Integral of the day XXI

Integral involving binomial series

Today we show the proof of this integral proposed by @BenriBot_ccbs. \[ \int_{0}^{1} \frac{\arcsin\left(\sqrt{x}\right)}{1+x} dx = \frac{\pi}{2}\ln(8) - \pi\ln(1+\sqrt{2}) \] The proof relies on the use of binomial series.

Proof

\begin{align*} I = \int_{0}^{1} \frac{\arcsin\left(\sqrt{x}\right)}{1+x} dx =& 2\int_{0}^{1} \frac{\arcsin(w)w}{1+w^2} dw \quad \left( w \mapsto \sqrt{x} \right)\\ \stackrel{IBP}{=}& \ln(w^2+1)\arcsin(w)\Big|_{0}^{1} - \int_{0}^{1} \frac{\ln(w^2+1)}{ \sqrt{1-w^2}}dw \\ =& \frac{\pi}{2}\ln(2) - \int_{0}^{1} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\frac{w^{2n}}{\sqrt{1-w^2}}dw \\ =& \frac{\pi}{2}\ln(2) - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1} w^{2n}(1-w^2)^{-\frac{1}{2}}dw \\ =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_{0}^{1} s^{n-\frac{1}{2}}(1-s)^{-\frac{1}{2}}ds \quad (s \mapsto w^2) \\ =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}B\left(n+\frac{1}{2},\frac{1}{2}\right) \\ =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\frac{\Gamma\left(n+\frac{1}{2}\right) \sqrt{\pi}}{n!} \end{align*} From the formulas \[ \Gamma\left(n+\frac{1}{2}\right) =\frac{(2n-1)!! \sqrt{\pi}}{2^n} \tag{1}\] \[ (2n)!! = 2^nn! \tag{2}\] \[ \binom{2n}{n} = \frac{2^{2n}(2n-1)!!}{(2n)!!} \tag{3}\] we have \begin{align*} I =& \frac{\pi}{2}\ln(2) - \frac{1}{2}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\frac{(2n-1)!! \pi}{2^nn!} \quad \textrm{ from } (1)\\ =& \frac{\pi}{2}\ln(2) + \frac{\pi}{2}\sum_{n=1}^{\infty} (-1)^n\frac{(2n-1)!!}{n(2n)!!} \quad \textrm{ from } (2)\\ =& \frac{\pi}{2}\ln(2) + \frac{\pi}{2}\sum_{n=1}^{\infty} (-1)^n\frac{\binom{2n}{n}}{2^{2n}n} \quad \textrm{ from } (3) \\ \end{align*} We previously proved that \[\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}x^n = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)\] If we put $x = -1$ \[\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}(-1)^n = 2\ln\left(\frac{2}{\sqrt{2}+1}\right) \] Hence, we can conclude \[ \boxed{\int_{0}^{1} \frac{\arcsin\left(\sqrt{x}\right)}{1+x} dx = \frac{\pi}{2}\ln(8) - \pi\ln(1+\sqrt{2})} \]