Prove $\frac{1}{a}\int_{0}^1 \frac{\sin{ax}}{e^x-1}dx=\frac{1}{2a^2}(1+a\pi\frac{e^{a\pi}+e^{-a\pi}}{e^{a\pi}-e^{a\pi}})$

Question

I found this interesting equation $$\sum_{n=0}^\infty \frac{1}{n^2+a^2}=\frac{1}{2a^2}\left(1+a\pi\frac{e^{a\pi}+e^{a\pi}}{e^{a\pi}-e^{a\pi}}\right)$$ at twitter. And I tried to prove it as below.

$$\begin{align} &\sum_{n=0}^\infty \frac{1}{n^2+a^2}\\ =&\frac{1}{2ai}\left(\sum_{n=0}^\infty \frac{1}{n-ai}-\frac{1}{n+ai}\right)\\ =&\frac{1}{2ai}\left(\sum_{n=0}^\infty \int_{0}^1 e^{(n-ai)x}-e^{(n+ai)x}dx\right)\\ =&\frac{1}{2ai}\int_{0}^1 \frac{e^{-aix}-e^{aix}}{1-e^x}dx\\ =&\frac{1}{a}\int_{0}^1 \frac{\sin(ax)}{1-e^x}dx\\ \end{align}$$

Now I need to evaluate $\frac{1}{a}\int_{0}^1 \frac{\sin(ax)}{1-e^x}dx$. According to wolfram alpha, this integration is very complicated.

Is there easier way to evaluate this integration?

Answer 1

Here is a proof of the sum. Recall that Euler gave $$\sin\pi x=\pi x\prod_{n\ge1}(1-x^2/n^2).$$ Then $$\ln\sin\pi x=\ln\pi+\ln x+\sum_{n\ge1}\ln(1-x^2/n^2).$$ Differentiating both sides, $$\pi \cot\pi x=\frac1x+2x\sum_{n\ge1}\frac{1}{n^2-x^2}$$ This is $$\sum_{n\ge0}\frac1{n^2-x^2}=\frac{\pi}{2x}\cot\pi x-\frac1{2x^2}.$$ This is your result after $x\mapsto ix$.