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# The numerical evaluation of the Riesz function

R. B. Paris
Division of Computing and Mathematics,
Abertay University, Dundee DD1 1HG, UK

###### Abstract

The behaviour of the generalised Riesz function defined by

$S_{m,p}(x)=\sum_{k=0}^{\infty}\frac{(-)^{k-1}x^{k}}{k!\zeta(mk+p)}\qquad(m\geq
1,\ p\geq 1)$

is considered for large positive values of $x$. A numerical scheme is given to
compute this function which enables the visualisation of its asymptotic form.
The two cases $m=2$, $p=1$ and $m=p=2$ (introduced respectively by Hardy and
Littlewood in 1918 and Riesz in 1915) are examined in detail. It is found on
numerical evidence that these functions appear to exhibit the $x^{-1/4}$ and
$x^{-3/4}$ decay, superimposed on an oscillatory structure, required for the
truth of the Riemann hypothesis.

Mathematics subject classification (2020): 26A12, 41A60, 30B10, 30E15, 33E20,
11M06

Keywords: Riesz function, asymptotic behaviour, numerical scheme, Riemann
hypothesis

1\. Introduction

The generalised Riesz function is defined by the sum

$S_{m,p}(x)=\sum_{k=0}^{\infty}\frac{(-)^{k-1}x^{k}}{k!\zeta(mk+p)}\qquad(x\geq
0),$ (1.1)

where $m\geq 1$, $p\geq 1$ and $\zeta(s)$ is the Riemann zeta function. The
original function considered by Riesz [8] took the form

$\sum_{k=1}^{\infty}\frac{(-)^{k-1}x^{k}}{(k-1)!\zeta(2k)},$

which corresponds to a case of (1.1) with $m=p=2$ since it is easily seen to
equal $-xS_{2,2}(x)$. A similar function corresponding to $m=2$, $p=1$ was
discussed in the famous memoir by Hardy and Littlewood [3]. The interest in
both these cases of (1.1) results from the fact that a necessary and
sufficient condition for the truth of the Riemann hypothesis is that [9, p.
382]

$S_{2,1}(x)=O(x^{-1/4+\epsilon}),\qquad S_{2,2}(x)=O(x^{-3/4+\epsilon})$ (1.2)

as $x\to+\infty$, where $\epsilon$ is an arbitrarily small positive quantity.
The results in (1.2) are superficially attractive as they are derived from
sums containing only values of $\zeta(s)$ at positive integer values of $s$.

The sum in (1.1) can also be viewed as an example of a perturbation of the
exponential series for $e^{-x}$ in the form $\sum_{k\geq 0}a_{k}(-x)^{k}/k!$,
where $a_{k}$ are coefficients that possess the property $a_{k}\to 1$ as
$k\to\infty$; in the case of the Riesz function we have $a_{k}=1/\zeta(mk+p)$.
The growth of this series for large (complex) $x$ is found to depend
sensitively on the decay of the perturbing coefficients $a_{k}$. A discussion
of this problem, together with several examples, is given in [6].

This paper is partly based on the earlier report by the author in [5]. We
present a computational scheme for the numerical evaluation of $S_{m,p}(x)$
for large positive values of $x$. In particular, we concentrate on the Hardy-
Littlewood case of $m=2$, $p=1$ and also on the Riesz case $m=p=2$ and
determine numerically their large-$x$ behaviour. Based on our numerical
results, we conclude that these cases are characterised by a damped
oscillatory structure for sufficiently large positive $x$ with an amplitude
that corresponds to the estimates in (1.2).

2\. A computational scheme

We use the result [9, p. 3]

$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}\qquad(\Re(s)>1),$

where $\mu(n)$ is the Möbius function defined by $\mu(n)=(-)^{r}$ if $n$ has
$r$ distinct primes (with $\mu(1)=1$) and $\mu(n)=0$ otherwise. Then we obtain
the expansion for the case $m=2$, $p\geq 1$

$\displaystyle S_{2,p}(x)\\!\\!$ $\displaystyle=$
$\displaystyle\\!\\!\sum_{k=0}^{\infty}\frac{(-)^{k-1}x^{k}}{k!\zeta(2k+p)}$
(2.1) $\displaystyle=$
$\displaystyle\\!\\!\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{p}}(1-e^{-X_{n}})-\frac{1}{\zeta(p)},\qquad
X_{n}:=\frac{x}{n^{2}}.$

The factor $1-e^{-X_{n}}$ has a “cut-off” when $n=N=\lceil x^{1/2}\rceil$; for
$n>N$, the decay of the terms in this series is slow with late terms
eventually behaving like $\mu(n)/n^{p+2}$.

This slow decay in the tail can be accelerated as follows. We write

$S_{2,p}(x)=\sum_{n=1}^{N-1}\frac{\mu(n)}{n^{p}}(1-e^{-X_{n}})+T_{N}(p;x)-\frac{1}{\zeta(p)},\qquad
T_{N}(p;x):=\sum_{n=N}^{\infty}\frac{\mu(n)}{n^{p}}(1-e^{-X_{n}}),$ (2.2)

where $N$ is chosen as above. In the tail $T_{N}(p;x)$ (where $X_{n}\leq 1$)
we put

$1-e^{-X_{n}}=X_{n}\biggl{(}1-\frac{X_{n}}{2!}+\frac{X_{n}^{2}}{3!}-\cdots\biggr{)}=X_{n}\,{}_{1}F_{1}(1;2;-X_{n})$
$=X_{n}e^{-X_{n}}{}_{1}F_{1}(1;2;X_{n})=X_{n}f_{1}(X_{n}),$

where

$f_{k}(z):=e^{-z}{}_{1}F_{1}(1;k+1;z)\qquad(k\geq 0).$ (2.3)

Here ${}_{1}F_{1}$ denotes the confluent hypergeometric function and Kummer’s
transformation [4, p. 325] has been used to change the argument from $-X_{n}$
to $X_{n}$.

From (A.2) in the appendix the following lemma is established:

Lemma 1. The function $zf_{1}(z)$ satisfies the recursion formula

$zf_{1}(z)=1-\frac{1}{e_{k}(z)}+\frac{z^{k}f_{k}(z)}{k!\,e_{k}(z)}\qquad(k\geq
1),$ (2.4)

where $e_{k}(z)$ denotes the sum of the first $k$ terms of the exponential
series

$e_{k}(z):=\sum_{r=0}^{k-1}\frac{z^{r}}{r!}.$ (2.5)

From the definition (2.5) it follows that

$\frac{1}{e_{k}(z)}=e_{k}(-z)+\frac{z^{k}g_{k}(z)}{k!\,e_{k}(z)},$ (2.6)

where $g_{k}(z)$ is a polynomial of degree $k-2$. Some routine algebra shows
that

$g_{2}(z)=2,\quad g_{3}(z)=-\frac{3}{2}z,\quad g_{4}(z)=2+\frac{2}{3}z^{2},$
$g_{5}(z)=-\frac{5}{3}z-\frac{5}{24}z^{3},\quad
g_{6}(z)=2+\frac{3}{4}z^{2}+\frac{1}{20}z^{4},\ldots\,.$

Then, using (2.4)–(2.6), the tail of the series can be written in the form

$T_{N}(p;x)=T_{N}^{(1)}(p;x)+T_{N}^{(2)}(p;x),$ (2.7)

where

$T_{N}^{(1)}(p;x)=\frac{1}{k!}\sum_{n=N}^{\infty}\frac{\mu(n)}{n^{p}}\,\frac{X_{n}^{k}\,\Delta_{k}(X_{n})}{e_{k}(X_{n})},\qquad\Delta_{k}(X_{n}):=f_{k}(X_{n})-g_{k}(X_{n})$
(2.8)

and

$T_{N}^{(2)}(p;x)=\sum_{n=N}^{\infty}\frac{\mu(n)}{n^{p}}\sum_{r=1}^{k-1}\frac{(-)^{r-1}X_{n}^{r}}{r!}=\sum_{r=1}^{k-1}\frac{(-X_{N})^{r}}{r!}\,\lambda_{r}$
(2.9)

with

$X_{N}:=\frac{x}{N^{2}},\qquad\lambda_{r}:=N^{2r}\biggl{\\{}\sum_{r=1}^{N-1}\frac{1}{n^{2r+p}}-\frac{1}{\zeta(2r+p)}\biggr{\\}}.$

Since $X_{n}\to 0$ as $n\to\infty$, we see that the decay of the late terms in
$T_{N}^{(1)}(x)$ is now controlled by $\mu(n)/n^{2k+p}$, which for $k\geq 2$
represents a modest improvement in the rate of convergence of the series.

Collecting together the results in (2.2), (2.7) – (2.9), we have

$S_{2,p}(x)=\sum_{n=1}^{N-1}\frac{\mu(n)}{n^{p}}(1-e^{-X_{n}})-\frac{1}{\zeta(p)}+T_{N}^{(1)}(p;x)+T_{N}^{(2)}(p;x)$
(2.10)

for $p\geq 1$.

3\. Numerical results for $m=2$ and $p=1,2$

We have employed the scheme (2.10) with $k=6$ to compute the Hardy-Littlewood
case $m=2$ $p=1$ for $x\geq 0$ up to $x=10^{8}$. The results are shown in the
sequence of plots in Fig. 1. It is found that $S(x)\equiv S_{2,1}(x)$
decreases once
$x\,\raisebox{-3.44444pt}{\mbox{$\stackrel{{\scriptstyle\textstyle>}}{{\sim}}$}}\,3$
down to values of the order $10^{-6}$, whereupon the graph commences to
oscillate about the zero line. The oscillations appear to be regular and have
a decreasing amplitude.

($a$) ($b$)

($c$) ($d$)

Figure 1: The graph of $S(x)\equiv S_{2,1}(x)$ against $\log_{10}x$ for
different ranges of $x$.

The successive maxima111We commence the enumeration of the maxima and minima
from the point where the graph of $S(x)$ first becomes negative. ($x_{k}^{+}$)
and minima ($x_{k}^{-}$) in the oscillatory region are determined by
calculating the zeros of the derivative $S_{2,1}^{\prime}(x)$, where

$S_{2,1}^{\prime}(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{3}}\,e^{-X_{n}}=\frac{1}{\zeta(3)}-\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{3}}\,(1-e^{-X_{n}}),$

by using (2.10) with $p=3$. The results of these calculations together with
the corresponding values of $S(x_{k}^{\pm})$ are shown in Table 1. Plots of
$\log\,S_{2,1}(x_{k}^{\pm})$ against $\log\,x$ are shown in Fig. 2, where it
is seen that they reveal a linear variation to a good approximation. The
dashed lines in these figures have slope equal to $-0.25$, thereby numerically
confirming the estimate in (1.2).

Table 1: Values of the maxima and minima $\log_{10}x_{k}^{\pm}$ and the corresponding values of $S_{2,1}(x_{k}^{\pm})$. $k$ | $\log_{10}x_{k}^{+}$ | $S_{2,1}(x_{k}^{+})$ | $\log_{10}x_{k}^{-}$ | $S_{2,1}(x_{k}^{-})$
---|---|---|---|---
1 | 4.83284 | $2.62044\times 10^{-6}$ | 4.65573 | $-4.78520\times 10^{-7}$
2 | 5.22278 | $1.65298\times 10^{-6}$ | 5.03476 | $-1.20507\times 10^{-6}$
3 | 5.61033 | $1.21568\times 10^{-6}$ | 5.41918 | $-1.18328\times 10^{-6}$
4 | 5.99669 | $9.38573\times 10^{-7}$ | 5.80406 | $-1.00735\times 10^{-6}$
5 | 6.38315 | $7.47310\times 10^{-7}$ | 6.19039 | $-8.20165\times 10^{-7}$
6 | 6.76905 | $5.93349\times 10^{-7}$ | 6.57596 | $-6.62603\times 10^{-7}$
7 | 7.15545 | $4.76797\times 10^{-7}$ | 6.96249 | $-5.30007\times 10^{-7}$
8 | 7.54124 | $3.80007\times 10^{-7}$ | 7.34817 | $-4.26057\times 10^{-7}$

($a$) ($b$)

Figure 2: Plots of (a) $\log_{10}S(x_{k}^{+})$ and (b) $\log_{10}S(x_{k}^{-})$
against $\log_{10}x$ (where $S(x)\equiv S_{2,1}(x)$). The dashed lines have
slope $-0.25$.

($a$) ($b$)

($c$) ($d$)

Figure 3: The graph of $S(x)\equiv S_{2,2}(x)$ against $\log_{10}x$ for
different ranges of $x$.

The case of the original Riesz function has $m=p=2$ and from (2.1) we have

$S_{2,2}(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}(1-e^{-X_{n}})-\frac{1}{\zeta(2)},$

where $\zeta(2)=\pi^{2}/6$. The results are shown in Fig. 3 for the range
$0\leq x\leq 10^{8}$ obtained using the scheme (2.10) with $p=2$. This
function presents a similar behaviour with its value decreasing until about
$10^{-8}$ before commencing to oscillate about the zero line. The first zero
has the value $x\doteq 1.15671$. A plot of $xS_{2,2}(x)$ is also given in [2].
The successive maxima and minima in the oscillatory region are computed as the
zeros of the derivative

$S_{2,2}^{\prime}(x)=\frac{1}{\zeta(4)}-\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{4}}(1-e^{-X_{n}}).$

The results of these calculations together with the corresponding values of
$S(x_{k}^{\pm})$ are shown in Table 2. Plots of $\log\,S_{2,2}(x_{k}^{\pm})$
against $\log\,x$ are shown in Fig. 4, where it is seen that they reveal a
linear variation to a good approximation. The dashed lines in these figures
have slope equal to $-0.75$, thereby numerically confirming the estimate in
(1.2).

Table 2: Values of the maxima and minima $\log_{10}x_{k}^{\pm}$ and the corresponding values of $S_{2,2}(x_{k}^{\pm})$. $k$ | $\log_{10}x_{k}^{+}$ | $S_{2,2}(x_{k}^{+})$ | $\log_{10}x_{k}^{-}$ | $S_{2,2}(x_{k}^{-})$
---|---|---|---|---
1 | 4.48752 | $4.97204\times 10^{-8}$ | 4.31797 | $-6.46896\times 10^{-9}$
2 | 4.87969 | $1.97351\times 10^{-8}$ | 4.69479 | $-1.67414\times 10^{-8}$
3 | 5.26779 | $9.09065\times 10^{-9}$ | 5.07699 | $-1.09071\times 10^{-8}$
4 | 5.65449 | $4.53355\times 10^{-9}$ | 5.46263 | $-5.99878\times 10^{-9}$
5 | 6.04080 | $2.28418\times 10^{-9}$ | 5.84775 | $-3.14719\times 10^{-9}$
6 | 6.42706 | $1.17580\times 10^{-9}$ | 6.23435 | $-1.62554\times 10^{-9}$
7 | 6.81307 | $5.98676\times 10^{-10}$ | 6.61979 | $-8.37753\times 10^{-10}$
8 | 7.19932 | $3.07392\times 10^{-10}$ | 7.00651 | $-4.29573\times 10^{-10}$

($a$) ($b$)

Figure 4: Plots of (a) $\log_{10}S(x_{k}^{+})$ and (b) $\log_{10}S(x_{k}^{-})$
against $\log_{10}x$ (where $S(x)\equiv S_{2,2}(x)$). The dashed lines have
slope $-0.75$.

It has been shown in [1] that $S_{2,2}(x)$ has an infinite number of zeros. It
is very probable that $S_{2,1}(x)$ also has an infinite number of zeros.

4\. An asymptotic expansion

An integral representation of $S_{m,p}(x)$ in the form of a Mellin-Barnes
integral is given by

$S_{m,p}(x)=-\frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty
i}\frac{\Gamma(s)x^{-s}}{\zeta(p-ms)}\,ds\qquad(0<c<c_{0}),$ (4.1)

where $c_{0}=(p-\mbox{${\textstyle\frac{1}{2}}$})/m$. The integrand possesses
simple poles at $s=0,-1-2,\ldots$ and at the trivial zeros of the zeta
function at $s=(p+2k)/m$, $k=1,2,\ldots\,$. On the assumption of the Riemann
hypothesis, there is also an infinite number of (simple) poles on the line
$\Re(s)=c_{0}$ given by $s=c_{0}\pm i\gamma_{k}/m$ ($k=1,2,\ldots$), where
$\zeta(\mbox{${\textstyle\frac{1}{2}}$}\pm i\gamma_{k})=0$. Assuming that it
is permissible to displace the integration path past this line, we obtain the
result

$S_{m,p}(x)=-\frac{2x^{-c_{0}}}{m}\Re\sum_{k=1}^{\infty}\frac{\Gamma(c_{0}-i\gamma_{k}/m)}{\zeta^{\prime}(\mbox{${\textstyle\frac{1}{2}}$}+i\gamma_{k})}\
x^{i\gamma_{k}/m}+O(x^{-(p+2)/m})$

as $x\to+\infty$. The details of the case $m=2$, $p=1$ are discussed in [3,
§2.5]; see also the account presented in [7, p. 143].

If we now set

$A_{k}=\frac{\Gamma(c_{0}-i\gamma_{k}/m)}{\zeta^{\prime}(\mbox{${\textstyle\frac{1}{2}}$}+i\gamma_{k})},\qquad\psi_{k}=\pi+\arg
A_{k},$

we find that

$S_{m,p}(x)=\frac{2x^{c_{0}}}{m}\sum_{k=1}^{\infty}|A_{k}|\cos\biggl{\\{}\frac{\gamma_{k}}{m}\,\log\,x+\psi_{k}\biggr{\\}}+O(x^{-(p+2)/m}).$
(4.2)

The convergence of the sum (4.2) is difficult to establish. The gamma function
present in the coefficients $A_{k}$ decays very rapidly for increasing $k$,
since from Stirling’s formula it contains the exponential factor
$\exp\,[-\pi\gamma_{k}/(2m)]$ for large $\gamma_{k}$. The magnitude of
$\zeta^{\prime}(\mbox{${\textstyle\frac{1}{2}}$}+i\gamma_{k})$ (which is non-
zero on the assumption that the non-trivial zeros are all simple) generally
increases with $k$, but it is possible that there are zeros for which this
quantity could become small.

When $m=p=2$, we obtain the expansion

$S_{2,2}(x)=x^{-3/4}\sum_{k=1}^{\infty}|A_{k}|\cos\,\biggl{\\{}\frac{\gamma_{k}}{2}\,\log\,x+\psi_{k}\biggr{\\}}+O(x^{-2})$
(4.3)

with
$A_{k}=\Gamma(\mbox{${\textstyle\frac{3}{4}}$}-\mbox{${\textstyle\frac{1}{2}}$}i\gamma_{k})/\zeta^{\prime}(\mbox{${\textstyle\frac{1}{2}}$}+i\gamma_{k})$.
The graph of $S_{2,2}(x)$ against $\log_{10}x$ compared with the expansion
(4.3) truncated after $k=5$ terms is shown in Fig. 5. It is seen that for
$x\,\raisebox{-3.44444pt}{\mbox{$\stackrel{{\scriptstyle\textstyle>}}{{\sim}}$}}\,10^{6}$
the curves are indistinguishable on the scale of the figure.

Figure 5: The graph of $S(x)\equiv S_{2,2}(x)$ against $\log_{10}x$ compared
with the asymptotic form (4.3) (dashed curve).

5\. Concluding remarks

The numerical results obtained in Section 3 are indicative only. It appears
from the numerical investigation – without any reference to the Riemann
hypothesis – that the large-$x$ behaviour of the functions $S_{2,1}(x)$ and
$S_{2,2}(x)$ possesses respectively the $x^{-1/4}$ and $x^{-3/4}$ decay
superimposed on an oscillatory structure. A striking feature of the plots in
Figs. 1 and 3 is the fact that the final decaying oscillatory structure is not
obtained until $x$ has attained the value of approximately $10^{5}$. This is
an unusual occurrence since most special functions begin to exhibit their
asymptotic structure for often surprisingly modest values of the variable.

The computation of $S_{m,p}(x)$ for $m\geq 3$ is made easier since the cut-off
value $N$ then scales like $x^{1/m}$ and the rate of decay of the various
series in (2.10) is correspondingly more rapid. As an example, the case $m=3$,
$p=2$ is shown in [5]. The behaviour is found to be similar to that depicted
in Figs, 1 and 3, with the maxima and minima following an approximate
$x^{-1/2}$ scaling predicted by (4.2).

Appendix: Derivation of Lemma 1

From the contiguous relation satisfied by the confluent hypergeometric
function ${}_{1}F_{1}(a;b,z)$ [4, (13.3.2)]

$b(b-1){}_{1}F_{1}(a;b-1;z)+b(1-b-z){}_{1}F_{1}(a;b;z)+z(b-a){}_{1}F_{1}(a;b+1;z)=0,$

we obtain, with $a=1$, $b=k+2$, the recursion formula satisfied by
$f_{k}(z):=e^{-z}{}_{1}F_{1}(1;k+1;z)$ in the form

$f_{k}(z)-\biggl{(}1+\frac{z}{k+1}\biggr{)}f_{k+1}(z)+\frac{z}{k+2}f_{k+2}(z)=0\qquad(k\geq
0).$

Repeated use of this result, combined with $f_{0}(z)=1$ and the partial sum of
the exponential series $e_{k}(z)$ defined in (2.5), shows successively that

$\displaystyle f_{1}(z)$ $\displaystyle=$
$\displaystyle\frac{1}{e_{2}(z)}+\frac{ze_{1}(z)}{2e_{2}(z)}\,f_{2}(z),$
$\displaystyle f_{2}(z)$ $\displaystyle=$
$\displaystyle\frac{1}{e_{3}(z)}+\frac{ze_{2}(z)}{3e_{3}(z)}\,f_{3}(z),$

and, in general,

$f_{k-1}(z)=\frac{1}{e_{k}(z)}+\frac{ze_{k-1}(z)}{ke_{k}(z)}\,f_{k}(z)\qquad(k\geq
1).$ (A.1)

Then, from (A.1) we find that

$\displaystyle zf_{1}(z)$ $\displaystyle=$
$\displaystyle\frac{z}{e_{2}(z)}+\frac{z^{2}f_{2}(z)}{2e_{2}(z)}=1-\frac{1}{e_{2}(z)}+\frac{z^{2}f_{2}(z)}{2e_{2}(z)}$
$\displaystyle=$ $\displaystyle
1-\frac{1}{e_{2}(z)}+\frac{z^{2}}{2e_{2}(z)}\biggl{(}\frac{1}{e_{3}(z)}+\frac{ze_{2}(z)}{3e_{3}(z)}\,f_{3}(z)\biggr{)}$
$\displaystyle=$ $\displaystyle
1-\frac{1}{e_{3}(z)}+\frac{z^{3}f_{3}(z)}{3!e_{3}(z)},$

where we have used $e_{1}(z)=1$, $e_{2}(z)=1+z$ and have written
$z^{2}/2!=e_{3}(z)-e_{2}(z)$. This procedure can be continued to produce the
final result

$zf_{1}(z)=1-\frac{1}{e_{k}(z)}+\frac{z^{k}f_{k}(z)}{k!e_{k}(z)}\qquad(k\geq
1).$ (A.2)

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