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

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

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.

• Do you know much about contour integration? Commented May 18, 2016 at 19:44
• Wolfram Alpha does give a closed form. It looks deeply unpleasant, but that means you can hope. Commented May 18, 2016 at 19:45

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 $$2\sum_{n\geq1}\frac{1}{\left(1+n^{2}\right)^{2}}=\frac{1}{2}\left(\pi^{2}\textrm{csch}^{2}\left(\pi\right)+\pi\coth\left(\pi\right)-2\right).$$

• Thank you twice @Marco Cantarini.
– user339807
Commented May 18, 2016 at 20:11
• @mahdi You're welcome. Commented May 18, 2016 at 20:16
• you found the closed form via the sum. How would you go about integrating the integral?
– user339807
Commented May 18, 2016 at 20:20

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

– user339807
Commented May 19, 2016 at 3:30
• @mahdi You're welcome. I'm glad it was useful to you. Commented May 19, 2016 at 3:40
• Felix is back!! Commented May 19, 2016 at 23:47