Can you give me some hints to prove equality:

$$\sum_{m,n=1}^{\infty} \frac1{(m^2+n^2)^2} =\zeta (2)\ G-\zeta(4)=\frac{\pi^2}{6}\ G-\frac{\pi^4}{90}$$

where $\zeta (t):= \sum\limits_{n=1}^{+\infty} \frac{1}{n^t}$ is the Riemann zeta function and $G := \sum\limits_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} \approx 0.915 965 594$ is Catalan's constant?

I tried with some reverse engineering, but I wasn't able to solve the problem at all. Even a good reference may be useful.

Thanks a lot in advance, guys.


This problem can be recast in terms of the famous problem of the number of ways to represent a positive integer as a sum of squares. With this perspective, we can see that the following more general statement is true for any $p > 1$ (so that each of the infinite series actually converges): $$\sum_{n=1}^{\infty} \frac{1}{(n^2)^p} + \sum_{m,n = 1}^{\infty} \frac{1}{(m^2+n^2)^p} = \left(\sum_{n=1}^{\infty} \frac{1}{n^p}\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^p}\right).$$

The left-hand side is $$\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p},$$ where $n_2(s)$ is the number of ways of representing $s$ as the sum of one or of two squares of positive integers, in which order is distinguished (i.e., $1^2 + 2^2$ is counted separately from $2^2+1^2$).

It is known that $n_2(s) = d_1(s) - d_3(s)$ (see eq. 24 in the site linked above), where $d_k(s)$ is the number of divisors of $s$ congruent to $k \bmod 4$.

The first sum on the right side of the equation has (the $p$th powers of) all positive integers as denominators and the second sum on the right has (the $p$th powers of) all odd numbers as denominators. After multiplying those sums together, then, $1/s^p$ (ignoring signs) appears on the right as many times as there are odd divisors of $s$. Each odd divisor of $s$ congruent to $1 \bmod 4$ contributes a $+1/s^p$, and each odd divisor of $s$ congruent to congruent to $3 \bmod 4$ contributes a $-1/s^p$. Thus the coefficient of $1/s^p$ on the right side is exactly $d_1(s) - d_3(s)$. Therefore, the right-hand side is also $$\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p}.$$

  • $\begingroup$ Very nice indeed! Thank you Mike. $\endgroup$ – Pacciu Mar 24 '11 at 14:52
  • $\begingroup$ @Pacciu: I'm glad you liked it. I just added a generalization you might be interested in, too. $\endgroup$ – Mike Spivey Mar 24 '11 at 16:21
  • $\begingroup$ @Mike: Thanks a lot. Really, shortly after reading your proof, the same generalization came also to my mind. I feel it is really interesting, but actually I can't realize if the $\sum_{s=1}^{+\infty} \frac{n_2(s)}{s^p}$ can be expressed in terms of $\zeta$ or related functions... I'm too new to this kind of things to tell! XD Maybe there's a chance when $p\in \mathbb{N}$ and $p\geq 2$... I have to think about it. $\endgroup$ – Pacciu Mar 25 '11 at 2:23
  • $\begingroup$ @Pacciu: The argument shows that, at least for real $p > 1$, $\sum_{s=1}^{\infty} \frac{n_2(s)}{s^p} = \zeta(p) \beta(p)$, where $\beta(p)$ is the Dirichlet beta function. I'm not sure about complex $p$. $\endgroup$ – Mike Spivey Mar 25 '11 at 3:15
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    $\begingroup$ just for a bit of perspective: the fact that $n_2=d_1-d_3$ is usually derived from decomposition of primes in $\mathbb{Z}[i]$. That's why your and my answer are, ultimately, the same. $\endgroup$ – user8268 Mar 25 '11 at 19:28

For more detail search for Dedekind zeta function (especially of quadratic fields) and for Dirichlet $L$-series.

Let us compute the sum over all $(m,n)\in\mathbb{Z}^2$ such that $(m,n)\neq(0,0)$. Let us denote that sum $S$. We have $(m^2+n^2)^2=|m+in|^4$. Every number $m+in\in\mathbb{Z}[i]$ decomposes uniquely into a product of primes in $\mathbb{Z}[i]$, up to multiplication by $\pm1,\pm i$. We therefore have $S=4\prod_\pi\frac{1}{1-|\pi|^{-4}}$ where the product runs over all primes of $\mathbb{Z}[i]$ (4 comes from $\pm1,\pm i$). Now $|\pi|^2$ is a either a prime $p$ (in $\mathbb{Z}$) $p\equiv 1 \bmod 4$ (we get $p$ twice, from $\pi$ and from $\bar\pi$) or (if $\pi=q\in\mathbb{Z}$) $|\pi|^2=q^2$, $q$ prime, $q\equiv 3 \bmod 4$, or $|\pi|^2=2$.

We thus get ($p$ runs over all primes 1 mod 4, $q$ over primes 3 mod 4) $$S=4\frac{1}{1-2^{-2}}\prod_p\frac{1}{(1-p^{-2})^2}\prod_q\frac{1}{1-q^{-4}}=$$ $$=4\frac{1}{1-2^{-2}}\prod_p\frac{1}{1-p^{-2}}\prod_q\frac{1}{1-q^{-2}}\times \prod_p\frac{1}{1-p^{-2}}\prod_q\frac{1}{1+q^{-2}}=4\zeta(2)\times G.$$ Your sum is $S/4-\zeta(4)$.


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