 Donald Johnson

2022-01-05

Evaluation of the integral:
${\int }_{0}^{1}\frac{\mathrm{ln}\left(1-x\right)}{1+x}dx$ boronganfh

You can use double integration:
${\int }_{0}^{1}\frac{\mathrm{log}\left(1-x\right)}{1+x}dx={\int }_{0}^{1}{\int }_{0}^{-x}\frac{du\cdot dx}{\left(1+u\right)\left(1+x\right)}$
${\int }_{0}^{1}{\int }_{0}^{x}\frac{dm\cdot dx}{\left(m-1\right)\left(1+x\right)}$
Now make
$m=ux$
${\int }_{0}^{1}{\int }_{0}^{1}\frac{x\cdot du\cdot dx}{\left(ux-1\right)\left(1+x\right)}={\int }_{0}^{1}{\int }_{0}^{1}\frac{du\cdot dx}{\left(ux-1\right)}-{\int }_{0}^{1}{\int }_{0}^{1}\frac{du\cdot dx}{\left(ux-1\right)\left(1+x\right)}$
We have that (partial fraction decomposition)
$\frac{1}{\left(ux-1\right)\left(x+1\right)}=\frac{u}{\left(u+1\right)\left(ux-1\right)}-\frac{1}{\left(x+1\right)\left(u+1\right)}$
So we get
${\int }_{0}^{1}{\int }_{0}^{1}\frac{du\cdot dx}{\left(ux-1\right)}-{\int }_{0}^{1}{\int }_{0}^{1}\frac{u\cdot du\cdot dx}{\left(ux-1\right)\left(u+1\right)}+{\int }_{0}^{1}{\int }_{0}^{1}\frac{du\cdot dx}{\left(x+1\right)\left(u+1\right)}$
Now:
${\int }_{0}^{1}{\int }_{0}^{1}\frac{du\cdot dx}{\left(ux-1\right)}={\int }_{0}^{1}\frac{\mathrm{log}\left(1-u\right)}{u}du=-\frac{{\pi }^{2}}{6}$ karton

Following is an elementary proof.
I assume only that

${\int }_{0}^{1}\frac{\mathrm{ln}x}{1-x}dx=-\frac{{\pi }^{2}}{6}\phantom{\rule{0ex}{0ex}}J={\int }_{0}^{1}\frac{\mathrm{ln}\left(1-x\right)}{1+x}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\frac{\mathrm{ln}\left(\frac{2y}{1+y}\right)}{1+y}dy\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\frac{\mathrm{ln}\left(\frac{2y}{1+y}\right)}{1+y}dy+{\int }_{0}^{1}\frac{\mathrm{ln}t}{1+t}dt\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\mathrm{ln}}^{2}2+{\int }_{0}^{1}\frac{\mathrm{ln}t}{1+t}dt\phantom{\rule{0ex}{0ex}}{\int }_{0}^{1}\frac{\mathrm{ln}t}{1+t}dt={\int }_{0}^{1}\frac{\mathrm{ln}x}{1-x}dx-{\int }_{0}^{1}\frac{2t\mathrm{ln}t}{1-{t}^{2}}dt\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}\frac{\mathrm{ln}x}{1-x}dx-\frac{1}{2}{\int }_{0}^{1}\frac{\mathrm{ln}w}{1-w}dw\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\int }_{0}^{1}\frac{\mathrm{ln}x}{1-x}dx\phantom{\rule{0ex}{0ex}}=-\frac{1}{12}{\pi }^{2}$ user_27qwe

You can use the integral you want to use, and the Dilogarithm function as mentioned in the comments.
Below we give a complete proof, including a derivation of the value of the integral you wanted to use.
The Dilogarithm function is defined as

The integral which you want to use is $-L{i}_{2}\left(1\right)$
Note that $L{i}_{2}\left(1\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}=\zeta \left(2\right)=\frac{{\pi }^{2}}{6}$. (For multiple proofs of that, see here: Different methods to compute $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}}$)
In your integral(whose value you want), make the substitution x=2t-1 and we get
${\int }_{\frac{1}{2}}^{1}\frac{\mathrm{log}\left(2\left(1-t\right)\right)}{t}dt={\mathrm{log}}^{2}2+{\int }_{\frac{1}{2}}^{1}\frac{\mathrm{log}\left(1-t\right)}{t}dt={\mathrm{log}}^{2}2+L{i}_{2}\left(\frac{1}{2}\right)-L{i}_{2}\left(1\right)$
Now the Dilogarithm function also satisfies the identity

This identity can easily be proven by just differentiating and using the value of $L{i}_{2}\left(1\right)$:
$L{i}_{2}^{\prime }\left(x\right)-L{i}_{2}^{\prime }\left(1-x\right)=-\frac{\mathrm{log}\left(1-x\right)}{x}+\frac{\mathrm{log}x}{1-x}$
$=\left(-\mathrm{log}x\mathrm{log}\left(1-x\right){\right)}^{\prime }$
and so

Taking limits as $x\to 1$ gives us $C=\frac{{\pi }^{2}}{6}$
Thus

Setting $x=\frac{1}{2}$ gives us the value of $L{i}_{2}\left(\frac{1}{2}\right)=\frac{{\pi }^{2}}{12}-\frac{{\mathrm{log}}^{2}2}{2}$
${\mathrm{log}}^{2}2+L{i}_{2}\left(\frac{1}{2}\right)-L{i}_{2}\left(1\right)=\frac{{\mathrm{log}}^{2}2}{2}-\frac{{\pi }^{2}}{12}$