TITLE: How to find the value of this integral? $\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$
QUESTION [9 upvotes]: $$I=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx=-\frac{G}{4}+\frac{1}{24}+\frac{\pi}{64}-\frac{\pi}{16}\ln2$$
 Where G is the Catalan's contant
 My try:
 Integrating by parts, we have:
 $$3I=\int_0^1x^4\ln(1-x)\frac{dx}{\sqrt{1-x^2}}+\int_0^1x^3\sqrt{1-x^2}\frac{dx}{1-x}=S+T$$
 $$\frac{x^3}{1-x}=-x^2-x-1+\frac{1}{1-x}$$
 It is easy to deduce:  $$\int_0^1x^2\sqrt{1-x^2}dx=\frac{\pi}{16},\int_0^1x\sqrt{1-x^2}dx=\frac{1}{3},\int_0^1\sqrt{1-x^2}dx=\frac{\pi}{4},\int_0^1\sqrt{1-x^2}\frac{dx}{1-x}=\frac{\pi}{2}+1$$
 $$3I=S+\frac{3\pi}{16}+\frac{2}{3}$$
 But how to deduct S?

REPLY [6 votes]: $\displaystyle J=\int_0^1x^2\sqrt{1-x^2}\log(1-x)dx$
Perform the change of variable $y=\sqrt{1-x}$,
$\displaystyle J=4\int_0^1 x^2(1-x^2)^2\sqrt{2-x^2}\ln x$
Perform the change of variable $x=\sqrt{2}y$,
$\begin{align}
J&=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln\left(\sqrt{2}x\right)dx\\
&=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln x dx+8\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln2 dx\\
&=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln x dx+\\
&8\ln 2\Big[\frac{\mathrm{arcsin}\left( x\right) }{32}-\frac{{{x}^{5}}\cdot {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{2}+\frac{{{x}^{3}}\cdot {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{4}-\frac{x\cdot {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{16}+\frac{x\cdot \sqrt{1-{{x}^{2}}}}{32}\Big]_0^{\tfrac{1}{\sqrt{2}}}\\
&=16\int_0^{\tfrac{1}{\sqrt{2}}}x^2(1-2x^2)^2\sqrt{1-x^2}\ln x dx+\dfrac{\pi}{16}\ln 2\\
&=16\left[\left(\frac{\mathrm{arcsin}\left( x\right) }{32}-\frac{{{x}^{5}}{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{2}+\frac{{{x}^{3}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{4}-\frac{x {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{16}+\frac{x\sqrt{1-{{x}^{2}}}}{32}\right)\ln x\right]_0^{\tfrac{1}{\sqrt{2}}}-\\
&16\int_0^{\tfrac{1}{\sqrt{2}}}\left(\frac{\mathrm{arcsin}\left( x\right) }{32}-\frac{{{x}^{5}}{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{2}+\frac{{{x}^{3}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{4}-\frac{x {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}}{16}+\frac{x\sqrt{1-{{x}^{2}}}}{32}\right)\dfrac{1}{x}dx+\\
&\dfrac{\pi}{16}\ln 2\\
&=-\int_0^{\tfrac{1}{\sqrt{2}}}\left(\frac{\mathrm{arcsin}\left( x\right) }{2 x}-8 {{x}^{4}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+4{{x}^{2}}{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}-{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+\frac{\sqrt{1-{{x}^{2}}}}{2}\right)dx\\
&=-\int_0^{\tfrac{1}{\sqrt{2}}}\left(-8{{x}^{4}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+4{{x}^{2}} {{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}-{{\left( 1-{{x}^{2}}\right) }^{\frac{3}{2}}}+\frac{\sqrt{1-{{x}^{2}}}}{2}\right)dx-\\
&\dfrac{1}{2}\int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{asin}\left( x\right) }{x}dx\\
&=-\left[\frac{\sqrt{1-{{x}^{2}}} \left( 48 {{x}^{7}}-104 {{x}^{5}}+74{{x}^{3}}-21 x\right) -3\mathrm{arcsin}\left( x\right) }{48}\right]_0^{\tfrac{1}{\sqrt{2}}}-\dfrac{1}{2}\int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx\\
&=\dfrac{\pi}{64}+\dfrac{1}{24}-\dfrac{1}{2}\int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx\\
\end{align}$
In the latter integral perform the change of variable $y=\arcsin x$,
$\begin{align} \int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx&=\int_0^{\tfrac{\pi}{4}}x\cot x\, dx\\
&=\Big[x\ln(\sin x)\Big]_0^{\tfrac{\pi}{4}}-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\
&=-\dfrac{\pi}{8}\ln 2-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{\pi}{2}-x$,
$\begin{align}\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx&=\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ln(\cos x)\,dx\\
&=\int_{0}^{\tfrac{\pi}{2}}\ln(\cos x)\,dx-\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx\\
&=-\dfrac{\pi}{2}\ln 2-\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx\\
\end{align}$
therefore,
$\begin{align}\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx&=\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)\,dx\\
&=\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx-\dfrac{\pi}{2}\ln 2-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx
\end{align}$
Therefore,
$\begin{align}\int_0^{\tfrac{\pi}{4}}\ln(\sin x)\,dx&=\dfrac{1}{2}\int_0^{\tfrac{\pi}{4}}\ln(\tan x)\,dx-\dfrac{\pi}{4}\ln 2\\
&=-\dfrac{G}{2}-\dfrac{\pi}{4}\ln 2\\
\end{align}$
Therefore,
$\begin{align} \int_0^{\tfrac{1}{\sqrt{2}}}\frac{\mathrm{arcsin}\left( x\right) }{x}dx=\dfrac{G}{2}+\dfrac{\pi}{8}\ln 2\\
\end{align}$
The result follows,
$\boxed{J=-\dfrac{G}{4}+\dfrac{1}{24}+\dfrac{\pi}{64}-\dfrac{\pi}{16}\ln2}$