_{t}$ and $C_{b}$ goes to zero as
$h\rightarrow\infty$. The largest issue is handling the denominator of the
tri-zeta term. For example, one of the problems with bounding the reciprocal
of the zeta function along horizontal contours through the critical strip is
the risk of passing arbitrarily close to a zeta zero. However, there exist
certain safe lines of passage through this region as the following Lemma
guarantees.

######  Lemma 3.

(See [3], Lemma 1). There exist positive constants $\lambda_{1}$,
$\lambda_{2}$, and $y_{0}$ such that for any $Y\geq y_{0}$, we can find a $y$
between $Y$ and $2Y$ so that,

$\left|\frac{1}{\zeta(x+iy)}\right|\leq\lambda_{2}\ln^{\lambda_{1}}y\mbox{ for
$-1\leq x\leq 3$}.$

To allow use of the lemma, define the sequence $\\{Y_{m}\\}$ as the sequence
of positive real values guaranteed by the lemma where we will choose
$Y>\max\\{y_{0},y_{1},2\\}$ (with $y_{0}$ as required by the lemma, and, as
we’ll see, the values $y_{1}$ and $2$ as required by Lemma 2) and each
$Y_{m}\in(2^{m-1}Y,2^{m}Y)$ so that the Lemma’s bound on $1/\zeta(jkz)$ holds
for all $Y_{m}$, independent of $m$.

We’ll split the horizontal contour $C_{t}$ (the contour $C_{b}$ is treated
similarly), into three sections and show how to make each section’s
contribution disappear as $|y|$ grows large. The sections are:
$-c_{c}/(jk)<x\leq-\delta$ , $-\delta<x<1/(jk)+\delta$, and $1/(jk)+\delta\leq
x<1/j+\delta$ (where, by definition, the original vertical contour integral of
$(\ref{I_jk_post_swap})$ was independent of the value of $x$, provided
$x>1/j$).

_The section $-\delta<x<1/(jk)+\delta$._

This ‘middle’ section requires the most care and will be dealt with first. To
keep the $1/\zeta(jkz)$ term in the integrand of $(\ref{I_jk_post_swap})$
under control, we’ll invoke Lemma 3. To do so, we position the contours
$C_{t}$ and $C_{b}$ at $\pm(Y_{m}/(jk))i$ so the lemma’s bounds apply. We’ll
also use the bounds for the zeta function from equations
$(\ref{zeta_bound_n1})-(\ref{zeta_bound_n3})$.

For the $|1/\zeta(jkz)|$ term, Lemma 3 produces

(3.12) $\left|\frac{1}{\zeta(jkz)}\right|<c_{0}ln^{c_{1}}(Y_{m})\mbox{ for
$-1/(jk)\leq x\leq 3/(jk)$}.$

(Notice the lower limit for $x$ in $(\ref{zeta_jkz_recip_bound})$ is
$-1/(jk)$, requiring, at least, $\delta\leq 1/(jk)$.)

For $|\zeta(kz+1)|$, the bound of equation $(\ref{zeta_bound_n2})$ gives

$|\zeta(kz+1)|=\mathcal{O}(Y_{m}^{1/2-(1+k\delta)/2+\epsilon}),$

provided $\delta<1/k$, and the bound $(\ref{zeta_bound_n3})$ gives

$|\zeta(jz)|=\mathcal{O}(Y_{m}^{1/2+\delta j}).$

Overall then,

$\left|\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}\right|=\mathcal{O}\left(ln^{c_{1}}(Y_{m})Y_{m}^{-1/2+\delta(k/2+j)+\epsilon}\right),$

and it suffices to make

$\delta<\frac{1}{3k+6j},$

which produces an overall bound on the integrand of

(3.13)
$\left|\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}\right|=\mathcal{O}\left(Y_{m}^{-1/3+\epsilon_{1}}\right)$

provided (combining all restrictions on $\delta$),

$\delta<\min\left\\{\frac{1}{k},\frac{1}{jk},\frac{1}{3k+6j}\right\\}=\min\left\\{\frac{1}{jk},\frac{1}{3k+6j}\right\\}.$

_The section $-c_{c}/(jk)\leq x\leq-\delta$._

This section follows from equations $(\ref{disp_bound_low})$ and the worst
case scenario at $j=1$, $k=2$, for an overall bound of

$\left|\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}\right|=\mathcal{O}\left(Y_{m}^{-3/4+\epsilon}\right)$

_The section $1/(jk)+\delta\leq x<1/j+\delta$._

Since in this section

$|\zeta(jz)|=\mathcal{O}(Y_{m}^{1/2+\epsilon}),$

by Lemma 2, we have the bound,

(3.14)
$\left|\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}\right|=\mathcal{O}(Y_{m}^{-1/2+\epsilon}).$

Combining the bounds $(\ref{disp_bound_mid})$-$(\ref{disp_bound_high})$ on the
integrand of $(\ref{I_jk_post_swap})$ shows that the integral along finite
contours $C_{t}$ (and $C_{b}$) vanishes as $Y_{m}\rightarrow\infty$,
justifying the displacement of the vertical contour of
$(\ref{I_jk_post_swap})$ to $x=-c_{c}/(jk)$, _provided_ the displaced integral
itself converges. That is, it has been shown,

(3.15) $\frac{1}{2\pi
i}\int_{z=x_{h}-i\infty}^{x_{h}+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}dz=\sum\mbox{Res
}f(z)+\frac{1}{2\pi
i}\int_{z=x_{l}-i\infty}^{x_{l}+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}dz,$

provided at least the improper integral on the right hand side converges, and
where $x_{l}=-c_{c}/(jk)$ (just skirting the $+\mbox{Re}$ side of the pole at
$-2/(jk)$), $1/j<x_{h}$, and $f(z)$ represents the integrand of
$(\ref{I_jk_post_swap})$.

#### Residue form of $\mathbf{I_{j,k}(w)}$

The residues in equation $(\ref{I_displaced_contour})$ are easily computed:

(3.16)
$\begin{split}I_{j,k}(w)=\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j}&+\lim_{m\to\infty}\stackrel{{\scriptstyle*}}{{\sum_{|\mbox{Im
}\rho|<Y_{m}/(jk)}}}\frac{\zeta(\rho/j+1)\zeta(\rho/k)}{\zeta^{\prime}(\rho)}\frac{{w}^{\rho/j}}{\rho}+\ln
w+\gamma-j\left(1-\frac{1}{k}\right)\ln 2\pi\\\ &+\frac{1}{2\pi
i}\int_{z=-c_{c}/(jk)-i\infty}^{-c_{c}/(jk)+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}dz,\end{split}$

where $\gamma$ is the Euler-Mascheroni constant. Two remarks are in order
about the sum over zeta zeros:

  1. (1)

The $Y_{m}$ sequence for the sum _is_ needed–it guarantees (by Lemma 3) the
limit implicit in the sum will converge to the proper value.

  2. (2)

The summands in the zeta zero sum only represent those for _simple_ zeros. The
full expression for a zeta zero of multiplicity $l_{\rho}$ is:

$\frac{1}{(l_{\rho}-1)!}\frac{d^{l_{\rho}-1}}{dz^{l_{\rho}-1}}\left.\left((z-\rho/jk)^{l_{\rho}}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}\right)\right|_{z=\rho/jk}.$

It is not known whether all the zeta zeros are simple, though the first
$1.5e9$ are (see [13]). The $*$ above the $\Sigma$ summation symbol will
indicate this simplification.

For equation $(\ref{I_residue})$ to be valid, it remains to show the remainder
integral converges which we’ll do next.

###  3.3. Evaluating the remainder integral, in general.

To ensure convergence of the remainder integral in $(\ref{I_residue})$,
namely,

(3.17) $\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}dz,$

we’ll evaluate it along $x=-c_{c}/(jk)$, recalling $c_{c}=3/2-\epsilon_{0}$.
There are two distinct cases. By equations $(\ref{I_remainder_abv_1_k})$ \-
$(\ref{I_a_remainder_bel_1_k})$ and the surrounding discussion, we know
$(\ref{I_remainder})$ converges absolutely along $x=-c_{c}/(jk)$ for all
allowed combinations of $j$ and $k$ _except_ the instance $j=1$, $k=2$. So the
cases are:

$\bf{j\geq 1,k\geq 3}$:

In this case, the integral $(\ref{I_remainder})$ converges absolutely along
$x=-c_{c}/(jk)$ and therefore has order bound,

(3.18) $\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(kz+1)\zeta(jz)}{\zeta(jkz)}\frac{{w}^{kz}}{z}dz=\mathcal{O}\left(w^{-c_{c}/j}\right).$

Note this order bound is not necessarily uniform over $j$ and $k$. Also note
the order bound is about the best we can do since it can’t be improved by
extending to a contour with $x<-2/(jk)$ because of the pole at $x=-2/(jk)$.

$\bf{j=1,k=2}$:

This case requires considerably more work. The approach will be to use the
post-reflected form of the integrand from equation $(\ref{I_jk_flipped})$,

(3.19)
$\frac{\zeta(-2z)\zeta(1-z)}{\zeta(1-2z)}\frac{1}{2z}\frac{\cos(z\pi)}{\cos(\frac{z\pi}{2})}\Gamma(-z)(\sqrt{2\pi}w)^{2z},$

and expand the tri-zeta term as a sum then evaluate the resulting sum of
integrals using steepest descent. The next sections develop this approach.

###  3.4. Evaluating the remainder integral, for the case $\mathbf{j=1,k=2}$.

The approach is by divide-and-conquer, expressing the tri-zeta term of
$(\ref{tri_zeta_neg_1_2})$ as a series, then evaluating this series of
integrals, and finally showing the integral of the series, and the series of
integrals are equivalent.

To develop an expansion for the negative tri-zeta term of equation
$(\ref{tri_zeta_neg_1_2})$, we’ll use the relation (see [14]),

$\frac{\zeta(z-1)}{\zeta(z)}=\sum_{n=1}^{\infty}\frac{\phi(n)}{n^{z}}$

which holds when $\mbox{Re }z=x>2$ and where $\phi$ is the Euler totient
function.

This gives

$\frac{\zeta(-2z)\zeta(1-z)}{\zeta(1-2z)}=\zeta(1-z)\sum_{n=1}^{\infty}\frac{\phi(n)}{n^{1-2z}}$

whenever $x<-\frac{1}{2}$. Expanding the $\zeta(1-z)$ term in its Dirichlet
series gives the double sum of two absolutely convergent series:

$\sum_{n,k}\frac{\phi(n)}{k^{1-z}n^{1-2z}}=\sum_{n,k}\frac{n\phi(n)}{(kn^{2})^{1-z}},$

where the reformulation on the right hand side has been made in the aim of
expressing this as a sum in a single index. As discussed, any $j$ in
$\mathbb{Z}^{+}$ can be factored uniquely as $ab^{2}$, where $a$ is a square-
free number. Let $\mu_{2}(k)$ as above be the square-free indicator function.
To consolidate terms in the double sum, consider the coefficient of the term
$\mu_{2}(k)/(kn^{2})^{1-z}$. For $j=kn^{2}$, $k$ square-free, if we remove the
restriction on $k$ being square-free, note $j$ can be written as
$k_{1}n_{1}^{2}$ once for every divisor of $n$ (for example, if $j=6\cdot
10^{2}$, $6$ square-free, it can also be written as $24\cdot 5^{2}$, $150\cdot
2^{2}$, and $600\cdot 1^{2}$). This allows reformulating the double sum as,

$\sum_{n,k}\frac{\mu_{2}(k)\sum_{d|n}d\phi(d)}{(kn^{2})^{1-z}}.$

So we have,

(3.20)
$\frac{\zeta(-2z)\zeta(1-z)}{\zeta(1-2z)}=\sum_{l=1}^{\infty}\frac{h_{l}}{l^{1-z}},\mbox{
provided $\mbox{Re }z<-1/2$},$

where $l$ has been factored uniquely as $l=nm^{2}$, $n$ square free, and where

$h_{l}=\sum_{d|m}d\phi(d).$

The sum is guaranteed to converge as the rearrangement of terms in the product
of two absolutely convergent sums.

Now noting that $(\ref{I_jk_flipped})$ simplifies with $j=1$ and $k=2$, and
applying the infinite sum reformulation of the tri-zeta of
$(\ref{hmj_v_zeta})$, we’ll exchange the order of summation and integration in
$(\ref{I_jk_flipped})$ to get,

(3.21) $\sum_{l=1}^{\infty}\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{h_{l}}{l^{1-z}}\frac{1}{2z}\frac{\cos(z\pi)}{\cos(\frac{z\pi}{2})}\Gamma(-z)(\sqrt{2\pi}w)^{2z}dz,$

with the goal of finding a bound on it and showing the order of integration
and summation can be swapped to prove its equivalence to
$(\ref{I_remainder})$.

To evaluate the $l^{th}$ summand of $(\ref{I_sum_over_j})$, choose
$M\geq\frac{3}{2}$, and split the contour into $C_{1}\equiv\\{x+iy\mid|y|\leq
2M\\}$, $C_{2}\equiv\\{x+iy\mid y\geq 2M\\}$ and $C_{3}\equiv\\{x+iy\mid
y\leq-2M\\}$.

The contour $C_{1}$ has a straightforward absolute bound,

$\left|\frac{1}{2\pi
i}\int_{C_{1}}\frac{h_{l}}{l^{1-z}}\frac{1}{2z}\frac{\cos(z\pi)}{\cos(\frac{z\pi}{2})}\Gamma(-z)(\sqrt{2\pi}w)^{2z}dz\right|<$
(3.22)
$\frac{1}{2\pi}\int_{C_{1}}\left|\frac{h_{l}}{l^{1-z}}\frac{1}{2z}\frac{\cos(z\pi)}{\cos(\frac{z\pi}{2})}\Gamma(-z)(\sqrt{2\pi}w)^{2z}\right|dz=c_{0}\frac{h_{l}}{l^{1-x}w^{-2x}}$

For contours $C_{2}$ and $C_{3}$ we’ll expand the integrand in this region
asymptotically. Generalize the lower limit to $2M\leq|y_{0}|$ to give, in the
case of $C_{2}$,

(3.23) $\frac{h_{l}}{l}\frac{1}{4\pi
i}\int_{x+y_{0}i}^{x+i\infty}\frac{1}{z}\frac{\cos(z\pi)}{\cos(\frac{z\pi}{2})}\Gamma(-z)(2\pi
lw^{2})^{z}dz,$

and similarly for $C_{3}$.

The asymptotic expansion for the gamma function (see, eg, [1]) and an
expansion of the cosine fraction in terms of its component exponentials gives
for those terms,

$\Gamma(-z)\frac{\cos(z\pi)}{\cos(\frac{z\pi}{2})}=\sqrt{2\pi}ie^{z}e^{i\frac{\pi}{2}z}z^{-(z+\frac{1}{2})}R_{0}\left(1+R_{1}(z)+R_{2}(z)\right)\mbox{
when $y\geq\delta>0$}.$

Assuming $\mbox{Im }z\geq M$, $R_{0}$ is the correction factor from the gamma
approximation and may be bounded by $e^{c_{1}/M}$, while $R_{1}$ and $R_{2}$
arise from the cosine fraction expansion and may be bounded by $e^{-c_{2}/M}$.

We can now rewrite $(\ref{hmj_simpler_integrand})$ as,

(3.24)
$\frac{h_{l}}{l}\frac{1}{2\sqrt{2\pi}}\int_{x+y_{0}i}^{x+i\infty}e^{i\frac{\pi}{2}z}z^{-(z+\frac{3}{2})}(2\pi
lw^{2}e)^{z}R_{0}(1+R_{1}(z)+R_{2}(z))dz.$

###  3.5. Steepest descent in general for the $\mathbf{j=1}$, $\mathbf{k=2}$
case.

We now evaluate the integral of equation $(\ref{asymptote_integrand})$,
ignoring for the moment its remainder terms:

(3.25)
$\frac{h_{l}}{l}\frac{1}{2\sqrt{2\pi}}\int_{x+iy_{0}}^{x+i\infty}e^{i\frac{\pi}{2}z}z^{-(z+\frac{3}{2})}(2\pi
lw^{2}e)^{z}dz.$

An estimate of this integral is straightforward by use of the method of
steepest descent. The change of variables $s=\frac{z}{lw^{2}}$ in
$(\ref{jth_summand_asymptote})$ produces

(3.26)
$\frac{h_{l}}{l^{\frac{3}{2}}w}\frac{1}{2\sqrt{2\pi}}\int_{s=\frac{c}{lw^{2}}+\frac{iy_{0}}{lw^{2}}}^{\frac{c}{lw^{2}}+i\infty}s^{-\frac{3}{2}}\exp\left[lw^{2}s\left(\frac{i\pi}{2}+1+\ln\left(\frac{2\pi}{s}\right)\right)\right]ds$

where the integral is now of saddle point form

(3.27) $K(\lambda)=\int_{C}g(s)e^{\lambda f(s)}ds.$

The saddle points of $(\ref{s_form})$ are found by solving $f^{\prime}(s)=0$
to give a unique saddle point, $s_{0}=2\pi i$; by using known evaluation
formulas for $(\ref{standard_SD})$ when $|\lambda|=lw^{2}$ is large (see, for
example, [7]), equation $(\ref{s_form})$ evaluates to

(3.28) $\approx-\frac{h_{l}}{l^{2}}\frac{1}{4\pi w^{2}}e^{i2\pi lw^{2}}$

to a first approximation whenever $0\leq y_{0}<<2\pi lw^{2}$ (since the
contour can be deformed to pass through the saddle point without modification
because the integrand has no intervening singularities).

Combining $(\ref{SP_estimate})$ with the expectation that $(\ref{s_form})$ is
$\approx 0$ when $2\pi lw^{2}<<y_{0}$, suggests that the individual integrals
as well as the overall sum of $(\ref{I_sum_over_j})$ at least converge, and
that with some additional requirements on the rate of convergence in $z$ of
the integrals, that exchanging the order of integration and summation is
justified. The next section will show that this is the case.

###  3.6. Steepest descent in particular for the $\mathbf{j=1}$,
$\mathbf{k=2}$ case.

To find precise bounds on $(\ref{jth_summand_asymptote})$, consider its
integrand:

(3.29)
$\left|\frac{h_{l}}{l}\frac{1}{2\sqrt{2\pi}}e^{i\frac{\pi}{2}z}z^{-(z+\frac{3}{2})}(2\pi
lw^{2}e)^{z}\right|=\frac{h_{l}}{l}\frac{1}{2\sqrt{2\pi}}r^{-3/2}e^{y(\theta-\pi/2)}\left(\frac{2\pi
lw^{2}e}{r}\right)^{x}$

where $z=x+iy=re^{i\theta}$, and the saddle point is $z=i\alpha_{l}=2\pi
ilw^{2}$.

The integral $(\ref{jth_summand_asymptote})$ can be divided into three cases,
depending on where the lower limit $x+iy_{0}$ lies with respect to the saddle
point $i\alpha_{l}$. Bounds can be found for each case, by evaluating the
integral of the magnitude of the integrand where knowledge of the gradient is
used in deforming the contour to provide a suitable bound. The computations to
find the bounds are not difficult but involve some algebra and have been put
in the Appendix because of their length. Assuming $w$ is fixed and greater
than zero and that $2M\leq y_{0}$, the three cases and their bounds are as
follows:

Case I:

In this case, $\alpha_{l}e<y_{0}$. The bound is,

(3.30)
$\frac{h_{l}}{2\sqrt{2\pi}l}\left(c^{|x|}e^{|x|}|x|(\alpha_{l}e)^{-3/2}+\frac{3}{2}y_{0}^{-3/2}\left(1-e^{-(2/3)y_{0}}\right)+\frac{2}{\ln
2}y_{0}^{-3/2}e^{-y_{0}(\pi/4+\ln(\sqrt{2}))}\right)$
$=\mathcal{O}\left(\frac{h_{l}}{l^{5/2}w^{3}}\right),$

which follows from Case I’s requirement that $2\pi lw^{2}e<y_{0}$.

Case II:

In this case, $\alpha_{l}<y_{0}\leq\alpha_{l}e$. The bound is

(3.31)
$\frac{h_{l}}{2\sqrt{2\pi}l}\left(c^{|x|}e^{|x|}|x|(\alpha_{l})^{-3/2}+\sqrt{\frac{\pi}{2}}y_{0}^{-1}+\frac{3}{2}y_{0}^{-3/2}\left(1-e^{-(2/3)y_{0}}\right)+\frac{2}{\ln
2}y_{0}^{-3/2}e^{-y_{0}(\pi/4+\ln(\sqrt{2}))}\right),$

The overall order bound for this Case is,

$=\mathcal{O}\left(\frac{h_{l}}{l^{2}w^{2}}\right),$

which similarly follows because of Case II’s requirement $2\pi lw^{2}<y_{0}$.

Case III:

In this case, $y_{0}\leq\alpha_{l}$. This bound is formed by summing three
parts. For the first, we’ll evaluate

$\frac{h_{l}}{2\sqrt{2\pi}l}\left(\left(\frac{\alpha_{l}e}{y_{1}-x}\right)^{x}(y_{1}+x)^{-3/2}e^{-x}|x|+\frac{\sqrt{\pi}}{2^{1/4}}y_{1}^{-1-x}{\alpha_{l}}^{x}\right)$

at $y_{1}=y_{0}$, then add to the result the same expression evaluated at
$y_{1}=\alpha_{l}$ . For the second, we have,

$\frac{h_{l}}{2\sqrt{2\pi}l}\left(2e^{M\pi/2}\left(\frac{\alpha_{l}e}{\sqrt{2}M}\right)^{-M}\left(\frac{1}{\sqrt{y_{0}-M}}-\frac{1}{\sqrt{\alpha_{l}-M}}\right)\right).$

And for the third, we use the bounds from Case II, setting $y_{0}=\alpha_{l}$
in $(\ref{case_II_part_bound})$. Given $M=3/2$, the overall order bound for
this case is

$\mathcal{O}\left(\frac{h_{l}}{l^{\mu}w^{\nu}}\right)$

where $\mu=1-x$ and $\nu=-2x$ if $-1<x<-1/2$ and $\mu=2$, $\nu=2$ otherwise.

The goal was to bound the asymptotic expansion of
$(\ref{hmj_simpler_integrand})$ along the $|y|\geq 2M$ contours $C_{2}$ and
$C_{3}$. We have bounded it with the Case bounds, the bounds on the remainder
terms (which at worst affect the Case bounds by a constant of scale), and the
fact though the Case bounds were derived for the $+\mbox{Im}$ contour
($C_{2}$), the function is analytic, so the bounds need only be multiplied by
a factor of at most $2$ to accomodate a bound on the sum of the symmetric
contours.

Given the goal was to evaluate the remainder integral
$(\ref{asymptote_integrand})$ along $x=-c_{c}/(jk)=-3/4+\epsilon$ (recalling
$c_{c}=3/2-\epsilon_{0}$), it is time to prove the following theorem:

######  Theorem 2.

The remainder integral, $(\ref{I_remainder})$, when $j=1$ and $k=2$ converges
and obeys order bound,

(3.32) $\frac{1}{2\pi
i}\int_{z=x-i\infty}^{x+i\infty}\frac{\zeta(2z+1)\zeta(z)}{\zeta(2z)}\frac{{w}^{2z}}{z}dz=\mathcal{O}\left(w^{-3/2+\epsilon}\right),$

for $x=-3/4+\epsilon$.

###### Proof.

The following basic proof rests on showing that the series of integrals in
$(\ref{I_sum_over_j})$ converges uniformly in $y$ (see for example [4],
section 3.5, for an almost identical proof). Let $g_{l}$ denote the integrand
of $(\ref{I_sum_over_j})$. First note that $\sum_{l}g_{l}$ converges uniformly
for all $y$ along finite vertical contours. This follows from the initial
integrand estimate $(\ref{hmj_simpler_integrand})$ and the subsequent absolute
bounds on $g_{l}$, $(\ref{jth_summand_absolute})$. We in fact have the bound

$\left|g_{l}\right|<\frac{h_{l}}{l^{1-x}}c_{0}r^{-3/2-x}e^{y(\theta-\pi/2)}<\frac{h_{l}}{l^{1-x}}c_{1},$

for all $y>2M$, and where recall the series in $l$ converges provided
$1-x>3/2$. Next claim

(3.33)
$\left|\int_{x+2Mi}^{x+iy}g_{l}(z)dz\right|<c\frac{h_{l}}{l^{3/2+\epsilon}w^{\kappa}},$

for $-1<x<-1/2$ independent of $y>2M$, and where if $w\geq 1$, $\kappa=1$,
otherwise, for $w\in(0,1)$, $\kappa=2$. To prove this, we’ll use the case
bounds, initially taking $y\to\infty$. Since $w$ can be any value $>0$, the
saddle point $2\pi lw^{2}i$ can be anywhere in relation to the lower integral
limit, and we must consider all case bounds. For case I, we had
$\mathcal{O}(h_{l}/(l^{5/2}w^{3}))$. For Case II,
$\mathcal{O}(h_{l}/(l^{2}w^{2}))$. Case III is the most restrictive case,
namely $\mathcal{O}(h_{l}/(l^{1-x}w^{-2x}))$, when $-1<x<-1/2$. The worst case
is when $x=-1/2-\epsilon$, giving the bound
$\mathcal{O}(h_{l}/l^{3/2+\epsilon}w^{\kappa})$, where the conditions on
$\kappa$ account for bounds on $w^{-2x}$ over all $x$ in the range. Now noting
the case bounds were constructed by using the absolute value of the integrand,
the integral in $(\ref{uniform_in_h})$ for finite $y$ would have a strictly
lower value than those considered in the case bounds. This proves the claim.
And since the series in $l$ converges over the range of interest for $x$, we
have that the sum of integrals in $(\ref{I_sum_over_j})$ converges in this
range.

From the claim, we have that the sum,

$\sum_{l=1}^{\infty}\int_{x+2Mi}^{x+iy}g_{l}(z)dz,$

converges uniformly in $y$. This suffices to justify swapping the sum and
integrals in $(\ref{I_sum_over_j})$, as follows. Consider,

i:

$\left|\sum_{l=1}^{\infty}\int_{x+2Mi}^{x+iy}g_{l}(z)dz-\sum_{l=1}^{N}\int_{x+2Mi}^{x+iy}g_{l}(z)dz\right|<\epsilon/3$

ii:

$\left|\sum_{l=1}^{N}\int_{x+2Mi}^{x+i\infty}g_{l}(z)dz-\sum_{l=1}^{\infty}\int_{x+2Mi}^{x+i\infty}g_{l}(z)dz\right|<\epsilon/3$

iii:

$\left|\sum_{l=1}^{N}\int_{x+2Mi}^{x+iy}g_{l}(z)dz-\sum_{l=1}^{N}\int_{x+2Mi}^{x+i\infty}g_{l}(z)dz\right|<\epsilon/3$

The inequality in (i) follows since,

$\left|\sum_{l=N+1}^{\infty}\int_{x+2Mi}^{x+iy}g_{l}(z)dz\right|\leq\left|\int_{x+2Mi}^{x+iy}g_{N+1}(z)dz\right|+...<c\frac{h_{N+1}}{(N+1)^{3/2}w^{\kappa}}+c\frac{h_{N+2}}{(N+2)^{3/2}w^{\kappa}}+....$

and we can choose $N$ sufficiently large that the sum from $N$ to $\infty$ of
the bounds on the summands (from $(\ref{uniform_in_h})$) meet the $\epsilon$
condition, _independent_ of $y$. By the same reasoning, allowing $y\to\infty$,
the inequality in (ii) is obtained.

Let $N$ be fixed from inequalities (i) and (ii). The inequality in (iii)
follows by choosing $y_{0}$ large enough that, as guaranteed by the
convergence of each of the $g_{l}$ integrals, for all $y>y_{0}$ and for all
$l\in[1,N]$,

$\left|\int_{x+2Mi}^{x+iy}g_{l}(z)dz-\int_{x+2Mi}^{x+i\infty}g_{l}(z)dz\right|<\epsilon/3N.$

Since the inequality holds for all $l\in[1,N]$, the sum, $\sum_{1}^{N}$ must
be less than $\epsilon/3$, as desired.

This justifies the inequalities. Transitively they give, for all $y$,

$\left|\sum_{l=1}^{\infty}\int_{x+2Mi}^{x+iy}g_{l}(z)dz-\sum_{l=1}^{\infty}\int_{x+2Mi}^{x+i\infty}g_{l}(z)dz\right|<\epsilon.$

Combined with the uniform convergence of $g_{l}$, which allows the exchange of
order between finite integrals and infinite sums, this proves the sum of
integrals, $(\ref{I_sum_over_j})$, equals the integral in
$(\ref{I_remainder_w_order_bound})$.

For the order bound, note for each $l$, the portion of the contour between
$x-2Mi$ and $x+2Mi$ has been shown
$\mathcal{O}\left(h_{l}/(l^{2}w^{3/2-\epsilon})\right)$ by
$(\ref{middle_portion_hmj})$ (with $x=-3/4+\epsilon$). For the $|y|\geq 2M$
portion, the overall bound from the case bounds, as in $(\ref{uniform_in_h})$,
gives $\mathcal{O}(h_{l}/(l^{1+3/4}w^{3/2-\epsilon}))$ for this $x$. This
gives, at worst, for the sum of the whole contours along $x=-3/4+\epsilon$, a
bound of

$\sum_{n=1}^{\infty}c_{1}\frac{h_{n}}{l^{1+3/4}w^{3/2-\epsilon}}=c_{2}w^{-3/2+\epsilon},$

proving the theorem. ∎

###  3.7. Summary of $\mathbf{I_{j,k}(w)}$ residue form.

Combining the results on the remainder integral for the case $j/k\neq 1/2$,
$(\ref{non_1_2_order_bound})$ with those for $j/k=1/2$,
$(\ref{I_remainder_w_order_bound})$, shows the validity of equation
$(\ref{I_residue})$ for all $j$ and $k$, $k>1$, $\gcd(j,k)=1$. Equation
$(\ref{I_residue})$ can be rewritten as,

(3.34)
$\begin{split}I_{j,k}(w)=&\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j}+\lim_{m\to\infty}\stackrel{{\scriptstyle*}}{{\sum_{|\mbox{Im
}\rho|<Y_{m}/(jk)}}}\frac{\zeta(\rho/j+1)\zeta(\rho/k)}{\zeta^{\prime}(\rho)}\frac{{w}^{\rho/j}}{\rho}+\ln
w\\\ &+\gamma-j\left(1-\frac{1}{k}\right)\ln
2\pi+\mathcal{O}\left(w^{-3/(2j)+\epsilon}\right)\end{split}$

(* indicates this form for the summands is valid for simple zeta zeros only;
note also the remainder $\mathcal{O}$ bound has not been shown to be uniform
over $j$ or $k$). This proves the first part of Theorem 1.

###  3.8. The general sum-of-$\mathbf{Q}$’s form, and a first order error
term; proof of Theorem 1.

Rewriting the residue sums for $I_{j,k}(w)$ (equation
$(\ref{I_residue_full})$) as,

(3.35) $I_{j,k}(w)=\frac{\zeta(1+k/j)}{\zeta(k)}{w}^{k/j}+R_{j,k}(w),$

it is possible to obtain overall estimates on the error term, $R_{j,k}(w)$.

To find a $\mathcal{O}$ bound on $R_{j,k}(w)$, we’ll use equation
$(\ref{I_jk_as_Q_sums})$:

$I_{j,k}(w)=\sum_{m=1}^{\lfloor
w\rfloor}\frac{1}{m}Q_{k}\left[\left(\frac{w}{m}\right)^{k/j}\right],$

and denote the order of the error term in $Q_{k}(x)$ by $\beta_{k}$. That is,

(3.36)
$Q_{k}(x)=\frac{x}{\zeta(k)}+R_{Q_{k}}(x)=\frac{x}{\zeta(k)}+\mathcal{O}(x^{\beta_{k}}).$

There has been a lot of work in the literature on finding a close upper bound
on $R_{Q_{k}}(x)$ (see, for example, [11] for a summary). It is known that
$\beta_{k}=1/k$ at worst, and if the Riemann hypothesis is true it is
conjectured that $\beta_{k}=1/(2k)+\epsilon$ (see [11]). The case $k=2$, where
$Q_{2}(x)$ is the square-free counting function, has received special
attention. The best known value for $\beta_{2}$ as of this writing is
$17/54+\epsilon$ (see [8]).

Substituting $(\ref{Q_k_Oh_bound})$ into equation $(\ref{I_jk_as_Q_sums})$
produces,

(3.37) $I_{j,k}(w)=\sum_{m=1}^{\lfloor
w\rfloor}\frac{1}{m}\left[\frac{w^{k/j}}{\zeta(k)m^{1+k/j}}+\mathcal{O}\left(\left(\frac{w}{m}\right)^{(k/j)\beta_{k}}\right)\right]=\sum_{m=1}^{\lfloor
w\rfloor}\frac{w^{k/j}}{\zeta(k)m^{1+k/j}}+\sum_{m=1}^{\lfloor
w\rfloor}\frac{1}{m}\mathcal{O}\left(\left(\frac{w}{m}\right)^{(k/j)\beta_{k}}\right).$

Looking at the second sum, by definition of the $\mathcal{O}$ bound we have

$\left|\sum_{m=1}^{\lfloor
w\rfloor}\frac{1}{m}\mathcal{O}\left(\left(\frac{w}{m}\right)^{(k/j)\beta_{k}}\right)\right|<\sum_{m=1}^{\lfloor
w\rfloor}\frac{1}{m}K\left(\frac{w}{m}\right)^{(k/j)\beta_{k}}<K\zeta((k/j)\beta_{k}+1)w^{(k/j)\beta_{k}}=\mathcal{O}(w^{(k/j)\beta_{k}})$

for some sufficiently large $K$, where the second inequality is conditional on
$\beta>0$.

Setting equations $(\ref{I_j_k_w_R})$ and $(\ref{I_as_m_sums})$ equal we
obtain

(3.38) $\begin{split}R_{j,k}(w)&=\sum_{m=1}^{\lfloor
w\rfloor}\frac{w^{k/j}}{\zeta(k)m^{1+k/j}}+\mathcal{O}(w^{(k/j)\beta_{k}})-\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j}\\\
&=-\left(\zeta(1+k/j)-\sum_{m=1}^{\lfloor
w\rfloor}\frac{1}{m^{1+k/j}}\right)\frac{w^{k/j}}{\zeta(k)}+\mathcal{O}(w^{(k/j)\beta}).\end{split}$

But by known order bounds on the partial sums of zeta Dirichlet series (see,
for example, [12] Theorem 4.11),

(3.39)
$\zeta(1+k/j)-\sum_{m=1}^{w}\frac{1}{m^{1+k/j}}=\frac{j}{k}w^{-k/j}+\mathcal{O}(w^{-1-k/j}),$

to produce, after substituting $(\ref{titch_4_11})$ into equation
$(\ref{R_jk_1})$,

$-\left(\frac{w^{-k/j}}{(k/j)}+\mathcal{O}(w^{-1-k/j})\right)\frac{w^{k/j}}{\zeta(k)}+\mathcal{O}(w^{(k/j)\beta_{k}})=c_{0}+\mathcal{O}(w^{-1})+\mathcal{O}(w^{(k/j)\beta_{k}}),$

to give,

(3.40) $R_{j,k}(w)=\mathcal{O}(w^{(k/j)\beta_{k}}).$

This proves the second of the two parts of Theorem 1, which, in addition to
the previous section’s proof of the first part, proves the theorem.

###  3.9. The particular case of $\mathbf{I_{1,2}(w)}$

Before getting to the estimates for $S_{j,k}(w)$, we’ll compare the estimates
of $I_{j,k}(w)$ with its actual values in the particular case of $I_{1,2}(w)$.
From equation $(\ref{I_residue_full})$ we have,

$I_{1,2}(w)=\frac{\zeta(3)}{\zeta(2)}{w}^{2}+\lim_{m\to\infty}\stackrel{{\scriptstyle*}}{{\sum_{|\mbox{Im
}\rho|<Y_{m}/2}}}\frac{\zeta(\rho+1)\zeta(\frac{\rho}{2})}{\zeta^{\prime}(\rho)}\frac{{w}^{\rho}}{\rho}+\ln
w+\gamma-\frac{1}{2}\ln(2\pi)+R_{I}(w)$

where $R_{I}(w)=\mathcal{O}(w^{-3/2+\epsilon})$.

This produces a first order approximation,

(3.41) $I_{1,2}(w)\approx\frac{\zeta(3)}{\zeta(2)}{w}^{2}\mbox{ when $w\geq
0$, $0$ otherwise}.$

And if we conjecture the contribution from the zeta zero sum has ‘mean
oscillation’ zero (for example if $X_{t}$ is the integral from $0$ to $t$ of
the zeta zero sum’s contribution, then
$\liminf_{t\to\infty}X_{t}=\mathcal{O}(1)$), then since the oscillatory
contribution from the remainder integral, $R_{I}$, has been shown to be small
we have the following ‘centerline’ approximation:

(3.42) $I_{1,2}(w)\approx\frac{\zeta(3)}{\zeta(2)}{w}^{2}+\chi_{[1,\infty)}\ln
w+\gamma-\frac{1}{2}\ln(2\pi)\mbox{ when $w\geq 0$, $0$ otherwise},$

where $\chi_{[1,\infty)}$ is the indicator function of value $1$ on
$[1,\infty)$, and $0$ elsewhere; this serves to truncate the natural logarithm
at $w=1$, to avoid large contributions in the region $[0,1)$.

Figures 3 and 3 show the first order and centerline estimation curves compared
with the actual computed step function. The first order estimate is within 3%
for $w\approx 15$ while the centerline estimate is entwined with the step
function itself in this same region.

Figure 2. _$I_{1,2}(w)$ stairs and estimates._

Figure 3. _Dashed box inset._

##  4\. Estimates for $\mathbf{S_{j,k}(w)}$

We are now ready to estimate $S_{j,k}(w)$. We’ll use the expression of
$S_{j,k}(w)$ as the integral of a convolution exponential from equation
$(\ref{S_as_conv_exp})$ and the preceding estimates of $I_{j,k}(w)$.

###  4.1. Basic results on convolution exponentials.

In order to develop bounding estimates on $S_{j,k}(w)$ by using estimates on
$I_{j,k}(w)$, the following results on convolution exponentials are needed.

######  Lemma 4.

Given $v^{a}$ supported on $[0,\infty)$, with $a\in(-1,\infty)$,

$e^{*cv^{a}}=\delta_{0}(t)+\sum_{k=1}^{\infty}\frac{c^{k}}{k!}\frac{(\Gamma(a+1))^{k}}{\Gamma(k(a+1))}t^{k(a+1)-1},$

and the integral, $\int_{t=\alpha}^{w}e^{*cv^{a}}dt$, where $\alpha$ is
$0^{+}$ or $0^{-}$, is finite.

###### Proof.

From the definition of the convolution exponential,
$e^{*cv^{a}}[t]=\delta_{0}(t)+ct^{a}+(cv^{a})^{*2}[t]/2!+...$. Each
convolution power $(cv^{a})^{*k}$ may be computed from repeated application of
the following beta function formula,

$\int_{0}^{t}v^{q-1}(t-v)^{p-1}dv=t^{p+q-1}B(p,q),$

where $B(u,v)$ is the beta function, and with the requirement $p,q>0$ (see for
instance [5], section 3.18). For the second convolution power we have,

$(cv^{a})^{*2}=c^{2}\int_{-\infty}^{\infty}\chi_{[0,\infty)}(v)v^{a}\chi_{[0,\infty)}(t-v)(t-v)^{a}dv=c^{2}\int_{0}^{t}v^{a}(t-v)^{a}dv=c^{2}t^{2a+1}B(a+1,a+1).$

Higher powers can be done recursively to give,

$\begin{split}e^{*cv^{a}}=&\delta_{0}(t)+\frac{c}{1!}t^{a}+\frac{c^{2}}{2!}B(a+1,a+1)t^{2a+1}+\frac{c^{3}}{3!}B(a+1,a+1)B(a+1,2a+2)t^{3a+2}+...\\\
&+\frac{c^{k}}{k!}B(a+1,a+1)...B(a+1,(k-1)a+k-1)t^{ka+k-1}+...\end{split}$

Since $B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)$, the product of
$B(a+1,a+1)...B(a+1,(k-1)a+k-1)$ collapses to
$(\Gamma(a+1))^{k}/\Gamma(k(a+1))$, proving the series formula.

For the finiteness of the integral, since for $\kappa$ sufficiently large,
$k(a+1)-1>0$ for all $k>\kappa$, making $t^{k(a+1)-1}$ monotone increasing in
$[0,w]$, we have for some sufficiently large $\kappa^{\prime}>\kappa$,

$\left|\sum_{k=\kappa^{\prime}}^{\infty}\frac{c^{k}}{k!}\frac{(\Gamma(a+1))^{k}}{\Gamma(k(a+1))}t^{k(a+1)-1}\right|<\sum_{k=\kappa^{\prime}}^{\infty}\frac{(|c|\Gamma(a+1))^{k}w^{k(a+1)-1}}{k!\Gamma(k(a+1))}<\epsilon,$

showing the sum converges uniformly in $[0,w]$. This allows the integral to be
passed inside. The integral of the first term of the series, the delta
function, will be $1$ or $0$, depending on the integral’s lower limit.
Integrating the remaining sum term by term gives,

$\sum_{k=1}^{\infty}\frac{c^{k}}{k!}\frac{(\Gamma(a+1))^{k}}{\Gamma(k(a+1))}\frac{w^{k(a+1)}}{k(a+1)},$

which clearly converges. ∎

######  Lemma 5.

Given the integral of a convolution exponential,

$\int_{t=0^{-}}^{w}e^{*f}dt,$

where $f$ has support on $[0,\infty)$, $f\geq 0$, $f(0)=0$, and $f$ is
integrable on any $[0,t_{0}]$, and suppose
$\int_{t=0^{-}}^{w}f(t)dt=F(w)<G(w)$ for all $w>0$ where $G\geq 0$, $G$ is
differentiable, $G$ has support on $[0,\infty)$, and is allowed to have a jump
discontinuity at $0$ of value $c>0$. Also assume the derivative of $G$, call
it $g$, obeys $g(t)\geq 0$. Then,

(4.1) $\int_{t=0^{-}}^{w}e^{*f}dt<\int_{t=0^{-}}^{w}e^{*g}dt.$

Under the same conditions except with $G(w)<F(w)$, the inequality may be
reversed.

###### Proof.

From the series expansion of $e^{*f(v)}$, we have terms
$\int_{t=0^{-}}^{w}(f(v))^{*m}/m!$. So it suffices to show

(4.2) $\int_{t=0^{-}}^{w}(f(v))^{*m}<\int_{t=0^{-}}^{w}(g(v))^{*m}$

for all $m\geq 1$.

The proof will be by induction on $m$. For compactness in notation, drop the
dependent variable $v$, and use $\int_{*}$ as shorthand for
$\int_{t=0^{-}}^{w}$.

First claim $\int_{*}f^{*m}=F*f^{*(m-1)}$. This follows since $f(0)=0$, so
that $F(0)=0$, and so $\int_{*}f=F$, and since for any $f_{1}$ and $f_{2}$
with support on the half-line, $[0,\infty)$, $f_{1}*f_{2}[t]$ will only have
support on $[0,t]$. This allows swapping the order of integration in
$\int_{*}\int_{-\infty}^{\infty}f(t-v)f^{*(m-1)}[v]dv$, and the intermediary
relation $\int_{0-}^{w}f(t-v)dt=\int_{0-}^{w-v}f(u)du=F(w-v)$, proving the
claim.

For $m=2$ we’ll show,

(4.3) $F*f<G*g=\int_{*}g^{*2}.$

By the monotonicity in the convolution operation for positive functions, note
whenever $0\leq p(t)<q(t)$ for all $t$ in the support of $p$ and $q$, it is
true that $p*p<q*q$ and that $p*r<q*r$ for all positive integrable $r(t)$.
This allows,

$F*f<G*f=g*F<g*G.$

The middle equality follows from $\int_{0-}^{w}f(t-v)dt=F(w-v)$ just above,
and the reasoning leading to it (noting as well that $\int_{*}g=G$ by
construction). This proves equation $(\ref{convo_exp_bound_3})$.

Now suppose $(\ref{convo_exp_bound})$ holds at $m-1$, then

$\int_{*}f^{*(m)}=F*f^{*(m-1)}<G*f^{*(m-1)}=g*\left(\int_{*}f^{*(m-1)}\right)<g*\left(\int_{*}g^{*(m-1)}\right)=\int_{*}g^{*(m)},$

where the second inequality follows from the induction hypothesis. This
completes the proof of $(\ref{convo_exp_bound})$.

So

$\int_{0-}^{w}e^{*f}<\int_{0-}^{w}\delta(w)+\sum_{m=1}^{\infty}\frac{1}{m!}\int_{0-}^{w}g^{*m}=\int_{0-}^{w}e^{*g}.$

The same steps may be repeated when $G(w)<F(w)$, reversing the inequality
signs. ∎

Finally note the additive property of convolution exponentials,
$e^{*(f+g)}=e^{*f}*e^{*g}$, allows the following relation, which will be
useful in discussions on estimates of $S_{j,k}(w)$:
$e^{*(c\delta(v)+g(v))}=e^{*c\delta}*e^{*g}=e^{c}e^{*g}$ (the second equality
follows from $\delta(v)*h(v)[t]=h(t)$, and from the series form for the
convolution exponential, noting $(c\delta(v))^{*m}=c^{m}\delta(v)$).

###  4.2. Estimates on $\mathbf{S_{j,k}}$, the general case.

Recalling that $I_{j,k}$ and $dI_{j,k}$ and all their estimates have support
on $[0,\infty)$, we have the following $S_{j,k}$ estimates.

#### First order estimate.

Combining equations $(\ref{S_as_conv_exp})$ and $(\ref{I_residue_full})$, and
noting the derivative of the first order estimate of $I_{j,k}$ in equation
$(\ref{I_residue_full})$ produces the form $e^{*cv^{a}}$ where $a=k/j-1>-1$,
Lemma 4 can be applied to show the validity of the convolution exponential and
the outer integral. The first order approximation of $S_{j,k}(w)$ then is,

$S_{j,k}(w)\approx\int_{t=0^{+}}^{w}\exp^{*}\left(\frac{k}{j}\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j-1}\right)dt.$

#### Centerline estimate.

Though again the justification is conjectural (as in the discussion leading to
$(\ref{I_centerline})$) the following centerline estimate is given (where to
obtain $dI$ from $(\ref{I_residue_full})$, we’ve set $d/dw(\ln w)=0$ in
$w\in[0,1]$, and, requiring the integrand to be zero at the lower integral
limit, we’ve treated the constant term as a jump discontinuity, giving a
$(\gamma-(1-j/k)\ln(2\pi))\delta_{0}$ term in $dI$; it emerges as a constant
of scale, for the reasons mentioned just below the proof of Lemma 5):

$S_{j,k}(w)\approx e^{\gamma-j(1-1/k)\ln
2\pi}\int_{t=0^{+}}^{w}\exp^{*}\left(\frac{k}{j}\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j-1}+\chi_{[1,\infty)}1/w\right)dt,$

Lemma 4 has been applied again as follows: note the argument of the
convolution exponential can be bounded by $c_{0}w^{k/j-1}$ for some $c_{0}>0$,
for all $w\in[0,\infty)$; Lemma 4 is then invoked with
$cv^{a}=c_{0}v^{k/j-1}$; by monotonicity of convolution powers for positive
functions (if $f>g$ then $f^{*k}>g^{*k}$), the convolution exponential and its
integral are then finite.

#### Order bounds.

$S_{j,k}(w)$ obeys the following $\mathcal{O}$ bound,

$S_{j,k}(w)=\int_{t=0^{-}}^{w}\exp^{*}\left(\frac{k}{j}\frac{\zeta(1+k/j)}{\zeta(k)}w^{k/j-1}+\frac{1}{j}\mathcal{O}(w^{1/j-1})\right)dt-1.$

First, we’ve used the error term formulation of $I_{j,k}$ in
$(\ref{I_j_k_w_R})$, where the error term is at worst of order $w^{1/j}$.
Lemma 5 is then applied as follows. From the definition of an order bound,
$I_{j,k}(w)$ can at worst be bounded below by
$\max\\{H(w)-c_{l}w^{1/j}-b_{l},0\\}$, and above by $H(w)+c_{h}w^{1/j}+b_{h}$,
where $H(w)=\zeta(1+k/j)w^{k/j}/\zeta(k)$. Since $k\geq 2$, both the lower and
upper bounds are then monotone increasing, making both bounds’ derivatives
$\geq 0$, completing the requirements of the lemma. (Note also the constant
term, $b_{h}$, when differentiating the bound, will produce $b_{h}\delta_{0}$
in the derivative, which has been accounted for in the lemma; there is no jump
discontinuity for the lower bound, since $H(0)=0$, and the bound is not
allowed to go below $0$.) To prove finiteness of the resulting convolution
exponential and its integral, apply Lemma 4 by noting that for any fixed
$w>0$, there is some $c_{w}>0$ such that,

$c_{w}v^{1/j-1}\geq v^{k/j-1}+\frac{c}{j}v^{1/j-1}\geq 0,$

for all $v\in[0,w]$. We can then apply the lemma with $cv^{a}=c_{w}v^{1/j-1}$.
The resulting finite bound, $e^{*c_{w}v^{1/j-1}}$, then applies by
monotonicity of convolution powers for positive functions as in the centerline
estimate.

###  4.3. Estimates in the particular case of $\mathbf{S_{1,2}(w)}$.

We’ll examine some of the differences between the estimates and the actual
step function in the case of the sum of square roots counting function,
$S_{1,2}(w)$.

#### First order estimate of $\mathbf{S_{1,2}(w)}$.

In the case of sums of square roots, a simple first order estimate can be
derived by using the first order estimate of $I_{1,2}(w)$ from equation
$(\ref{I_first_order})$. Lemma 4 gives,

$e^{*\chi_{[0,\infty)}cw}=\delta(w)+\sum_{m=1}^{\infty}c^{m}\frac{w^{2m-1}}{(2m-1)!m!}.$

This produces

(4.4)
$S_{1,2}(w)=\int_{t=0^{+}}^{w}e^{*dI_{1,2}}dt\approx\int_{t=0^{+}}^{w}e^{*2t\zeta(3)/\zeta(2)}dt=\sum_{m=1}^{\infty}\left(\frac{2\zeta(3)}{\zeta(2)}\right)^{m}\frac{w^{2m}}{(2m)!m!}.$

#### Centerline estimate of $\mathbf{S_{1,2}(w)}$.

Because of the sensitivity of $S_{1,2}(w)$ to errors in estimating
$I_{1,2}(w)$, the first order estimate of $S_{1,2}(w)$ can be improved (at
least for values of $w$ up to around $30$) by using the so-called centerline
estimate for $I_{1,2}(w)$ above (equation $(\ref{I_centerline})$). It gives
for $dI_{1,2}$,

$dI_{1,2}\approx\frac{2\zeta(3)}{\zeta(2)}w+\chi_{[1,\infty)}\frac{1}{w}+\left(\gamma-\frac{1}{2}\ln(2\pi)\right)\delta(w).$

This produces,

$S_{1,2}(w)\approx\int_{t=0^{+}}^{w}\exp^{*}\left(\frac{2\zeta(3)}{\zeta(2)}w+\chi_{[1,\infty)}\frac{1}{w}+(\gamma-\frac{1}{2}\ln(2\pi))\delta(w)\right)dt$
(4.5)
$=e^{\gamma-\frac{1}{2}\ln(2\pi)}\left(\sum_{n=1}^{\infty}\left(\frac{2\zeta(3)}{\zeta(2)}\right)^{n}\frac{w^{2n}}{(2n)!n!}\right)*\left(e^{*\chi_{[1,\infty)}1/w}\right).$

The righthand factor, $e^{*\chi_{[1,\infty)}1/w}[t]$, appears to have no
analytical expression, though it can be closely bounded between constants
using Stirling numbers of the first kind and the point mass approximation:
$\sum_{n=1}\delta(w-n)/n$; numerical tests suggest it is quickly asymptotic
(by $t\approx 3$) to $\approx 0.59$.

#### Bounds on $\mathbf{S_{1,2}(w)}$.

The $\mathcal{O}$ bound on $I_{1,2}(w)$ from equation $(\ref{R_jk_O_bound})$
with the help of Lemma 5 can produce order bounds on $S_{1,2}(w)$:

$S_{1,2}(w)=\left(1+\sum_{n=1}^{\infty}\left(\frac{2\zeta(3)}{\zeta(2)}\right)^{n}\frac{w^{2n}}{(2n)!n!}\right)*\left(e^{*\mathcal{O}(w^{2\beta_{2}-1})}\right)-1.$

#### Discussion of estimates of $\mathbf{S_{1,2}(w)}$.

Figures 5 and 5 show comparisons between the actual curve of $S_{1,2}(w)$ and
the estimates of equations $(\ref{S_first_order})$ and $(\ref{S_centerline})$.
These curves show the rapid increase of $S_{1,2}(w)$ as it exceeds the order
of any polynomial but remains sub-exponential.

The centerline estimate stays within 10% of the actual curve while the first
order estimate shows a wider variation.

Figure 4. _Estimates of $S_{1,2}(w)$._

Figure 5. _Dashed box inset._

###  4.4. Remarks on accuracy in the general case, $\mathbf{S_{j,k}(w)}$.

Though in general the first order and centerline $I_{j,k}$ estimates initially
do not deviate from the $I_{j,k}$ staircase by more than a few percent, the
effect of the error is magnified when $dI_{j,k}$ is put through the
convolution exponential; consider that shift upward in the estimate of
$I_{j,k}$ of magnitude $1$ translates to a point mass of weight $1$ for $dI$’s
estimate at $w=0$, which in turn results in an additional factor of $e^{1}$ in
the estimate of $S_{j,k}$ (to help see why, see just below the proof of Lemma
5). In other words, estimates for $S_{j,k}$ are very sensitive to errors in
the $I_{j,k}$ estimates; offset errors in the second translate into errors of
_scale_ for the first. From numerical checks when $\\{j,k\\}\neq\\{1,2\\}$,
both the first order and centerline estimates for example can be quite poor in
terms of relative error.

If greater accuracy is required in estimating $S_{j,k}$, one approach that has
worked in practice are hybrid estimates to $I_{j,k}$. One involves using the
first order estimate in $[0,w_{0}]$ for some suitable $w_{0}$, then the
centerline estimate is used for the remainder of the interval of support.
(Another improvement may result from truncating the $I_{j,k}$ estimate below
$w=1$, so that both the estimate to $I$ and its derivative are zero in
$[0,1]$.) Another form of hybrid estimate would be to hard code the actual
$I_{j,k}$ and $dI_{j,k}$ function up to some threshold $w_{0}$, then resume
with the first order or centerline estimates for the rest, since on a fixed
interval $t\in[0,w]$, with $f(t)$ supported in $[0,\infty)$, the convolution
exponential of $f$ is especially sensitive to errors in $f$ at low values of
$t$.

## Appendix

### Case bounds.

This section is devoted to finding order bounds in $l$ (and the accompanying
$w$ order) on the following integral (equation
$(\ref{jth_summand_asymptote})$):

$\frac{h_{l}}{l}\frac{1}{2\sqrt{2\pi}}\int_{x+y_{0}i}^{x+i\infty}e^{i\frac{\pi}{2}z}z^{-(z+\frac{3}{2})}(2\pi
lw^{2}e)^{z}dz,$

where $l\geq 1$ and $y_{0}\geq 2M=3$ and the requirement of $-3/2<x<-1/2$
(from the requirement on the expansion of the negative tri-zeta term (equation
$(\ref{hmj_v_zeta})$) and from the low $x$ limit for convergence of vertical
contours (equation $(\ref{I_remainder_bel_1_k})$) ).

The method for a given $l$ and $y_{0}$ is to take the absolute value of the
integrand and follow the path(s) of steepest descent, which allows finding a
tight bound while avoiding the complications that can arise from oscillatory
terms. The Cauchy integral theorem then guarantees the equivalence between the
original vertical contour of $(\ref{jth_summand_asymptote})$ and the steepest
descent paths since the integrand has no off-axis singularities in the
negative-real half-plane when $|\mbox{Im }z|>0$.

Recall the location of the saddle point in the $+\mbox{Im }$ half-plane:
$i\alpha_{l}=2\pi ilw^{2}$. Under the scaling $s=z/(lw^{2})$ equation
$(\ref{jth_summand_asymptote})$ became

$\frac{h_{l}}{l^{\frac{3}{2}}w}\frac{1}{2\sqrt{2\pi}}\int_{s=\frac{x}{lw^{2}}+\frac{iy_{0}}{lw^{2}}}^{\frac{x}{lw^{2}}+i\infty}s^{-\frac{3}{2}}\exp\left[lw^{2}s\left(\frac{i\pi}{2}+1+\ln\left(\frac{2\pi}{s}\right)\right)\right]ds.$

Rewriting the exponential portion as $e^{lw^{2}(u(a,b)+iv(a,b))}$, the
negative of the gradient of $u(a,b)$ is shown superimposed on the complex
s-plane in Figure 6. The saddle point is shown at $2\pi i$ along with the
proposed (scaled) contours that will be used below. The contours have been
chosen according to the method of steepest descent to minimize the error in
estimating bounds for equation $(\ref{jth_summand_asymptote})$ when evaluating
it by taking the absolute value of its integrand.

Figure 6. Contours in the scaled s-plane used for the Case bounds. The vector
field depicts the negative gradient of
$u(a,b)=a\ln(2\pi)+a(1-\ln(r))+b(\theta-\pi/2)$. The triangle denotes the
saddle point and the horizontal dashed lines are at $iM$, $2iM$, $2\pi i$ and
$2\pi ei$.

The following bounds will be derived by considering the contours in Figure 6.
We’ll use the absolute value of the integrand in
$(\ref{jth_summand_asymptote})$ without the leading $h_{l}/(2\sqrt{2\pi}l)$
term (for simplicity of notation):

$\left|e^{i\frac{\pi}{2}z}z^{-(z+\frac{3}{2})}(2\pi
lw^{2}e)^{z}\right|=r^{-3/2}e^{y(\theta-\pi/2)}\left(\frac{\alpha_{l}e}{r}\right)^{x}\equiv
F(z)$

Also for clarity in notation, denote $x$ in the vertical contour of
$(\ref{jth_summand_asymptote})$ by $x_{0}=x$.

Examining the following three cases of $iy_{0}$ in reference to saddle point
$i\alpha_{l}$ will provide the needed bounds:

  1. Case I:

$\alpha_{l}e\leq y_{0}$

  2. Case II:

$\alpha_{l}\leq y_{0}\leq\alpha_{l}e$

  3. Case III:

$y_{0}\leq\alpha_{l}$

In all Cases, the following bounds on the arctangent will be useful:

(A.6)
$\frac{u}{v}-\frac{1}{3}\left(\frac{u}{v}\right)^{3}<\arctan\left(\frac{u}{v}\right)\leq\frac{u}{v}\mbox{when
$0\leq u\leq v$}$

along with the well-known identity
$\arctan\left(u/v\right)=\pi/2-\arctan\left(v/u\right)$

#### Case I

In this Case, $\alpha_{l}e<y_{0}$. As suggested by $u(a,b)$ and Figure 6, the
contour will be changed from the line connecting $x+iy_{0}$ and $x+i\infty$ to
the line connecting $x+iy_{0}$ and $\infty+iy_{0}$. (At the end of the Cases
below, these horizontal and vertical contours in the original integral,
$(\ref{hmj_simpler_integrand})$, will be shown to be equal.)

The horizontal contour will be split into three parts. Part I will be the line
segment connecting $x_{0}+iy_{0}$ and $iy_{0}$, Part II will be the segment
beteween $iy_{0}$ and $y_{0}+iy_{0}$ and Part III will be the segment
connecting $y_{0}+iy_{0}$ and $\infty+iy_{0}$.

Part I: By $(\ref{arctan_bounds})$ combined with
$e^{y_{0}(\theta-\pi/2)}=e^{y_{0}(\arctan(|x|/y_{0}))}<e^{|x|}$, and since for
any $x$ on the contour,

$\left(\frac{\alpha_{l}e}{r}\right)^{x}\leq\left(\frac{\alpha_{l}e}{r}\right)^{x_{0}}<\left(\frac{\alpha_{l}e}{cy_{0}}\right)^{x_{0}}$

(where the last inequality follows by choosing $c$ such that $r<cy_{0}$ for
all $y_{0}>2M=3$ and $x_{0}\in(-3/2,-1/2)$),

$\int_{x_{0}+iy_{0}}^{iy_{0}}F(z)dz<\int_{x_{0}+iy_{0}}^{iy_{0}}y_{0}^{-3/2}e^{|x_{0}|}\left(\frac{\alpha_{l}e}{cy_{0}}\right)^{x_{0}}dz<y_{0}^{-3/2}e^{|x_{0}|}\left(\frac{\alpha_{l}e}{cy_{0}}\right)^{x_{0}}|x_{0}|.$

Writing $y_{0}=t\alpha_{l}e$, $t>1$, the bound becomes,

(A.7)
$t^{-3/2-x_{0}}c^{-x_{0}}e^{|x_{0}|}|x_{0}|(\alpha_{l}e)^{-3/2}<c^{-x_{0}}e^{|x_{0}|}|x_{0}|(\alpha_{l}e)^{-3/2}$

where the inequality follows since $-3/2-x_{0}<0$ and $t>1$.

Part II: By $(\ref{arctan_bounds})$ again,

$\textstyle
e^{y_{0}(\theta-\pi/2)}<e^{y_{0}(-x/y_{0}+(1/3)(x/y_{0})^{3})}<e^{-(2/3)x}$

where the last inequality results from applying a bounding chord between the
contour endpoints at $x=0$ and $x=y_{0}$ of the curve of positive concavity in
this region, $y_{0}(-x/y_{0}+(1/3)(x/y_{0})^{3})$. Since
$r^{-3/2}<y_{0}^{-3/2}$, and $\left(\alpha_{l}e/r\right)^{x}\leq 1$,

(A.8)
$\int_{iy_{0}}^{y_{0}+iy_{0}}F(z)dz<\frac{3}{2}y_{0}^{-3/2}\left(1-e^{-(2/3)y_{0}}\right)$

Part III: On this contour, $e^{y(\theta-\pi/2)}\leq e^{-y_{0}\pi/4}$,
$r^{-3/2}<y_{0}^{-3/2}$ and
$\left(\alpha_{l}e/r\right)^{x}\leq\left(1/\sqrt{2}\right)^{x}$ since
$\alpha_{l}e\leq y_{0}$. So,

$\int_{y_{0}+iy_{0}}^{\infty+iy_{0}}F(z)dz<y_{0}^{-3/2}e^{-y_{0}\pi/4}\int_{x=y_{0}}^{\infty}\left(\frac{1}{\sqrt{2}}\right)^{x}dz$
(A.9) $<\frac{2}{\ln 2}y_{0}^{-3/2}e^{-y_{0}(\pi/4+\ln(\sqrt{2}))}$

#### Case II

In this Case, the integral of $(\ref{jth_summand_asymptote})$ has lower limit
$y_{0}$ between $\alpha_{l}$ and $\alpha_{l}e$. The contour also will consist
of three parts: the path $x_{0}+iy_{0}$ to $iy_{0}$, the path along a diagonal
of slope $+1$ to $\alpha_{l}e-y_{0}+i\alpha_{l}e$, and the horizontal contour
from $\alpha_{l}e-y_{0}+i\alpha_{l}e$ to $\infty+i\alpha_{l}e$.

Part I: By reworking Case I with $\alpha_{l}$ in place of $\alpha_{l}e$, the
bound for the section $x_{0}+iy_{0}$ to $iy_{0}$ is
$c^{-x_{0}}e^{|x_{0}|}|x_{0}|(\alpha_{l})^{-3/2}$.

Part II:

Call this diagonal contour $C_{d}$. Rewrite $F(z)$ as
$r^{-3/2}e^{y(\theta-\pi/2)+x(\ln(\alpha_{l}e)-\ln r)}$, then noting along
$C_{d}$, $y=x+y_{0}$ and $r=\sqrt{x^{2}+(x+y_{0})^{2}}$, and that
$\theta=\tan^{-1}(-x/(x+y_{0}))+\pi/2$, define

$h(x)=(x+y_{0})\tan^{-1}\left(\frac{-x}{x+y_{0}}\right)+x\left[\ln(\alpha_{l}e)-\frac{1}{2}(x^{2}+(x+y_{0})^{2})\right].$

Next factor out the $\alpha_{l}$ term, and let $v=x/y_{0}$ to get
$h(x)=(\alpha_{l}/y_{0})^{x}e^{g(v)}$, where,

$g(v)=y_{0}\left((1+v)\arctan\left(-v/(1+v)\right)+v\left(1-(1/2)\ln(v^{2}+(1+v)^{2})\right)\right).$

This gives $F(z)=r^{-3/2}\left(\frac{\alpha_{l}}{y_{0}}\right)^{x}e^{g(v)}$.
The following bound applies to $g(v)$: $g(v)\leq-(y_{0}/2)v^{2}$ on
$v\in[0,e-1]$ (shown at the end of the Cases below). Also, along the contour,
$r^{-3/2}\leq{y_{0}}^{-3/2}$ and $\left(\alpha_{l}/y_{0}\right)^{x}\leq 1$.
So,

(A.10)
$\int_{C_{d}}F(z)dz<{y_{0}}^{-3/2}\int_{x=0}^{\alpha_{l}e-y_{0}}e^{-x^{2}/(2y_{0})}dx<\sqrt{\frac{\pi}{2}}y_{0}^{-1}$

Part III: Lastly, for the portion from $\alpha_{l}e-y_{0}+\alpha_{l}ei$ to
$\infty+\alpha_{l}ei$, Case I, equations $(\ref{Case_I_b})$ and
$(\ref{Case_I_c})$ with $y_{0}=\alpha_{l}e$ provide the needed bound since the
horizontal contour integral is monotone in its lower limit.

#### Case III

In this Case, $y_{0}<\alpha_{l}$. The path will be as follows: starting at
$x_{0}+iy_{0}$, go in the $-\mbox{Im}$ direction to meet the diagonal
connecting $iy_{0}$ and $-y_{0}$ then follow the diagonal down to
$-y_{0}+M+iM$, then proceed in the $-\mbox{Re}$ direction to
$-\alpha_{l}+M+iM$ then along a return diagonal to $x_{0}+i(\alpha_{l}+x_{0})$
and finally end at $x_{0}+\alpha_{l}i$ after going in the $+\mbox{Im}$
direction. The bounds from Case II will then complete the contour from
$x_{0}+\alpha_{l}i$ to $x_{0}+i\infty$.

Part I: First, to treat the vertical portions of the contour along the lines
connecting $x_{0}+iy_{1}$ and $x_{0}+i(y_{1}+x_{0})$, where
$y_{1}\in\\{\alpha_{l},y_{0}\\}$, the following bounds apply:
$r^{-3/2}<(y_{1}+x_{0})^{-3/2}$, $e^{y(\theta-\pi/2)}<e^{|x_{0}|}$ (by
$(\ref{arctan_bounds})$) and

$\textstyle\left(\frac{\alpha_{l}e}{r}\right)^{x}<\left(\frac{\alpha_{l}e}{y_{1}-x_{0}}\right)^{x_{0}}.$

This produces,

(A.11)
$\int_{x_{0}+iy_{1}}^{x_{0}+i(y_{1}+x_{0})}F(z)dz<(y_{1}+x_{0})^{-3/2}e^{|x_{0}|}\left(\frac{\alpha_{l}e}{y_{1}-x_{0}}\right)^{x_{0}}|x_{0}|.$

Part II: Next, we will find a bound on the diagonal portions. The procedure is
very similar to that of Case II, part II. The contour, call it $C_{d}$, will
be $x_{0}+i(y_{1}+x_{0})$ to $-y_{1}+M+Mi$ where $y_{1}=y_{0}$ for the portion
descending to $\mbox{Im }z=M$ and where $y_{1}=\alpha_{l}$ for the return.
Rewrite the integrand as,

$r^{-3/2}e^{y(\theta-\pi/2)}\left(\frac{\alpha_{l}e}{r}\right)^{x}=r^{-3/2}\left(\frac{\alpha_{l}}{y_{1}}\right)^{x}e^{g(v)}$

where, similar to Case II,
$g(v)=y_{1}\left((1+v)\arctan\left(-v/1+v\right)+v\left(1-(1/2)\ln(v^{2}+(1+v)^{2})\right)\right)$
with $v=x/y_{1}$. Here, $g(v)$ can be bounded: $g(v)\leq-y_{1}v^{2}$ on
$v\in[-1,0]$ (shown at the end of the Cases below). Also, along the contour,
$r^{-3/2}\leq\left(y_{1}/\sqrt{2}\right)^{-3/2}$ and
$\left(\alpha_{l}/y_{1}\right)^{x}\leq\left(\alpha_{l}/y_{1}\right)^{x_{0}}$.
So,

(A.12)
$\int_{C_{d}}F(z)dz<\left(\frac{y_{1}}{\sqrt{2}}\right)^{-3/2}\left(\frac{\alpha_{l}}{y_{1}}\right)^{x_{0}}\int_{x=x_{0}}^{-y_{1}+M}e^{-x^{2}/y_{1}}dx<\frac{\sqrt{\pi}}{2^{1/4}}y_{1}^{-1-x_{0}}{\alpha_{l}}^{x_{0}}$

Part III: Next, we will find a bound on the horizontal portion from
$-y_{0}+M+Mi$ to $-\alpha_{l}+M+Mi$; call this contour $C_{h}$. First, since
$\theta\in[3\pi/4,\pi)$, $e^{y(\theta-\pi/2)}<e^{M\pi/2}$. Next, to bound
$\left(\alpha_{l}e/r\right)^{x}$, rewrite it as $\exp[x(\ln(\alpha_{l}e)-\ln
r)]=e^{h(x)}$. Since $x$ is at most $-M$ (this occurs in the event
$y_{0}=2M$), take for the interval of interest $I_{x}:x\in[M-\alpha_{l},-M]$.
We’ll show $h(x)$ is negative and monotone increasing in this interval. This
allows bounding the whole term by its value at its rightmost (most positive)
possible endpoint. Since $0<r<\alpha_{l}<\alpha_{l}e$, $\ln(\alpha_{l}e/r)>0$.
This proves $h(x)<0$ on the interval in question. Next, take the derivative of
$h(x)$:

$\frac{d}{dx}h(x)=1-\frac{x^{2}}{x^{2}+M^{2}}+\ln\left(\frac{\alpha_{l}}{r}\right).$

Since both $1-x^{2}/(x^{2}+M^{2})$ and the $\ln$ term are greater than zero on
$I_{x}$, this proves $h(x)$ is monotone increasing, as desired. This allows
the bound,

$\textstyle\left(\frac{\alpha_{l}e}{r}\right)^{x}\leq\left(\frac{\alpha_{l}e}{\sqrt{2}M}\right)^{-M},$

valid on $I_{x}$. The whole integral can then be bounded,

(A.13) $\displaystyle\int_{C_{h}}F(z)dz$ $\displaystyle<$ $\displaystyle
e^{M\pi/2}\left(\frac{\alpha_{l}e}{\sqrt{2}M}\right)^{-M}\int_{C_{h}}r^{-3/2}dz$
$\displaystyle<$ $\displaystyle
2e^{M\pi/2}\left(\frac{\alpha_{l}e}{\sqrt{2}M}\right)^{-M}\left(\frac{1}{\sqrt{y_{0}-M}}-\frac{1}{\sqrt{\alpha_{l}-M}}\right)$

(where, recall, $M=3/2$).

Part IV: For the remainder of the contour, $x_{0}+i\alpha_{l}$ to
$x_{0}+i\infty$, the bounds of Case II can be used by setting
$y_{0}=\alpha_{l}$.

### Supporting bounds.

This section addresses the open items from the Cases.

#### Horizontal vs. vertical contours.

In Cases I and II, a bound was developed for a horizontal contour. To show
this is equivalent to the original vertical contour in
$(\ref{hmj_simpler_integrand})$, we will first show that a contour from
$z=K+iy_{0}$ to $K+iy_{1}$, $y_{0}<y_{1}$ and $K>>0$ approaches zero as
$K\to\infty$. Since $e^{y(\theta-\pi/2)}<1$, $F(z)$, the integrand of the
principal term in the asymptotic expansion, $(\ref{asymptote_integrand})$, is
less than $(\alpha_{l}e)^{K}(K^{2}+y^{2})^{-(1/2)(K+3/2)}$, producing the
bound

$\int_{K+iy_{0}}^{K+iy_{1}}F(z)dz<c(y_{1}-y_{0})\frac{(\alpha_{l}e)^{K}}{(K^{2}+y_{0}^{2})^{(1/2)(K+3/2)}}$

which goes to zero in the limit $K\to\infty$. Since the bound on the integrand
also holds after including the remainder terms $R_{k}$ (within the mentioned
constant factor), the integral of $(\ref{hmj_simpler_integrand})$ must go to
zero for this modified contour in the limit of $K$ as well.

Next, for a contour from $x_{0}+iy_{1}$ to $K+iy_{1}$ where
$y_{1}>\alpha_{l}e$, Case I shows a bound of
$\mathcal{O}\left(y_{1}^{-3/2-x_{0}}\right)$ for fixed $j$ and so must vanish
as $y_{1}\to\infty$ for all $K>0$. Since, again, the bound applies equally
well when the remainder terms are included, $(\ref{hmj_simpler_integrand})$
must vanish along this contour in the limit $y_{1}\rightarrow\infty$.

Since the integrand of $(\ref{hmj_simpler_integrand})$ has no singularities in
the $y>0$ half-plane, we can apply the Cauchy integral theorem to the
rectangle formed by $x+iy_{0}$, $K+iy_{0}$, $K+iy_{1}$ and $x+iy_{1}$. Taking
the limit in $K$ and $y_{1}$ shows the equivalence of horizontal and vertical
contours in $(\ref{hmj_simpler_integrand})$.

#### The parabolic bounds.

In Cases II and III, parabolic bounds were used within the integrand for the
diagonal portions of the contours. To justify these bounds, recall the
function in question was

$\textstyle
g(v)=y_{1}\left((1+v)\arctan\left(\frac{-v}{1+v}\right)+v\left(1-\frac{1}{2}\ln(v^{2}+(1+v)^{2})\right)\right)$

and a bound of $-y_{1}v^{2}$ was claimed on $v\in[-1,0]$ and a bound of
$-(y_{1}/2)v^{2}$ was claimed on $v\in[0,e-1]$. For the bound when $v\leq 0$,
notice the bound and $g(v)$ both are equal to zero and have a derivative of
zero at $v=0$ as well as share a point of intersection at $v=-1$. Two
applications of the mean value theorem will show for there to be another point
of intersection in the interior $v\in(-1,0)$, the second derivatives of $g(v)$
and the bound must agree in at least two points within the interval. Since
$g^{\prime\prime}(v)=-2y_{1}(v+1)/(v^{2}+(1+v)^{2})$, it is easy to verify
this does not occur. Checking an arbitrary point in the interior that the
bound is greater than $g(v)$ completes the proof. Similar reasoning applies
for the bound when $v\geq 0$, noting that a point of intersection occurs
within $v\in(e-1,4)$.

## References

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