Integral $\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx$

I have a trouble with this integral $$I=\int_0^1\frac{\log(1-x)}{\sqrt{x-x^3}}dx.$$ Could you suggest how to evaluate it?

Solution 1:

Here is a proof of Cleo's answer.

Rewrite the integral as

$$ \begin{align*} I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \end{align*} $$ The first integral can be computed by calculating a derivative of the beta function. $$ \begin{align*} &\;\int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &\stackrel{y=x^2}{=}\frac{1}{2}\int_0^1 \frac{\log(1-y)}{y^{\frac{3}{4}}\sqrt{1-y}}dy \\ &= \frac{1}{2}\frac{\partial}{\partial \alpha}\left\{B\left(\frac{1}{4},\alpha \right)\right\}_{\alpha=\frac{1}{2}}\\ &= \frac{\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}{2\Gamma\left(\frac{3}{4} \right)}\left\{ \psi_0\left(\frac{1}{2} \right)-\psi_0\left(\frac{3}{4} \right)\right\} \\ &= \frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(2\log 2-\pi \right) \end{align*} $$

The other integral can be evaluated by using equation $(3.22)$ of this paper.

$$ \begin{align*} &\;\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &\stackrel{x=\sin^2 t}{=}2\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 t)}{\sqrt{1+\sin^2 t}}dt\\ &= \log(2) K(\sqrt{-1}) \\ &= \log(2)\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}} \end{align*} $$ where $K(k)$ is the complete elliptic integral of the first kind. After combining everything, one gets

$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$