# Analytic properties of Euler sums

Introduction

As far as I know this topic has not been discussed before. I have found interesting results which I wish to share with you in the standard MSE manner by asking questions.

Consider the Euler sum

$$S(1,s) =\sum _{k=1}^{\infty } \frac{H_k}{k^s}\tag{1}$$

where $H_k = \sum_{j=1}^k 1/j$ is the harmonic number and $s$ is a parameter which in the field of Euler sums is considered to assume integer values $\ge 2$.

Here we let $s$ be a complex number and ask for the analytic properties of $S(1,s)$ as a function of $s$.

In parallel and for comparison we consider the similar sum with the harmonic number replaced by unitiy. This is the well known zeta function studied first by Euler

$$\zeta(s) = \sum _{k=1}^{\infty } \frac{1}{k^s}\tag{2}$$

The analytic properties of this function have been studied first by Bernhard Riemann in his famous paper on the distribution of prime numbers of 1859 [1].

The zeta function is analytic in the whole complex $s$ plane except for a simple pole at $s=1$ with residue $1$. It has trivial simple zeroes at negative even integers and non trivial zeroes on the line $\Re[s] = 1/2$ possibly even all non trivial zeroes are located on this line (Riemann's hypothesis) [2],[3].

Question 1

What are the singularities of $S(1,s)$ in the complex s-plane? What can be said about the zeroes? Compare the results with those of $\zeta(s)$

First solution steps: derive an integral representation of $S(1,s)$, and split is into an holomorphic and a meromorphic part. Study the meromorphic part to find the singularities.

Question 2

$S(1,q)$ has a well known closed form for integer argument $q = 2, 3, 4, ..$ which was already found by Euler.

It is given by

$$S_E(1,q) = \left(\frac{q}{2}+1\right) \zeta (q+1)-\frac{1}{2} \sum _{k=1}^{q-2} \zeta (k+1) \zeta (q-k)\tag{3}$$

The question asks for a possible closed form for real $q \ge 2$

First solution steps: Here I have no idea how to find a better solution than the integral representation.

Question 3

Consider the analogue questions 1 and 2 also for the general linear harmonic sum including the generating function via the parameter $x$

$$S_H(p, q, x)=\sum _{k=1}^{\infty } \frac{x^k H_k^{(p)}}{k^q}\tag{4}$$

Here $H_k^{(p)}=\sum_{n=1}^k 1/n^p$ is the generalized harmonic number.

As there are three quantities in the game, to be definite, we consider some simplifying cases.

Case 1: $x=1$ and $x=-1$ (alternating sum) for fixed $p$ as a function of $q$

Case 2: fixed $x$ and $p=q$ as a function of $p$. Notice that there is an analogous formula to (3) for integer $p$

Case 3: $X$ fixed somewhere within the interval $(-1,1)$, e.g. $x=\frac{1}{2}$ and, say, $p=1$.

References

[1] http://www.claymath.org/sites/default/files/zeta.pdf (Bernhard Riemann, Über die Anzahl der Primzahlen unter einer gegebenen Größe, Berlin November 1859)
[2] https://en.wikipedia.org/wiki/Riemann_zeta_function
[3] https://de.wikipedia.org/wiki/Riemannsche_%CE%B6-Funktion

• (+1) You are not the first to wonder if $(3)$ leads to an approximated functional equation for the $\zeta$ function with some non-trivial consequences. This essentially is Exercise 38 of my notes, page 22. – Jack D'Aurizio Apr 15 '18 at 19:43
• Thanks for the hint. Very interesting. I'll take it look at it tomorrow.This question appeared to me just in passing. The main focus is on question 1, of course. – Dr. Wolfgang Hintze Apr 15 '18 at 19:59
• Thank you for the hint. Congratulations! There is plenty of valuable material in your notes. And I would appreciate to see your answer to question 2. – Dr. Wolfgang Hintze Apr 16 '18 at 7:15
• Well, by the integral representation below we already have that $S(1,s)$ is given by a derivative of the $\zeta$ function and a convolution of values of the $\zeta$ function, for any $s\not\in\mathbb{Z}$. – Jack D'Aurizio Apr 16 '18 at 9:56

Let us start collecting something.

$$\begin{eqnarray*} S(1,s) = \sum_{n\geq 1}\frac{\psi(n+1)+\gamma}{n^s} &=& \sum_{n\geq 1}\frac{1}{n^s}\left[\log n+\gamma+\frac{1}{2n}-\sum_{j\geq 1}\frac{B_{2j}}{2j n^{2j}}\right]\\&=&\sum_{n\geq 1}\frac{1}{n^s}\left[\log n+\gamma+\frac{1}{2n}-\sum_{j\geq 1}\frac{(-1)^{j+1}(2j)! \zeta(2j)}{j (2\pi n)^{2j}}\right]\\&\stackrel{\mathcal{L}^{-1}}{=}&\int_{0}^{+\infty}\frac{1}{e^u-1}\left(\frac{u^{s-1}}{\Gamma(s)}(\psi(s)-\log u+\gamma)-\sum_{j\geq 1}\frac{(-1)^{j+1}(2j)!\zeta(2j)u^{s+2j-1}}{j(2\pi)^{2j}\Gamma(2j+s)}\right)\,du\end{eqnarray*}$$

and if $s\in\{2,3,4,\ldots\}$ the series on $j$ simplifies in terms of the $\log\Gamma$ function, such that an explicit evaluation of the last integral is allowed by the residue theorem and the reflection formula. The similarity between $S(1,s)$ and $-\zeta'(s)$ is encoded by the term $\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{u^{s-1}\log(u)}{e^u-1}$, and the integral representation allows an analytic continuation of $S(1,s)$ to the complex plane. On the other hand, due to the presence of many spurious terms it is not clear how the roots of $S(1,s)$ and $-\zeta'(s)$ are related.

Still $S(1,s)$ has a reflection formula. We may derive it in the same way we derive the reflection formula for the $\zeta$ function, i.e. by enforcing the substitution $u\mapsto v^2$ in the integral representation and by exploiting the Poisson summation formula. The involved computations are pretty heavy so I need some time for performing an update of this (partial) answer.

• @ Jack D'Aurizio: Just a specific question: I have shown in my solution that S(1,q) has a double pole at $q=1$ and no simple pole in the same position. Using the asymptotic expansion of $H_k$ leads to an addtional a simple pole with residue $\gamma$. I consider this latter pole spurious. What is your opinion? – Dr. Wolfgang Hintze Apr 18 '18 at 13:39

Here we first give a step-by-step solution of the basic question 1, and then turn to other questions.

Integral representation of $S(p,q,x)$

Writing

$$k^{-q}=\frac{1}{\Gamma (q)}\int_0^{\infty } e^{-k \;t} t^{q-1} \, dt\tag{s1}$$

and observing the formula for the generating function of the generalized harmonic number

$$\sum _{k=1}^{\infty } z^k H_k^{(p)}=\frac{\text{Li}_p(z)}{1-z}\tag{s2}$$

we can write (4) as

$$S(p,q,x) =\frac{1}{\Gamma (q)}\int_0^{\infty } \frac{t^{q-1} \text{Li}_p\left(e^{-t} x\right)}{1-e^{-t} x} \, dt\tag{s3}$$

Hence $S(p,q,x)$ has the form of a Mellin tranformation defined as

$$M(f(t),t,q) = \int_{0}^\infty t^{q-1} f(t) \, dt$$

$$S(p,q,x) =\frac{1}{\Gamma (q)} M(f(p,x,t),t,q)\tag{s3a}$$

with the kernel

$$f(p,x,t) =\frac{ \text{Li}_p\left(e^{-t} x\right)}{1-e^{-t} x} \tag{s3b}$$

For $x = 1$ and $p=1$ this simplifies to

$$S(1,q) =\frac{1}{\Gamma (q)} M(f(t),t,q)\tag{s4}$$

The kernel simplifies to

$$f(t) = \frac{-\log(1-e^{-t})}{1-e^{-t}}\tag{s4a}$$

Notice that for $t>>1$ this kernel approaches the kernel for the $\zeta$-function:

$$f(t \to \infty) =f_{\zeta}(t) = \frac{1}{e^{t}-1}\tag{s5}$$

The method will now be illustrated by examining the simplified expression (s4).

In order to find singularities in $q$, we split the integral in (s4) into two parts $F=\int_0^1 f \, dt$ and $G=\int_1^\infty f \, dt$ and notice that the integral $G$ is always convergent so that $G$ is a holomorphic in $q$.

The singularities must therefore come from $F$, in particular from the vicinity of $t=0$ of the integration. Hence they can be found by expanding the integrand of $F$ into a series about $t=0$.

Singularities of $S(1,q)$

To lowest order in $t$ the integrand (s4a) is given by

$$t^{q-1} \left(-\frac{\log (t)}{2}-\frac{\log (t)}{t}+\frac{1}{2}\right)$$

Integrating over $t$ from $0$ to $1$ an taking into account the $\Gamma$ function gives

$$F_0 = \frac{1}{2 \Gamma (q)}\left( \frac{1}{q^2}+\frac{1}{q}+\frac{2}{(q-1)^2}\right) = \frac{1}{2 \Gamma(q) q^2}+\frac{1}{2 \Gamma(q) q}+\frac{1}{\Gamma(q) (q-1)^2} \\=\frac{1}{2 \Gamma(q+1) q}+\frac{1}{2 \Gamma(q+1)}+\frac{1}{\Gamma(q) (q-1)^2}$$

From this we can easily identify the following basic singularities:

1. a double pole at $q=1$ with residue $r=1$
2. a simple pole at $q=0$ with residue $r=\frac{1}{2}$

This is in contrast to the zeta function which has a simple pole at $q=1$ with residue $r=1$ and no singularity at $q=0$.

The next order gives for the integrand

$$t^{q-1} \left(t \left(\frac{5}{24}-\frac{\log (t)}{12}\right)-\frac{\log (t)}{2}-\frac{\log (t)}{t}+\frac{1}{2}\right)$$

which after integrating an taking into account the $\Gamma$ function gives

$$F_1 = F_0 + \frac{5}{24 (q+1) \Gamma (q)}+\frac{1}{12 (q+1)^2 \Gamma (q)}$$

A new pole appears here at $q=-1$. It comes from the last term and the observation that $(q+1)^2 \Gamma (q)=\frac{(q+1)^2 \Gamma (q+1)}{q}=\frac{(q+1) \Gamma (q+2)}{q}$ which goes $\to -(q+1)$ for $q\to -1$.

The last but one term is regular at $q=-1$. In summary we have found a new simple pole at $q=-1$ with a residue $r=-\frac{1}{12}$.

Continuing this procedure leads to the following structure of the poles besides the basic ones:

$S(1,q)$ has simple poles at negative odd integers $q=-(2k-1)$. Their residues turn out to be

$$r(k) =- \frac{B_{2 k}}{2 k}$$

where $B_{n}$ is the n-th Bernoulli number.

Here are the first few pole locations and residues

$$\left( \begin{array}{cc} -1 & -\frac{1}{12} \\ -3 & \frac{1}{120} \\ -5 & -\frac{1}{252} \\ -7 & \frac{1}{240} \\ -9 & -\frac{1}{132} \\ -11 & \frac{691}{32760} \\ -13 & -\frac{1}{12} \\ \end{array} \right)$$

For comparision: $\zeta(q)$ has just one simple pole $\frac{1}{q-1}$ in the whole complex $q$-plane.

$S(p,q)$ for $p\gt1$

For a partial answer to question 3 the same method can be applied for $p\gt 1$.

The results for $p=1$ through $p=4$ are written, for each $p$, as a list of poles, their possible multiplicity, and their residues. The list starts with the pole at $q=1$ and proceeds in the direction of the negative real $q$ axis. The last entry is the general expression from that point on, where for each $p$ we let $k=1,2,3,...$

$p=1\; \left( (1^2 , 1), (0, \frac{1}{2} ), ( -(2 k-1) , -\frac{B_{2 k}}{2 k}) \right)$

$p=2\; \left((1, \zeta(2)),(0,-1), (-1,\frac{1}{2}),(-2k, -B_{2 k})\right)$

$p=3\; \left((1, \zeta(3)),(0,0), (-1,-\frac{1}{2}),(-2,\frac{1}{2}),(-(2k+1), -(k+\frac{1}{2})B_{2 k})\right)$

$p=4\; \left((1, \zeta(4)),(0,0), (-1,0),(-2,-\frac{1}{3}),(-3,\frac{1}{2}),(-(2k+2), -\frac{1}{3}(k+1)(2k+1)B_{2 k})\right)$

Observations

1. The only double pole appears for the case $p=1$ at $q=1$, all other poles are simple ones

2. with increasing $p$ an increasing gap appears between the pole at $q=1$ and the next pole on the negative real $q$-axis. This corresponds to the fact that the generalized harmonic number $H_k^{(p)}$ approaches $1$ for large $p$ which in turn means that $S(p,q)$ approaches $\zeta(q)$ which has only one pole at $q=1$. The residue of the pole of $S(p,q)$ at $q=1$ is $\zeta(p)$ which for $p\to\infty$ goes to $1$ as it is with $\zeta(q)$.

Singularities of $S(1,q)$ using asymptotic expansion for $H_{k}$

A much simpler way to find the pole structure of $S(1,q)$ consists in using the asymptotic expansion

$$H_k = \log(k) +\gamma + \frac{1}{2k} - \sum_{m\ge 1} \frac{B_{2m}}{2m k^{2m}}$$

Inserting this in the definition of (1) and interchanging the summation gives

$$S(1,q) = \sum_{k\ge 1}\frac{\log(k)}{k^q} +\gamma \sum_{k\ge 1}\frac{1}{k^q}+ \frac{1}{2k} - \sum_{m\ge 1} \frac{B_{2m}}{2m} \sum_{k\ge 1}\frac{1}{ k^{2m+q}}\\= \zeta'(q) +\gamma \zeta(q) + \frac{1}{2}\zeta(q+1) - \sum_{m\ge 1} \frac{B_{2m}}{2m} \zeta(2m+q)$$

All we need to know is that $\zeta(q)$ has a simple pole at $q=1$ with residue $1$.

$\zeta'(q)$ then obviously has a double pole (the derivative of the simple pole) at $q=1$ with residue $-1$.

The second term has a simple pole at $q=1$ with residue $\gamma$ which did not appear previously, and which I consider therefore to be "spurious".

Third term: pole at $(1+q)=1$, i.e. $q=0$ with residue $\frac{1}{2}$.

Fourth term: pole at $2m+q=1$, i.e. $q=1-2m$ ($=-1, -3, -5, ...$) and residues $-\frac{B_{2m}}{2m}$.

Summing up: except for the term with $\gamma$ we find the previously obtained pole structure.