# 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? – Ben Sheller May 18 '16 at 19:44
• Wolfram Alpha does give a closed form. It looks deeply unpleasant, but that means you can hope. – Clement C. May 18 '16 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 $$2\sum_{n\geq1}\frac{1}{\left(1+n^{2}\right)^{2}}=\frac{1}{2}\left(\pi^{2}\textrm{csch}^{2}\left(\pi\right)+\pi\coth\left(\pi\right)-2\right).$$