Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ 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.
 A: It is not necessary to use complex analysis or to use power series to compute. From @Random Variable, we have
$$ I'(a) =\int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}=\int_{-1}^1 \frac{1}{1+ax}d\arcsin x. $$
Let $x=\sin t$. Then
$$ I'(a) =\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin t}dt. $$
Let $u=\tan\frac{t}{2}$. Then
\begin{eqnarray}
I'(a)&=&2\int_{-1}^{1} \frac{1}{u^2+1+2au}du=2\int_{-1}^{1} \frac{1}{(u+a)^2+1-a^2}du\\
&=&\frac{2}{\sqrt{1-a^2}}(\arctan\frac{1+a}{\sqrt{1-a^2}}-\arctan\frac{-1+a}{\sqrt{1-a^2}})\\
&=&\frac{2}{\sqrt{1-a^2}}(\arctan\frac{1+a}{\sqrt{1-a^2}}+\arctan\frac{1-a}{\sqrt{1-a^2}})\\
&=&\frac{2}{\sqrt{1-a^2}}\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt{1-a^2}}.
\end{eqnarray}
But $I(0)=0$ and so $ I(a)=\pi\arcsin a. $
A: It's easier to take the derivative of both sides according to $a$, than to perform the integration:
$$\int_0^1 \frac{2}{1-a^2x^2}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{\sqrt{1-a^2}}$$
The left hand side giving:
$$2 \text{arctanh}\left(\frac{\sqrt{a^2-1}}{\sqrt{1 - x^2}}x\right)\frac{1}{\sqrt{a^2-1}}+C$$
Which, when applying the limits, gives:
$$\frac{\pi}{\sqrt{1-a^2}}+C$$
as desired. Now all that needs to be done is compare at a single point to prove that the $C=0$, which you already have done.
A: View $I$ as a function of $a$, differentiate under integral sign and let 
$x = \sin\theta$, we have
$$\begin{align}
I'(a) &= \int_0^1 \left( \frac{x}{1+ax} - \frac{-x}{1-ax}\right) \frac{dx}{x\sqrt{1-x^2}}
= \int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}\\
&= \int_{-\pi/2}^{\pi/2} \frac{d\theta}{1+a\sin\theta}
= \frac12 \int_0^{2\pi}\frac{d\theta}{1+a\sin\theta}
= \frac12 \int_0^{2\pi}\frac{d\theta}{1+a\cos\theta}
\end{align}
$$
Introduce $z = e^{i\theta}$ and convert above integral to a contour integral over the unit
circle in $z$, we get
$$I'(a) = \frac{1}{2i}\oint_{|z|=1} \frac{dz}{z+\frac{a}{2}(z^2+1)}
= \frac{1}{ai}\oint_{|z|=1} \frac{dz}{(z - \lambda_{+})(z - \lambda_{-})}
$$
where $\displaystyle\;\lambda_{\pm} = -\frac{1}{a} \pm \sqrt{\frac{1}{a^2}-1}.\;$
When $|a| \le 1$, only the root $\lambda_{+}$ lies inside the unit circle, we have
$$I'(a) = \frac{1}{ai}\frac{2\pi i}{\lambda_{+} - \lambda_{-}}
= \frac{2\pi}{2a\sqrt{\frac{1}{a^2}-1}} = \frac{\pi}{\sqrt{1-a^2}}
$$
Since $I(0) = 0$, we get
$$I(a) = \pi \int_0^a \frac{dt}{\sqrt{1-t^2}} = \pi \arcsin(a)$$
A: The integral in question,
\begin{align}
I = \int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}}
\end{align}
can be separated into the two integrals
\begin{align}
I = \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx - \int_{0}^{1} \frac{ \ln(1-ax)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
which will be labled $I_{1}$ and $I_{2}$. Now
\begin{align}
I_{1} &= \int_{0}^{1} \frac{ \ln(1+ax)}{x \sqrt{1-x^{2}}} \ dx \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} a^{n}}{n} \ \int_{0}^{1} \frac{x^{n-1} \ dx}{\sqrt{1-x^{2}}} \\
&= - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-a)^{n}}{n} \ B(n/2, 1/2) \\
&= - \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-2a)^{n} \ \Gamma^{2}(n/2)}{n!} \\
&= - \frac{1}{4} \left[ \sum_{k=0}^{\infty} \Gamma^{2}(k+1/2) \frac{(-2a)^{2k+1}}{(2k+1)!} + \sum_{k=0}^{\infty} \frac{(k!)^{2} (-2a)^{2k+2}}{(2k+2)!} \right] \\
&= - \frac{1}{4} \left[ -2\pi a \sum_{k=0}^{\infty} \frac{(1/2)_{k} (1/2)_{k} a^{2k}}{ k! (3/2)_{k}} + 4 a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}} \right] \\
&= \frac{\pi}{2} \ \sin^{-1}(a) - a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}}.
\end{align}
In a similar manor, 
\begin{align}
I_{2} &= - \frac{\pi}{2} \ \sin^{-1}(a) - a^{2} \sum_{k=0}^{\infty} \binom{2k+2}{k+1}^{-1} \frac{(2a)^{2k}}{(k+1)^{2}}.
\end{align}
Since $I = I_{1} - I_{2}$ then 
\begin{align}
\int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}} = \pi \ \sin^{-1}(a)
\end{align}
which is the desired value. 
A: $\newcommand{\+}{^{\dagger}}
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$\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}
