# Infinite Sum $\sum_{n=1}^\infty 2x\arctan(x/n)-n\log(1+x^2/n^2)$

I ran into the infinite sum $\sum_{n=1}^\infty 2x\arctan(x/n)-n\log(1+x^2/n^2)$, where $x$ is a positive real number. Mathematica can't do the sum, but shows that it's very well approximated by $1.923\cdot10^5x^2$ (tested for values of $x$ between $10^{-7}$ and $10^{10}$). Anyone got an idea how to evaluate the sum? What's the constant $1.923\cdot10^5\simeq\exp(12.17)$?

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How did this sequence arise in your research? –  Fernando Muro Jun 10 '13 at 9:00
Notice that F.Muro asks the question because you didn't follow the "how to ask" guidelines, and it could be someone asking for help on homework... –  Julien Puydt Jun 10 '13 at 9:07
I tried to evaluate the integral $\int_0^\infty d\omega\,\coth(\beta\omega/2)\frac{\sin^2(\omega t/2)}{(\omega/2)^2}$, which arised from quantum thermodynamics. The integral seems not solvable in closed form, so I used the expansion $\coth(z)=1+\sum_{n=1}^\infty2e^{-2n}$, then integrated each summand separately and defined $x=t/\beta$. –  Aaron Harper Jun 10 '13 at 9:28
Ok, I realized now myself that the sum diverges since for $n\gg x$ the summand is $2x^2/n-x^2/n$, which agrees with the fact that the integral diverges like $1/\omega$ for $\omega\rightarrow0$. Sorry for wasting your time... –  Aaron Harper Jun 10 '13 at 10:53
Are you interested in large $\beta$ ? If so, you might be able to approximate the integral. –  Tom Dickens Jun 11 '13 at 1:17