I've been struggling a bit with a double sum that arose as the trace of an operator:


$$\sum\_{(j,k)\in Z^2 \setminus (0,0)} \frac{(j+k)^n}{(j^2+k^2)^n},$$


where $n$ is an even natural number. **Is there a closed form for this sum?**


Even in the case of $n=2$, I'm a bit miffed. If there were no numerator, I would use the Mellin transform and rewrite the sum in terms of theta functions, which is a standard approach in lattice sums (cf. MR3135109). However, there is a numerator, and I can't seem to be clever enough to make some change of variables work well.


I am also interested in the corresponding $d$-dimensional case.


Thanks for any thoughts/suggestions.



2D lattice sum with numerator
The $n=4$ sum is accessible as follows. Expand $(j+k)^4$; by symmetry
the odd terms $4j^3k$ and $4jk^3$ cancel out so we need only sum
$(j^4 + 6j^2k^2 + k^4) / (j^2+k^2)^4$. You say you already know
the sum of $1 / (j^2+k^2)^2$, which is $(j^4 + 2j^2k^2 + k^4) / (j^2+k^2)^4$ [see also the "P.S." paragraph below];
so we only need one more sum $\sum^\prime (j^4 + c j^2 k^2 + k^4) / (j^2+k^2)^4$.
Well, we know the sum of $1 / (j+ik)^4$ from the theory of the Weierstrass
$\wp$-function; if I did this right it comes to
$\Gamma(1/4)^8 / (960 \pi^2) = 3.1512120+$. But
$$
\frac1{(j+ik)^4} = \frac{(j-ik)^4}{j^2+k^2)^4}
= \frac{(j^4 - 4i j^3 k - 6 j^2 k^2 + 4i j k^3 + k^4}{(j^2+k^2)^4},
$$
and the odd terms again cancel out, so this gives us the sum of
$(j^4 - 6j^2k^2 + k^4) / (j^2+k^2)^4$ — and we're done because
$$
2 (j^4 + 6j^2k^2 + k^4) = 3 (j^2+k^2)^2 - (j^4 - 6j^2k^2 + k^4).
$$


P.S. In general the sum of $1/(j^2+k^2)^n$ is $4 \zeta(n) L(n,\chi\_4)$
[because it's 4 times the value at $n$ of the zeta function for ${\bf Q}(i)$].
Thus when $n=2$ that's $2\pi^2 {\bf G} / 3$ where ${\bf G}$ is Catalan's constant
$$
L(2,\chi\_4) = 1 - \frac1{3^2} + \frac1{5^2} - \frac1{7^2} + - \cdots
 = .915965594\ldots
$$
so $2\pi^2 {\bf G} / 3 = 6.026812\ldots$; no further simplification is known
or expected, though there are various ways known to compute the numerical
value of ${\bf G}$ efficiently to any desired accuracy.


