Determine the value of the integral $I=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx$

Question

Determine the value of the integral $$I(a)=\int_{0}^{1}\frac{\ln\left(1-a^2x^2\right)}{\sqrt{1-x^2}}dx, \: |a|\leq 1$$

My try:

$\to I'(a)=\int_{0}^{1}\frac{-2ax^2}{\left(1-a^2x^2\right)\sqrt{1-x^2}}dx$

Set $x=\cos t\to dx=-\sin tdt$

Hence $I'(a)=\int_{0}^{\frac{\pi}{2}}\frac{-2a\cos^2t}{1-a^2\cos^2t}dt=\pi\left(\frac{1}{a}-\frac{1}{\sqrt{1-a^2}}\right)\to I(a)=\pi\left(\ln|a|-\arcsin a\right)+C$

Question: Find C?

Answer 1

$$\begin{align}I(a) &= -\sum_{k=1}^{\infty} \frac{a^{2 k}}{k} \int_0^1 dx \frac{x^{2 k}}{\sqrt{1-x^2}}\\ &= -\frac{\pi}{2}\sum_{k=1}^{\infty} \frac1{k} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} \end{align} $$

$$I'(a) = -\frac{\pi}{2} \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k-1} $$

$$a I'(a) = -\pi \sum_{k=1}^{\infty} \binom{2 k}{k} \left (\frac{a}{2} \right )^{2 k} = -\pi \left (\frac1{\sqrt{1-a^2}} -1\right )$$

$$\implies \begin{align}I(a) &= -\pi \int da \left (\frac1{a\sqrt{1-a^2}}-\frac1{a} \right )\\ &= \pi \log{\left (1+\sqrt{1-a^2}\right )} +C \end{align}$$

$$I(0)=0 \implies C=-\pi \log{2}$$

$$\therefore I(a) = \pi \log{\left (\frac{1+\sqrt{1-a^2}}{2} \right )}$$

To derive the result for the integral in the first line, see here.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\,{\rm I}\pars{a} =\int_{0}^{1}{\ln\pars{1 - a^{2}x^{2}} \over \root{1 - x^{2}}}\,\dd x: \ {\large ?}.\qquad\verts{a} \leq 1}$.

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\begin{align}\color{#66f}{\large\,{\rm I}\pars{a}} &=-a^{2}\int_{0}^{1}{x^{2} \over \root{1 - x^{2}}} \int_{0}^{1}{\dd t \over 1 - a^{2}x^{2}t}\,\dd x \\[5mm]&=-a^{2}\int_{0}^{1} \int_{0}^{1}{x^{2} \over \root{1 - x^{2}}\pars{1 - a^{2}t\,x^{2}}}\,\dd x\,\dd t \\[5mm]&=\int_{0}^{1} \int_{0}^{1} {\pars{1 - a^{2}t\,x^{2}} - 1 \over \root{1 - x^{2}}\pars{1 - a^{2}t\,x^{2}}} \,\dd x\,{\dd t \over t} \\[5mm]&=\int_{0}^{1}\ \overbrace{\int_{0}^{1}\bracks{ {1 \over \root{1 - x^{2}}} - {1 \over \root{1 - x^{2}}\pars{1 - a^{2}t\,x^{2}}}} \,\dd x} ^{\ds{\dsc{x}\ \equiv\ \dsc{\cos\pars{\theta}}}}\ \,{\dd t \over t} \\[5mm]&=\int_{0}^{1}\bracks{{\pi \over 2} -\int_{0}^{\pi/2}{\dd\theta \over 1 - a^{2}t\,\cos^{2}\pars{\theta}}} \,{\dd t \over t} \\[5mm]&=\int_{0}^{1}\bracks{{\pi \over 2} -\int_{0}^{\pi/2} {\sec^{2}\pars{\theta}\,\dd\theta \over \sec^{2}\pars{\theta} - a^{2}t}} \,{\dd t \over t} \\[5mm]&=\int_{0}^{1}\bracks{{\pi \over 2}\ -\ \overbrace{\int_{0}^{\pi/2} {\sec^{2}\pars{\theta}\,\dd\theta \over \tan^{2}\pars{\theta} + 1 - a^{2}t}}^{\ds{\dsc{\tan\pars{\theta}}\ \equiv\ \dsc{\root{1 - a^{2}t}\ \xi}}}}\ \,{\dd t \over t} \\[5mm]&=\int_{0}^{1}\pars{{\pi \over 2} -{1 \over \root{1 - a^{2}t}}\ \overbrace{\int_{0}^{\infty}{\,\dd\xi \over \xi^{2} + 1}}^{\dsc{\pi \over 2}}} \,{\dd t \over t} ={\pi \over 2}\ \overbrace{ \int_{0}^{1}\pars{1 - {1 \over \root{1 - a^{2}t}}}\,{\dd t \over t}} ^{\ds{\dsc{t}\ \equiv \dsc{1 - y^{2} \over a^{2}}}} \\[5mm]&={\pi \over 2}\int_{1}^{\root{1 - a^{2}}}\pars{1 - {1 \over y}} \,{-2y\,\dd y/a^{2} \over \pars{1 - y^{2}}/a^{2}} =\pi\int^{\root{1 - a^{2}}}_{1}{\dd y \over 1 + y} \\[5mm]&=\color{#66f}{\large\pi\ln\pars{1 + \root{1 - a^{2}} \over 2}} \end{align}