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Consider the integral $(1)$

$$\int_{0}^{\infty}x\sin{x}\ln{(1-e^{-x})}\mathrm dx=I\tag1$$ How can we show that $$I=1-{\pi\over 2\tanh\pi}-{\pi^2\over 2\sinh^2{\pi}}$$

An attempt: Dealing with indefinite integral

$$\int x\sin{x}\ln{(1-e^{-x})}\mathrm dx=J\tag2$$

Apply integration by parts

$u=\ln{(1-e^{-x})}$ then $du={e^{-x}\over 1-e^{-x}}\mathrm dx$

$v=-\int x\sin{x}\mathrm dx=-x\cos{x}+\sin{x}$

$$J=(-x\cos{x}+\sin{x})\ln{(1-e^{-x})}-\int{e^{-x}\over 1-e^{-x}}(\sin{x}-x\cos{x})\mathrm dx\tag3$$

$$J=(-x\cos{x}+\sin{x})\ln{(1-e^{-x})}-\int\sum_{n=0}^{\infty}e^{x(1-n)}(\sin{x}-x\cos{x})\mathrm dx\tag4$$

$$J=(-x\cos{x}+\sin{x})\ln{(1-e^{-x})}-\sum_{n=0}^{\infty}\int e^{x(1-n)}(\sin{x}-x\cos{x})\mathrm dx\tag5$$

Let

Applying integration by parts

$$J_1=\int e^{x(1-n)}\sin{x}\mathrm dx={e^{x(1-n)}[(1-n)\sin{x}-\cos{x}]\over (1-n)^2+1}$$

$$J_2=\int xe^{x(1-n)}\cos{x}\mathrm dx={xe^{x(1-n)}[(1-n)\cos{x}+\sin{x}]\over (1-n)^2+1}-{e^{x(1-n)}[(n^2-2n)\cos{x}-2(1-n)\sin{x}]\over ((1-n)^2+1)^2}$$

So far applying integration by parts seem bit hard to resolve problem $(1)$, how else can we tackle $(1)?$

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    we can also use the taylor expansion of sine: Using parts, the integral in question becomes equivalent to $$ I=J(1)-\partial_{\alpha}J(\alpha)|_{\alpha=1} $$ with $$ J(\alpha)=\int_0^{\infty}\frac{\sin(\alpha x)}{e^x-1}=\sum_{n=0}^{\infty}\frac{(-1)^j\alpha^{2j+1}}{(2j+1)!}\int_0^{\infty}\frac{x^{2j+1}}{e^x-1}=\sum_{j=0}^{\infty}(-1)^j\alpha^{2j+1}\zeta(2j+2)\\=-\frac{1}{\alpha}\sum_{j=1}^{\infty}(i)^{2j}\alpha^{2j}\zeta(2j)=\frac{\pi}{2}\coth(\alpha\pi)-\frac{1}{2 \alpha^2} $$ where we used the generating function of the $\zeta$-function... – 2017-02-18

1 Answers 1

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Hint. First one may expand the integrand in the following way, $$ \ln{(1-e^{-x})}=-\sum _{n=1}^{\infty } \frac{e^{-n x}}{n},\qquad x>0,\tag1 $$ which leads to $$ \int_0^\infty x\sin{x}\ln{(1-e^{-x})}\:dx=-\sum _{n=1}^{\infty } \frac{1}{n}\int_0^\infty x\sin{x}e^{-n x}\:dx,\tag2 $$ now, one may differentiate the standard evaluation, $$ \int_0^\infty \sin{x}\:e^{-n x}\:dx=\frac{1}{n^2+1}, \quad n>0,\tag3 $$ to get $$ \int_0^\infty x\sin{x}\:e^{-n x}\:dx=\frac{2n}{(n^2+1)^2}, \quad n>0,\tag4 $$ then $(2)$ rewrites

$$ \int_0^\infty x\sin{x}\ln{(1-e^{-x})}\:dx=-2\sum _{n=1}^{\infty } \frac{1}{(n^2+1)^2}.\tag5 $$

One may recall that $$ \sum _{n=1}^{\infty } \frac{1}{(n^2+a^2)}=\frac{\pi \coth (\pi a)}{2 a}-\frac{1}{2 a^2},\qquad a>0,\tag6 $$ which, by differentiating, gives $$ \sum _{n=1}^{\infty } \frac{1}{(n^2+a^2)^2}=-\frac{1}{2 a^4}+\frac{\pi \coth (\pi a)}{4 a^3}+\frac{\pi ^2 \text{csch}^2(\pi a)}{4 a^2}\tag7 $$ leading to the announced result by putting $a=1$.

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    A proof of $(6)$ can be found here: http://math.stackexchange.com/questions/141470/find-the-sum-of-sum-1-k2-a2-when-0a1 – 2017-02-17
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    Thank you @OLivier, a quick question is there a closed form for $\sum_{n=1}^{\infty}{(-1)^{n-1}\over (n^2+1)^2}$? – 2017-02-17
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    @Algebra You are welcome. To your quick question the answer is yes, if one writes $\sum_{n=1}^{\infty}{(-1)^{n-1}\over (n^2+1)^2}=\sum_{p=1}^{\infty}{1\over ((2p-1)^2+1)^2}-\sum_{p=1}^{\infty}{1\over ((2p)^2+1)^2}$, then one gets a result. Try also Wolfy I think it gives directly the initial sum. – 2017-02-17
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    Sorry to bother you again @Olivier, can you possibly help me to work out the close form for $\sum_{n=1}^{\infty}{(b)^{n-1}\over (n^2+a^2)^2}$. I try on wolfram sum but it's doesn't seem to be able to evaluate it. – 2017-02-17
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    @Algebra In this case, a partial fraction decomposition of ${1\over (n^2+a^2)^2}$ leads to consider $\sum_{n=1}^{\infty}{b^{n}\over (n \pm ai)^s}$, $s=1,2$, which one may express in terms of the Lerch transcendent function:http://mathworld.wolfram.com/LerchTranscendent.html – 2017-02-17