series involving Catalan and Zeta I ran across another challenging and interesting series, and I am wondering if someone could shed some light on how to evaluate it.
$$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\zeta(2)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2}}-\zeta(4)$$
This has turned out to be rather challenging. At first glance, I thought it may be somewhere along the lines of the famous $$ \sum_{n=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi}{2k}\coth(\pi k)-\frac{1}{2k^{2}}$$ that is often seen in Complex Analysis.
So, I ran the first sum through Maple and it gave me:
$$ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{\coth^2(\pi n)}{n^{2}}+\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}-\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^{4}}$$
Of course, two of these are the very familiar $\zeta(2), \;\ \zeta(4)$.  I managed to evaluate $\displaystyle \sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}=\frac{7{\pi}^{3}}{180}$ using Complex Analysis.
The one that has given me the fit is $\displaystyle\sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.
This evaluates to $\displaystyle\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$.  But, how?.
I tried a Complex Analysis method, but had trouble finding the residues at $ni$, which are the zeroes of $\sinh^{2}(\pi n)$ .
Anyone have any good ideas on how to evaluate the original sum at the top or even just this 'coth-squared' one?. Complex analysis or other wise.  I thought maybe a clever use of Fourier would work, but maybe not.
 A: A more general form is known :
$$\tag{1}f(c)=\sum_{n,m}'\frac 1{(n^2+m^2)^c}=4\,\zeta(c)\,L_{-4}(c)$$
with $\ L_{\sigma}(c):=\sum_{n\ge 1} \,\left(\frac {\sigma}n\right)\,n^{-c}\quad$ (where $\ \displaystyle\left(\frac {\sigma}n\right)$ is the Kronecker symbol and the sum is on all signed integers $n,m$ excluding $n=m=0$ as indicated by ' ) 
For $\;\sigma=-4\ $ only the odd terms remain with alternating signs :
$$\tag{2}L_{-4}(c)=\sum_{n\ge 1} \,\left(\frac {-4}n\right)\,n^{-c}=\sum_{n\ge 1} \frac {(-1)^{n-1}}{(2n-1)^c}=\beta(c)$$
with $\beta$ the Dirichlet beta function
so that (distinguishing the values $n=0$ and $m=0$) : 
$$4\,\zeta(c)\,\beta(c)=\sum_{n,m}'\frac 1{(n^2+m^2)^c}=2\sum_{m\ge 1}\frac 1{(0^2+m^2)^c}+2\sum_{n\ge 1}\frac 1{(n^2+0^2)^c}+4\sum_{n,m\ge 1}\frac 1{(n^2+m^2)^c}$$
and (dividing by $4$) a generalization of your nice result :
$$\tag{3}\sum_{n,m\ge 1}\frac 1{(n^2+m^2)^c}=\,\zeta(c)\,\beta(c)-\zeta(2c)$$

The (rather non-trivial) derivation of a further generalization of $(1)$ is provided in Borwein, Bailey and Girgensohn's "Experimentation in Mathematics" (page $167-169$ of my $2004$ edition).
They define :
$$\tag{4}\zeta(a,b,c) = \sum_{n,m}' \frac{n^{2a}\,m^{2b}}{(n^2+m^2)^c}$$ 
($a,\;b,\;c\;$ are nonnegative integers)
(the sum was for $n,m\ge 0$ but I think that "$\ge 0$" is a typo that doesn't appear page $169$)  
The page $168$ of the book contains (I hope that a one page re-transcription will not be a problem, extracts of the book may be viewed at Amazon) :
"The following identity expresses $\zeta(a,0,c)$ by using a Bessel function expansion of the normalized Mellin transform
$$M_c(f) = \frac 1{\Gamma(c)} \int_0^\infty f(t)\,t^{c-1}\,dt$$
of the function
$$t \to \sum_{n,m} n^{2a}q^{n^2+m^2} =\sum_n n^{2a}q^{n^2} \theta_3(q)$$
with $q=e^{-t}$ after using the theta transform $(2.15)$
$$\theta_3\bigl(e^{\pi t}\bigr)= \sqrt{\frac{\pi}t}\,\theta_3\left(e^{\frac{\pi}t}\right)$$
This leads to the identity
$$\tag{5}\zeta(a,0,c)= 2\delta_{0a}\zeta(2c)+ 2\beta\left(c-\frac12,\frac12\right) \zeta(2c-2a-1)+4\sum_{p\ge 1} \sigma_{[2a+1-2c]}(p)\,{\cal E}_c(p)$$
This is valid for real $ac$ with $d=c-a>1$ and presumably provides an analytic continuation of $\zeta(a,0,c)$ for the sum over positive integers. Here $\sigma$ is a divisor function
$$\sigma_{[d]}(p) =\sum_{n|p} n^d,$$
and
$${\cal E}_c(p)= \frac{\sqrt{\pi}}{\Gamma(c)} 2(\pi p)^{c-1/2}K_{\bigl(c-\frac 12\bigr)}(2\pi p)$$
is derived from the modified Bessel function of half integer order, $K_{\bigl(c-\frac 12\bigr)}$, of the second kind.
When $c=N$ is integer, ${\cal E}_N(p)$ is of the form
$$\pi\,e^{-2\pi p} P_N(\pi p)$$ where $P_N$ is a rational polynomial of degree $N-1$ with positive coefficients :  $P_1(x)=1,\
P_2(x)=x+1/2,\ P_3(x)=x^2/2+3/4x+3/8,\ P_4(x)=1/6x^3+1/2x^2+5/8x+5/16$. 
In general
$$P_N(x)=\sum _{k=0}^{N-1}\frac{{{N+k-1}\choose{N-1}}x^{N-1-k}}{(N-k-1)!4^k}$$
This allows one to very efficiently compute $\zeta(a,0,c)$ via $(5)$, using roughly $D/4$ terms for $D$ digits. 
Note also that for fixed $c$ and variable $a$ only the powers in
$\sigma_{[2a+1-2c]}$ vary so most of the computation can be saved.
Then the general integer case follows from
$$\tag{6}\zeta(a,b,c)=\sum_{k=0}^e (-1)^{e-k}{{e}\choose{k}}\zeta(a+b-k,0,c-k)$$
where $\ e=\min(a,b)$."
The formula at the beginning was $\zeta(0,0,c)$ and you asked merely for $\zeta(0,0,2)$. I think there is a much simpler answer for the second one (so that the problem remains...) but generalizations are of value too!

Let's add some keywords and references I just found (MathWorld and others) :


*

*(Chemical) 'Lattice Sum' 

*

*Zucker and McPhedran start with the history of your problem in this paper

*Borwein's book 'Pi and the AGM' has a §9.2 with title '(Chemical) Lattice Sum' : search Madelung at Amazon and your problem is left as an exercise on page 291 

*another paper of Borwein 'Convergence of Madelung-like lattice sums'

*a paper of Shail 


*'Madelung Constants' 

*

*Finch's Mathematical constants (1.10)

*Zucker's paper on the evaluation of Madelung constants


*Grenz-Formel (for 3 squares) for example page $378$ of Crandall's 'New Representations for the Madelung Constant'
Hoping this helped too, 
A: I managed to evaluate the series by beginning with:
$\displaystyle \sum_{k=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi n\coth(\pi n)-1}{2n^{2}}$, and using some known sums/identities. 
Then, differentiate, do some manipulating and it works out to what Maple gave.
Then, sum these four series separately.
It works out to $
\frac{-{\pi}^{2}}{4}\zeta(2)+\frac{{\pi}^{2}}{4}\left(\frac{2}{3}K+\frac{19{\pi}^{2}}{180}\right)-\frac{1}{2}\zeta(4)+\frac{\pi}{4}\cdot \frac{7{\pi}^{3}}{180}$
This then simplifies down to $\zeta(2)K-\zeta(4)$.  K being the Catalan constant represented by the series in the problem. 
The trickiest sum was $\displaystyle \sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.
$=\displaystyle\sum_{n=1}^{\infty}\frac{csch^{2}(\pi n)}{n^{2}}+\sum_{n=1}^{\infty}\frac{1}{n^{2}}$
$=\displaystyle\frac{2}{3}K-\frac{11{\pi}^{2}}{180}+\frac{{\pi}^{2}}{6}=\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$
A: An answer for a ten years old question. We want to evaluate infinite series of the type
\begin{align*}
    S=\sum_{(n,m) \neq (0,0)}\frac{1}{(n^2+m^2)^s} 
\end{align*}
To this end we will rely on the follwing result
\begin{align}
     \left(\sum_{n=-\infty}^\infty q^{n^2} \right)^2=1+4 \sum_{n=1}^\infty \frac{q^n}{1+q^{2n}} \tag{1}
\end{align}
A self content proof of $(1)$ can be found here in my blog.
Another result that we need is
\begin{align}
        \frac{1}{x^s}&=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}e^{-xt}\,dt \tag{2}
    \end{align}
Then:
\begin{align}
    \sum_{(n,m) \neq (0,0)}\frac{1}{(n^2+m^2)^s}&=4 \zeta(s)\beta(s) \tag{3}
\end{align}
Proof:
\begin{align*}
   \sum_{\substack{n,m=-\infty \\ n,m \neq (0,0)}}^\infty \frac{1}{(n^2+m^2)^s}&=\frac{1}{\Gamma(s)} \sum_{\substack{n,m=-\infty \\ n,m \neq (0,0)}}^\infty  \int_0^\infty t^{s-1}e^{-(n^2+m^2)t}\,dt\\
    &=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\left( \sum_{\substack{n,m=-\infty \\ n,m \neq (0,0)}}^\infty  e^{-(n^2+m^2)t}\right)\,dt\\
     &=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\left(\sum_{n=-\infty}^\infty \sum_{m=-\infty}^\infty e^{-(n^2+m^2)t}-1\right)\,dt\\
      &=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\left(\left(\sum_{n=-\infty}^\infty  e^{-n^2 t}\right)^2-1\right)\,dt\\
        &=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\left(\left(\sum_{n=-\infty}^\infty  q^{n^2 }\right)^2-1\right)\,dt &(\text{for}\,\,q=e^{-t})\\
        &=\frac{1}{\Gamma(s)}\int_0^\infty t^{s-1}\left(\left(1+4\sum_{n=1}^\infty \frac{q^n}{1+q^{2n}}\right)-1\right)\,dt &(\text{by}\,\, (1))\\
         &=\frac{4}{\Gamma(s)}\int_0^\infty t^{s-1}\left(\sum_{n=1}^\infty q^n \sum_{m=0}^\infty (-1)^m q^{2 n m}\right)\,dt\\
          &=\frac{4}{\Gamma(s)}\int_0^\infty t^{s-1}\left(\sum_{m=0}^\infty  (-1)^m\sum_{n=1}^\infty  q^{(2m+1)n}\right)\,dt\\
          &\frac{4}{\Gamma(s)}\sum_{m=0}^\infty  (-1)^m\sum_{n=1}^\infty \int_0^\infty t^{s-1} e^{-(2m+1)n t}\,dt\\
          &=\frac{4}{\Gamma(s)}\sum_{m=0}^\infty  (-1)^m\sum_{n=1}^\infty \frac{\Gamma(s)}{(2m+1)^s n^s}\\
          &=4 \sum_{m=0}^\infty  \frac{(-1)^m}{(2m+1)^s} \sum_{n=1}^\infty \frac{1}{n^s}\\
          &=4\zeta(s)\beta(s)
\end{align*}
We also have that
\begin{align*}
  \sum_{\substack{n,m=-\infty \\ n,m \neq (0,0)}}^\infty \frac{1}{(n^2+m^2)^s}&=4\sum_{m=1}^\infty\sum_{m=1}^\infty  \frac{1}{(n^2+m^2)^s}+2\sum_{m=1}^\infty  \frac{1}{m^{2s}}+2\sum_{n=1}^\infty  \frac{1}{n^{2s}}\\
&=4\sum_{m=1}^\infty\sum_{m=1}^\infty  \frac{1}{(n^2+m^2)^s}+4\zeta(2s)
\end{align*}
Thus
\begin{align*}
\sum_{m=1}^\infty\sum_{m=1}^\infty  \frac{1}{(n^2+m^2)^s}=\zeta(s)\beta(s)-\zeta(2s)
\end{align*}
In particular for $s=2$ we obtain
\begin{align*}
\sum_{m=1}^\infty\sum_{m=1}^\infty  \frac{1}{(n^2+m^2)^2}&=\zeta(2)\beta(2)-\zeta(4)\\
&=\frac{\pi^2}{6}G-\frac{\pi^4}{90} \qquad \blacksquare
\end{align*}
where $G$ is Catalan´s constant
