Integral $\int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\frac{\mathrm dx}{(1-x)^{2/3}(1+x)^{1/3}}$

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*}

• Are you sure that your integral does converge on the given interval? Sep 18, 2019 at 13:44
• @Dr.SonnhardGraubner yes, it converges
– Egor
Sep 18, 2019 at 14:00
• From where did you get this integral? Btw the function you gave help, the answer is $\frac{2\pi^2}{3}$, but please give some context for this integral. Sep 18, 2019 at 14:08
• @カカロット This integral I get from "collection of problems in mathematical analysis, Kudryavtsev. Book 3.". Under theme "Euler functions"
– Egor
Sep 18, 2019 at 14:22

$$\text{let }\ \frac{1-x}{1+x}=t\Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt\ \text{ then}$$ $$\int_{-1}^1 \ln\left(\frac{1+x}{1-x}\right)\left(\frac{1+x}{1-x}\right)^{2/3}\frac{dx}{1+x}=-\int_0^\infty \frac{\ln t}{t^{2/3}(1+t)}dt$$ $$=-\lim_{p\to \frac13}\frac{d}{dp}\int_0^\infty\frac{t^{p-1}}{1+t}dt= -\lim_{p\to \frac13}\frac{d}{dp}\phi(p)=-\phi'\left(\frac13\right)=\frac{2\pi^2}{3}$$
Substitute $$x\mapsto2x-1$$ to get
$$I=\int_0^1\ln\left(\frac x{1-x}\right)\frac{\mathrm dx}{x^{1/3}(1-x)^{2/3}}$$
$$I=B^{(1,0)}\left(\frac23,\frac13\right)-B^{(0,1)}\left(\frac23,\frac13\right)$$