how to integrate $\;I(a) = \int_0^1 \frac{\ln(1-a^2x^2)}{ x^2\sqrt{1-x^2}}dx$

Question

How to solve this integral?

$$I(a) = \int_0^1 \frac{\ln(1-a^2x^2)}{x^2\sqrt{1-x^2}}\,dx$$

Answer 1

EDIT

There was a mistake in the original - I forgot a factor in the expansion of the log. My apologies.

Expand the log in a Taylor series (assuming $|a| \lt 1$):

$$I(a) = -\sum_{k=0}^{\infty} \frac{a^{2 k+2}}{k+1} \int_0^1 dx \frac{x^{2 k}}{\sqrt{1-x^2}} = -\frac{\pi}{2} \sum_{k=0}^{\infty} \frac1{2^{2 k}}\binom{2 k}{k} \frac{a^{2 (k+1)}}{k+1}$$

(The derivation of the integral may be found here.)

The derivative of the sum is a well-known Taylor expansion in its own right. Thus,

$$I'(a) = -\frac{\pi}{2} \frac{2 a}{\sqrt{1-a^2}} $$

$$I(a) = 0 \implies I(a) = -\pi \left (1-\sqrt{1-a^2} \right ) $$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \,{\rm I}\pars{a}&\ =\ \overbrace{\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 - a^{2}x^{2}} \over x^{2}\root{1 - x^{2}}}} \,\dd x}^{\ds{\color{#c00000}{x\ \equiv\ \cos\pars{\theta}}}}\ =\ \int_{\pi/2}^{0}{\ln\pars{1 - a^{2}\cos^{2}\pars{\theta}} \over \cos^{2}\pars{\theta}\root{1 - \cos^{2}\pars{\theta}}} \,\bracks{-\sin\pars{\theta}\dd\theta} \\[5mm]&=\overbrace{\int_{0}^{\pi/2} \ln\pars{1 - {a^{2} \over \tan^{2}\pars{\theta} + 1}}\sec^{2}\pars{\theta} \,\dd\theta}^{\ds{\color{#c00000}{\tan\pars{\theta}\ \equiv t}}}\ =\ \int_{0}^{\infty}\ln\pars{1 - {a^{2} \over t^{2} + 1}}\,\dd t \\[5mm]&=\int_{0}^{\infty} \bracks{\ln\pars{t^{2} + 1 - {a^{2}}} - \ln\pars{t^{2} + 1}}\,\dd t =2\int_{0}^{\infty} \pars{{t^{2} \over t^{2} + 1} - {t^{2} \over t^{2} + 1 - a^{2}}}\,\dd t \\[5mm]&=2\int_{0}^{\infty} \pars{-\,{1 \over t^{2} + 1} + {1 - a^{2} \over t^{2} + 1 - a^{2}}}\,\dd t =-2\int_{0}^{\infty}{\dd t \over t^{2} + 1} +2\root{1 - a^{2}}\int_{0}^{\infty}{\dd t \over t^{2} + 1} \\[5mm]&=-2\pars{1 - \root{1 - a^{2}}}\ \underbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}} _{\ds{\color{#c00000}{\pi \over 2}}}\ =\ \color{#66f}{\large -\pi\pars{1 - \root{1 - a^{2}}}} \end{align}

Answer 3

Hint: Alternately, try differentiating under the integral sign with regard to a.