Evaluating $\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$ I have the following integral:
$$\int_{0}^{1} \frac{ \ln x \ln (1-x)}{\sqrt{x} \sqrt{1-x}} dx$$
I think I may be to evaluate this with beta and gamma functions but I am not quite sure how. Any help?
 A: Let $I$ denote the integral in question. With the substitution $x = \sin^{2} \theta$, we have
$$ I = 8 \int_{0}^{\frac{\pi}{2}} \log \cos \theta \log \sin \theta \, d \theta = \frac{\partial^2}{\partial z \partial w} \beta \left( \frac{1}{2}, \frac{1}{2} \right), $$
where
$$\beta(z, w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)} = 2 \int_{0}^{\frac{\pi}{2}} \cos^{2z-1}\theta \sin^{2w-1}\theta \, d\theta \tag{1} $$
is the beta function. This gives
$$ I = 4\pi \log^{2} 2 - \frac{\pi^3}{6}. $$
A: An alternative way, using elementary methods, which begins with the same substitution as sos440's solution:
$$\int_0^1 \frac{\ln x\ln (1-x)}{\sqrt{x(1-x)}}\,dx=8\int_0^{\frac{\pi}{2}}\ln \sin x\ln \cos x\,dx$$
Note that:
$$\begin{aligned}2\int_0^{\frac{\pi}{2}}\ln \sin x\ln \cos x\,dx &=\int_0^{\frac{\pi}{2}}(\ln \sin x+\ln \cos x)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\int_0^{\frac{\pi}{2}}(\ln \sin 2x-\ln 2)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\int_0^{\frac{\pi}{2}}(\ln \sin x-\ln 2)^2\,dx-2\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\ln^2 2\int_0^{\frac{\pi}{2}}\,dx-2\ln 2\int_0^{\frac{\pi}{2}}\ln \sin x\,dx-\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\\&=\frac{3\pi\ln^2 2}{2}-\int_0^{\frac{\pi}{2}}\ln^2\sin x\,dx\end{aligned} $$
Now we are left to evaluate the latter integral, for which a little preliminary work must be done:
$$ \sum_{k=1}^n \sin 2kz =\frac{1}{2\sin z} \Big( \cos z-\cos (2n+1)z\Big)$$ which upon integration with $0<x<\frac{\pi}{2}$ gives:
$$ \int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\sum_{k=1}^n \frac{\cos 2k x}{k}+\sum_{k=1}^n \frac{(-1)^{k+1}}{k}+\ln \sin x$$
Furthermore, $$\int_x^{\frac{\pi}{2}} \frac{\cos (2n+1) z}{\sin z}\,dz=\frac{(-1)^{n+1}}{2n+1}+\frac{\sin (2n+1) x}{(2n+1)\sin x}- \int_x^{\frac{\pi}{2}} \frac{\sin (2n+1) z\cos z}{(2n+1)\sin^2 z}\,dz$$
Which $\to 0$ as $n\to\infty$. Hence we obtain the relation: $$ \ln \sin x= -\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}-\ln 2$$
$$ \ln^2 \sin x= \ln^2 2+\left(\sum_{a=1}^{\infty} \frac{\cos 2a x}{a}\right)\left(\sum_{b=1}^{\infty} \frac{\cos 2b x}{b}\right)+2\ln 2\sum_{k=1}^{\infty} \frac{\cos 2k x}{k}$$
Integrating over $(0,\frac{\pi}{2})$ the latter sum clearly vanishes and all the paired cosines for which $a\neq b$ as well. Thus we have:
$$\begin{aligned}\int_0^{\frac{\pi}{2}}\ln^2 \sin x\,dx &=\ln^2 2\int_0^{\frac{\pi}{2}}dx+\sum_{a=1}^{\infty}\frac{1}{a^2}\int_0^{\frac{\pi}{2}}\cos^2 2a x\,dx\\&=\frac{\pi\ln^2 2}{2}+\sum_{a=1}^{\infty}\frac{\pi}{4a^2}\\&=\frac{\pi\ln^2 2}{2}+\frac{\pi^3}{24}\end{aligned}$$
Combining everything and multiplying by $4:$
$$\int_0^1 \frac{\ln x\ln (1-x)}{\sqrt{x(1-x)}}\,dx=4\pi\ln^2 2-\frac{\pi^3}{6}$$
