TITLE: Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$
QUESTION [17 upvotes]: Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \log\left(\frac{1+ x}{1-x}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}
=\frac{\pi^2}{2}.
$$
I tried to write
$$
I=\int_0^1 \frac{\log(1+ax)}{x\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1-ax)}{x\sqrt{1-x^2}}dx
$$
If we work with one of these integrals we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1} a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx-\sum_{n=1}^\infty \frac{a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx,
$$
simplifying this I get an infinite sum of Gamma functions. which i'm not sure how to relate to the $\arcsin$ Thanks.

REPLY [10 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{{\rm I}\pars{a}\equiv\int_{0}^{1}\ln\pars{1 + ax \over 1 - ax}\,
     {\dd x \over x\root{1 - x^{2}}} = \pi\arcsin\pars{a}:\ {\large ?}
     \,,\qquad\verts{a}\leq 1}$.

\begin{align}
\color{#c00}{{\rm I}'\pars{a}} &=2\
\overbrace{\int_{0}^{1}{\dd x \over \pars{1 - a^{2}x^{2}}\root{1 - x^{2}}}}
^{\ds{\mbox{Set}\ x \equiv \cos\pars{\theta}}}\ =\
2\int_{0}^{\pi/2}{\dd\theta \over 1 - a^{2}\cos^{2}\pars{\theta}}
\\[3mm]&=2\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta\over
\sec^{2}\pars{\theta} - a^{2}}\
=2\
\overbrace{\int_{0}^{\pi/2}{\sec^{2}\pars{\theta}\,\dd\theta\over
\tan^{2}\pars{\theta} + 1 - a^{2}}}^{\ds{\mbox{Set}\ t \equiv \tan\pars{\theta}}}
\\[3mm] & =2\int_{0}^{\infty}{\dd t \over t^{2} + 1 - a^{2}}
={2 \over \root{1 - a^{2}}}\
\overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}^{\ds{=\ {\pi \over 2}}}
\\[3mm] & \qquad\imp\qquad \color{#c00000}{{\rm I}'\pars{a} = {\pi \over \root{1 - a^{2}}}}
\end{align}

Since $\ds{{\rm I}\pars{0} = 0}$:
\begin{align}
{\rm I}\pars{a} & =\color{#66f}{\large\int_{0}^{1}\ln\pars{1 + ax \over 1 - ax}\,
{\dd x \over x\root{1 - x^{2}}}}
=\pi\int_{0}^{a}{\dd t \over \root{1 - t^{2}}}
\\[3mm] & =\color{#66f}{\large \pi\ \arcsin\pars{a}}
\end{align}