Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$

Question

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.

Answer 1

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

$$I(a) = \pi \int_0^a \frac{dt}{\sqrt{1-t^2}} = \pi \arcsin(a)$$

Answer 2

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.

Answer 3

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

\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}

Answer 4

UPDATE: To make my answer less confusing, I changed the contour to a standard keyhole contour.

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Assume that $|a| <1$.

From achille hui's answer, we have

$$I'(a) =\int_{-1}^1 \frac{\mathrm dx}{(1+ax)\sqrt{1-x^2}}.$$

Making the substitution $u = \frac{1-x}{1+x}$,

$$I'(a) = \frac{1}{1-a}\int_{0}^{\infty} \frac{\mathrm du}{\sqrt{u} \left(u+ \frac{1+a}{1-a} \right) }. $$

Using the branch of the logarithm where $0 \le \arg(z) < 2 \pi$, let's integrate the function $$f(z) = \frac{1}{1-a}\frac{1}{\sqrt{z} \left(z+\frac{1+a}{1-a}\right)}$$ around a keyhole contour with the "keyhole" on the positive real axis.

We get $$ \begin{align} I'(a) &= \frac{2 \pi i }{1-e^{- \pi i}} \operatorname{Res} \left[f(z), \frac{1+a}{a-1} \right] \\ &= \frac{2 \pi i }{1-e^{-\pi i}} \frac{1}{1-a} \frac{1}{\sqrt{\left|\frac{1+a}{a-1}\right|e^{\pi i}}} \\ &= \frac{ \pi}{\sqrt{1-a^{2}}}. \end{align} $$

Integrating back with respect to $a$ and using the fact that $I(0) = 0$, we have

$$ I(a) = \pi \arcsin (a) .$$

Answer 5

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. $

Answer 6

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.

Answer 7

Substitute $x=\sin t$ \begin{align} &\int_0^1 \ln\frac{1+ax}{1-ax}\,\frac{1}{x\sqrt{1-x^2}}dx\\ =&\ 2\int_0^{\pi/2}\frac{\tanh^{-1}(a\sin t)}{\sin t}dt = 2\int_0^{\pi/2} \int_0^a \frac1{1- y^2\sin^2t}dy\ dt\\ = &\ \pi \int_0^a \frac1{\sqrt{1-y^2}}dy =\pi\sin^{-1}a \end{align}