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?
 A: You may not like this, but it is coincidentally similar:
$$1/4\int_{0}^{1}\frac{log(x)}{x^{3/4}(1-x)^{1/2}}dx-1/4\int_{0}^{1}\frac{log(x)}{x^{1/2}(1-x)^{1/2}}dx$$
These can be solved with the Beta function without a whole lot of effort.
They evaluate to $$-\frac{\pi}{4}\beta(1/4,1/2)+\frac{\pi}{2}log(2)$$
$$=\frac{-\pi^{5/2}\sqrt{2}}{4\Gamma^{2}(3/4)}+\frac{\pi}{2}log(2)$$
$$\approx -3.029925329.....$$
A: 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
$$I=\frac{\Gamma\left(\frac{1}{4} \right)^2}{4\sqrt{2\pi}}\left(\log 2-\pi \right)\approx -3.20998$$
A: $$I=\frac{\Gamma\left(\frac14\right)^2}{4\,\sqrt{2\,\pi}}\Big(\ln2-\pi\Big).$$
A: There is a closed form expression for this integral and I am sure you will enjoy it $$\frac{3}{2} \pi  \left(2
   \text{Hypergeometric2F1}^{(0,0,1,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)+\text
   {Hypergeometric2F1}^{(0,1,0,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)+\text
{Hypergeometric2F1}^{(1,0,0,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)-\,
   _2F_1\left(\frac{3}{4},\frac{3}{4};1;-8\right) \log (4)\right)$$ This was found by a CAS (do not ask me how to arrive to this marvel !). The numerical value is $-3.209982474415127728669875$
