Beta Dirichlet function

Integral involving the beta Dirichlet function

We repost here the proof of this result:
\[\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx = 2\beta(2)\]
Proof

Do the following change of variable \[w=\ln \cot(\frac{x}{2}) \Longrightarrow e^w = \cot(\frac{x}{2}) \Longrightarrow 2\cot^{-1}(e^w)=x\] \[w\to \infty \textrm{ if } x \to 0, \quad w\to 0 \textrm{ if } x \to \frac{\pi}{2}\] Then \[dw = -\csc(x)dx = -\frac{1}{\sin(x)}dx \] Therefore \[\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx =-\int_{\infty}^{0} 2\cot^{-1}(e^w)dw = \int_{0}^{\infty} 2\cot^{-1}(e^w)dw \] Recall the series expansion of $\cot^{-1}$ \[ \cot^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)}x^{-2n-1} \quad x\geq 0\] Therefore \[\int_{0}^{\infty} 2\cot^{-1}(e^w)dw = \int_{0}^{\infty} 2\sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)}(e^w)^{-2n-1} dw= 2\sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)}\int_{0}^{\infty}(e^w)^{-2n-1} dw\] We have that \[\int (e^w)^{-2n-1} dw = \frac{(e^w)^{-2n-1}}{-2n-1}+C \Longrightarrow \int_{0}^{\infty}(e^w)^{-2n-1} dw = \left[\frac{(e^w)^{-2n-1}}{-2n-1} \right]_{0}^{\infty} = \frac{1}{2n+1}\] \[\therefore \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx = \int_{0}^{\infty} 2\cot^{-1}(e^w)dw = 2\sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)^2} =2\beta(2)\] \[\boxed{\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin(x)}dx = 2\beta(2)}\]