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How would you tackle this series by real analysis?

$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m^4(m^2+n^2)}$$

user 1591719
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  • Assuming that you want to compute the sum (the convergence is rather obvious), what makes you think that the sum can be expressed in a simpler way? – ajotatxe Jun 18 '15 at 14:16
  • @ajotatxe a long time ago I saw it in a paper I don't recollect now in which one, and as far as I can remember it had a closed form. – user 1591719 Jun 18 '15 at 14:18
  • Value 1.1260088410668 is not found by ISC http://isc.carma.newcastle.edu.au/ – GEdgar Jun 18 '15 at 14:36
  • @r9m I'd like too. Actually, I'd like to see any solution by means of real analysis. :-) – user 1591719 Jun 18 '15 at 15:00
  • @GEdgar as regards my series and integrals, for major part of them ISC is not helpful unfortunately, that is for more than 99% of them. I stopped using it a long time ago. – user 1591719 Jun 18 '15 at 15:07
  • The Plouffe's article linked in my answer shows that there are nice closed forms if we replace $m^4$ by $m^2$ or $m^6$. The case $m^4$ is not so easy to deal with. – Jack D'Aurizio Jun 18 '15 at 15:11
  • @JackD'Aurizio Yeah, I know, that's why I was particularly interested in it. Thanks for the job (+1). – user 1591719 Jun 18 '15 at 15:13
  • @Chris'ssistheartist: ok, I think I managed to prove that no "nice" closed form exists, unless we find one for $$\sum_{n\geq 1}\frac{\log n}{n^5},$$ highly unlikely. – Jack D'Aurizio Jun 18 '15 at 15:26

1 Answers1

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Since: $$\sum_{n\geq 1}\frac{1}{m^2+n^2}=\frac{-1+m\pi\coth(m\pi)}{2m^2}\tag{1}$$ our sum equals: $$ -\frac{\zeta(6)}{2}+\frac{\pi}{2}\sum_{m\geq 1}\frac{\coth(m\pi)}{m^5}=-\frac{\pi\zeta(5)-\zeta(6)}{2}+\pi\sum_{m\geq 1}\frac{1}{m^5(e^{2m\pi}-1)}\tag{2}$$ and maybe the last series has a nice closed form. Here there is a related paper.

Following Simon Plouffe, $$ \sum_{m\geq 1}\frac{1}{m^5(e^{2m\pi}-1)}=\sum_{m\geq 1}\text{Li}_5(e^{-2\pi m})=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(z)}{(2\pi)^z}\zeta(z+5)\zeta(z)\,dz.\tag{3}$$ If now we set $f(z)=\frac{\Gamma(z)}{(2\pi)^z}\zeta(z+5)\zeta(z)$, we have: $$\begin{eqnarray*} \text{Res}\left(f(z),z=0\right)&=&-\frac{\zeta(5)}{2},\\\text{Res}\left(f(z),z=-1\right)&=&\frac{\pi^5}{540}\\\text{Res}\left(f(z),z=-3\right)&=&-\frac{\pi^5}{540}\\\text{Res}\left(f(z),z=-4\right)&=&\frac{2\pi^4}{3}\zeta'(-4)\\\text{Res}\left(f(z),z=-5\right)&=&-\frac{\pi^5}{1890}\tag{4}\end{eqnarray*}$$ hence a closed form for $(2)$ just depends on a closed form for $\zeta'(-4)$, or, by the reflection formula, on a closed form for: $$\zeta'(5)=-\sum_{n\geq 1}\frac{\log n}{n^5}.\tag{5}$$

Jack D'Aurizio
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