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$$-\int_{0}^{\infty}x\sin(x)\ln(1-e^{-x})dx=2\sum_{n=1}^{\infty}\frac{1}{(n^2+1)^2}$$

Is there a closed form for this integral?

I was only able to find the equivalent sum to it(How? using wolfram integrator or I did some other calculations but can't remembered)

To integrate this integral we can use integration by parts, but I don't think that is suitable for this situation. There must be a short way to evaluate this. Can anyone give a hand here?


Let see

$$-\int_{0}^{\infty}x\sin(x)\ln(1-e^{-x})dx=I$$

$$I=[-x\cos(x)+\sin(x)]\ln(1-e^{-x})-\int\frac{e^{-x}}{1-e^{-x}}\cdot[-\cos(x)+\sin(x)]dx$$

As you can see, it is enormous.

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  • $\begingroup$ Do you know much about contour integration? $\endgroup$ – Ben Sheller May 18 '16 at 19:44
  • $\begingroup$ Wolfram Alpha does give a closed form. It looks deeply unpleasant, but that means you can hope. $\endgroup$ – Clement C. May 18 '16 at 19:45
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We can use the identity $$ \frac{1}{2}+\frac{\pi}{2x}\coth\left(\frac{\pi}{x}\right)=\sum_{n\geq0}\frac{1}{1+n^{2}x^{2}} $$ so if we assume that $x>0 $ we get $$\frac{1}{2}+\frac{\pi}{2\sqrt{x}}\coth\left(\frac{\pi}{\sqrt{x}}\right)=\sum_{n\geq0}\frac{1}{1+n^{2}x} $$ and so $$\frac{x}{2}+\frac{\pi\sqrt{x}}{2}\coth\left(\frac{\pi}{\sqrt{x}}\right)=\sum_{n\geq0}\frac{1}{\frac{1}{x}+n^{2}} $$ then taking the derivative and taking $x=1 $ we get $$\left(\frac{x}{2}+\frac{\pi\sqrt{x}}{2}\coth\left(\frac{\pi}{\sqrt{x}}\right)\right)_{x=1}^{'}=\sum_{n\geq0}\frac{1}{\left(1+n^{2}\right)^{2}}$$ so $$2\sum_{n\geq1}\frac{1}{\left(1+n^{2}\right)^{2}}=\frac{1}{2}\left(\pi^{2}\textrm{csch}^{2}\left(\pi\right)+\pi\coth\left(\pi\right)-2\right).$$

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  • $\begingroup$ Thank you twice @Marco Cantarini. $\endgroup$ – user339807 May 18 '16 at 20:11
  • $\begingroup$ @mahdi You're welcome. $\endgroup$ – Marco Cantarini May 18 '16 at 20:16
  • $\begingroup$ you found the closed form via the sum. How would you go about integrating the integral? $\endgroup$ – user339807 May 18 '16 at 20:20
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ First, we expand the $\ln$ function: \begin{align} \color{#f00}{-\int_{0}^{\infty}x\sin\pars{x}\ln\pars{1 -\expo{-x}}\,\dd x} & = \Im\sum_{n = 1}\int_{0}^{\infty}x\expo{\ic x}\,{\expo{-nx} \over n}\,\dd x = \sum_{n = 1}{1 \over n}\,\Im\ \overbrace{\int_{0}^{\infty}x\expo{-\pars{n - \ic}x}\,\dd x} ^{\ds{{1 \over \pars{n - \ic}^{2}}}} \\[3mm] & = \sum_{n = 1}{1 \over n}\,\Im \bracks{{\pars{n + \ic}^{2} \over \pars{n^{2} + 1}^{2}}} = \color{#f00}{2\sum_{n = 1}{1 \over \pars{n^{2} + 1}^{2}}} \end{align}

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  • $\begingroup$ This is above my head, elegant answer thank you. $\endgroup$ – user339807 May 19 '16 at 3:30
  • $\begingroup$ @mahdi You're welcome. I'm glad it was useful to you. $\endgroup$ – Felix Marin May 19 '16 at 3:40
  • $\begingroup$ Felix is back!! $\endgroup$ – cgiovanardi May 19 '16 at 23:47

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