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

Question

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 $(1)?$

Answer 1

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 $a=1$.