How to solve the following integral, using Euler functions (Gamma, Beta, and Phi):
$$\int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\frac{\mathrm dx}{(1-x)^{2/3}(1+x)^{1/3}} $$
I got this integral from "Collection of problems in mathematical analysis, Kudryavtsev. Book 3.". Under the theme "Euler functions".
There are also functions, which can help:
\begin{align*}
\phi(p) &= \beta(p,1-p)= \frac{\pi}{\sin(p\pi)}\\
\phi'(p) &= – \frac{\pi^2\cos(p\pi)}{\sin^2(p\pi)}\\
\phi''(p) &= \frac{\pi^3}{\sin(p\pi)}\left(\cot^2(p\pi)+\frac1{\sin^2(p\pi)}\right)
\end{align*}
Best Answer
$$\text{let }\ \frac{1-x}{1+x}=t\Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt\ \text{ then}$$ $$\int_{-1}^1 \ln\left(\frac{1+x}{1-x}\right)\left(\frac{1+x}{1-x}\right)^{2/3}\frac{dx}{1+x}=-\int_0^\infty \frac{\ln t}{t^{2/3}(1+t)}dt$$ $$=-\lim_{p\to \frac13}\frac{d}{dp}\int_0^\infty\frac{t^{p-1}}{1+t}dt= -\lim_{p\to \frac13}\frac{d}{dp}\phi(p)=-\phi'\left(\frac13\right)=\frac{2\pi^2}{3}$$