# Double sum and zeta function

This is a personal research that came to an end , since the results were not those which were being anticipated. I was unable to come up with a solution therefore I post the topic here:

Prove (it certainly holds because it was checked on a computer) that the following identity holds:

$$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\left ( n^2+k^2 \right )^2}=\zeta(2)\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\left ( 2n-1 \right )^2}-\zeta(4)$$

wheras $\zeta$ represents the zeta function defined as $\displaystyle \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}, \;\; \mathfrak{Re}(s)>1$. Of course both values of $\zeta$ appearing here are known. For complete of sakeness i quote them :

$$\zeta(2)=\frac{\pi^2}{6}, \;\; \zeta(4)=\frac{\pi^4}{90}$$

Now, one can also see (not trivial ) that:

$$\sum_{n=1}^{\infty}\frac{1}{n^2 \sinh^2 \pi n}=\frac{4}{\pi^2}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\left ( n^2+k^2 \right )^2} -\frac{\pi^2}{60}$$

This equation also holds (checked by computer). The following result was extracted by using the known formulae $\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{z+n}=\frac{\pi}{\tan \pi z}$ and the known (?) Fourier series: $$\displaystyle \frac{1}{\sinh^2 \pi z}=\frac{1}{\pi^2 z}+\frac{4z^2}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{\left ( z^2+k^2 \right )^2}-\frac{2}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{z^2+k^2}$$

Of course the last sum at the last equation can easily be computed via residues. What remains now is the proof for the first equation in the post. No-one can guarantee that is going to be an easy task.

1. I came across the identity on a book. I checked the validity with a computer and yes it does hold.

2. I have ckecked enough books to see if there is in there somewhere , but unfortunately it was not. So, I speculate that it is not that famous.

3. It can also be linked with other sums (single or double). Unfortunately I don't have my papers in front of me in order to write them down. So, i think it is an interseting identity.

• "it certainly holds because it was checked on a computer": have you ever heard of the law of small numbers? :) – user207868 Jan 17 '15 at 20:46
• @Leon Aragones: No.. I have not.. By that way I would like to inform that I corrected a typo in my post. – Tolaso Jan 17 '15 at 20:47
• See the formula $(38)$ at MathWorld for a more general formula (the generic term is "lattice sums"). – Raymond Manzoni Jan 17 '15 at 20:49
• @Tolaso Take a look at this article. – user207868 Jan 17 '15 at 20:49
• @RaymondManzoni So, what basically you are saying here is that this sum actually relates to Beta Diriclet? Well, wow! Yes, one way it should.. because that alternating sum evaluates to $G$ (catalan constant) which is actually $\beta(2)$. Sounds interesting.. However I'm interested for a proof here... since all efforts failed. – Tolaso Jan 17 '15 at 20:51

Let's consider the double sum over all $\,(n,k)\in\mathbb{Z}^2\,$ except the origin $(0,0)$ : $$\tag{1}S(s):=\sum_{(n,k)\neq (0,0)}\frac{1}{\left ( n^2+k^2 \right )^s}=4\left(\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\left ( n^2+k^2 \right )^s}+\zeta(2\,s)\right)$$ This is a "Lorenz–Hardy sum" (ref. Lorenz L. $1871$ "Bidrag tiltalienes theori" and Hardy G. H. $1919$ "On some deﬁnite integral considered by Mellin" in his Collected papers vol $7$).

## Analytical method

The Mellin transform of a function $f$ is defined by : $$\tag{2}\{\mathcal{M}f\}(s):=\int_0^\infty t^{s-1}f(t)\;dt$$ applied to $\;f:t\mapsto e^{\large{-mt}}\;$ and from the definition of the Gamma function gives : $$\tag{3}\frac{\Gamma(s)}{\left(m \right)^s}=\int_0^\infty t^{s-1}\,e^{-m\,t}\;dt$$ Supposing that $\Re(s)>1$ we may rewrite our double-sum as : \begin{align} \Gamma(s)\,S(s)&=\sum_{(n,k)\neq (0,0)}\frac{\Gamma(s)}{\left ( n^2+k^2 \right )^s}\\ &=\sum_{(n,k)\neq (0,0)}\int_0^\infty t^{s-1}\,e^{-(n^2+k^2)\,t}\;dt\\ &=\int_0^\infty t^{s-1}\sum_{(n,k)\neq (0,0)}\,e^{-(n^2+k^2)\,t}\;dt\\ &=\int_0^\infty t^{s-1}\left(\sum_{n\in\mathbb{Z}}\,e^{-n^2\,t}\sum_{k\in\mathbb{Z}}\,e^{-k^2\,t}-1\right)\;dt\\ \tag{4}&=\int_0^\infty t^{s-1}\left(\theta_3(0,e^{-t})^2-1\right)\;dt\\ \end{align} i.e. the Mellin transform of $\,f:t\mapsto \theta_3(0,e^{-t})^2-1\,$ where the Jacobi theta function $\theta_3$ is defined by (in this answer we will implicitly suppose that $z=0$ and $q=e^{-t}$) : $$\tag{5}\theta_3(z,q):=\sum_{n=-\infty}^\infty q^{n^2}e^{2niz}$$

In $1829$ Jacobi published his deep elliptic functions book "Fundamenta nova theoriae functionum ellipticarum" where he obtained numerous identities including the formula $4$) of page $103$ : $$\tag{6}\theta_3(0,q)^2=\frac {2\,K}{\pi}=1+4\sum_{n=1}^\infty \frac{q^n}{1+q^{2n}}$$ ($K$ is the Complete elliptic integral of the first kind but we won't use it here)

The series at the right of $(6)$ is a Lambert series. The expansion of the denominator in power series and substitution of the double-sum in $(4)$ gives :

\begin{align} \Gamma(s)\,S(s)&=4\int_0^\infty t^{s-1}\sum_{n=1}^\infty \sum_{m=0}^\infty (-1)^m q^n\,q^{2mn}\;dt\\ &=4\sum_{n=1}^\infty \sum_{m=0}^\infty (-1)^m \int_0^\infty t^{s-1}e^{-n(2m+1)t}\;dt\\ &=4\,\sum_{n=1}^\infty \sum_{m=0}^\infty (-1)^m \frac{\Gamma(s)}{(n(2m+1))^s}\\ &=4\,\Gamma(s)\sum_{n=1}^\infty \frac 1{n^s}\sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)^s}\\ &=4\,\Gamma(s)\,\zeta(s)\,\beta(s)\\ \\ &\text{so that }\\ \\ \tag{7}\sum_{(n,k)\neq (0,0)}\frac{1}{\left ( n^2+k^2 \right )^s}&=4\,\zeta(s)\,\beta(s),\quad\Re(s)>1\\ &\text{and}\\ \tag{8}\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{\left ( n^2+k^2 \right )^s}&=\zeta(s)\,\beta(s)-\zeta(2s),\quad\Re(s)>1\\ \\\end{align}

$\qquad\qquad\qquad$with $\beta$ the Dirichlet beta function.

This powerful method allows to deduce many closed forms of lattice sums from the known theta identities.

I'll try to add one (or more) alternative derivations when time allows...

References :

• Glasser and Zucker 1980 "Lattice sums" In "Theoretical Chemistry, Advances and Perspectives Vol $5$".
• Borwein and Borwein 1987 "Pi and the AGM".
• Borwein, Glasser, McPhedran, Wan and Zucker 2013 "Lattice Sums Then and Now" (and older papers from these authors)
• Really excellent explanations! – yarchik Jun 25 '20 at 12:03
• Thanks @yarchik! You may be interested by the deeper algebraic solutions provided here and here. Excellent continuation, – Raymond Manzoni Jun 25 '20 at 12:12