Integral $\int_0^{\infty} \frac{x\cos^2 x}{e^x-1}dx$

Try to use the following expansion: $$ \frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{k=1}^{+\infty}e^{-kx}. $$ Then switch order of summation and integration, integrate, and you will end up with the series $$ \sum_{k=1}^{+\infty}\biggl(\frac{1}{2k^2}-\frac{4}{(4+k^2)^2}+\frac{1}{2(4+k^2)}\biggr). $$ I'm sure you can handle it.

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Integration

Riemann Zeta

Improper Integrals

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