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Hi I am trying to solve this integral $$ I:=\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a,\qquad |a|\leq 1. $$ It gives beautiful result for $a=1$ $$ \int_0^1 \log \frac{1+x}{1-x}\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi^2}{2}. $$ 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.

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  • $\begingroup$ This question takes much time to load on my browser! $\endgroup$ – xxx--- Aug 9 '14 at 14:55
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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.

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  • $\begingroup$ +1 this is a brilliant solution BUT you also continued with the method I was trying. Thank you and this is why I am checking this as the answer! $\endgroup$ – Jeff Faraci May 16 '14 at 3:51
  • $\begingroup$ What is the notation $(1/2)_k$? thanks @Leucippus $\endgroup$ – user1111 Jun 5 '14 at 5:29
  • $\begingroup$ @user1111 \begin{align} (a)_{n} = \frac{\Gamma(n+a)}{\Gamma(a)} \end{align} is Pochhammer's notation. $\endgroup$ – Leucippus Jun 5 '14 at 14:05
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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

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

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  • $\begingroup$ That's how I'd have done it. (+1) $\endgroup$ – Ron Gordon May 15 '14 at 15:10
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm I}\pars{a}\equiv\int_{0}^{1}\ln\pars{1 + ax \over 1 - ax}\, {\dd x \over x\root{1 - x^{2}}} = \pi\arcsin\pars{a}:\ {\large ?} \,,\qquad\verts{a}\leq 1}$.

\begin{align} {\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}}} =2\int_{0}^{\infty}{\dd t \over t^{2} + 1 - a^{2}} \\[3mm]&={2 \over \root{1 - a^{2}}}\ \overbrace{\int_{0}^{\infty}{\dd t \over t^{2} + 1}}^{\ds{=\ {\pi \over 2}}} \qquad\imp\qquad \color{#c00000}{{\rm I}'\pars{a} = {\pi \over \root{1 - a^{2}}}} \end{align}

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

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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. $

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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.

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  • $\begingroup$ I would check this as the solution however you show no proof as to how to do the integral on the left hand side. The point of the problem is to do the integral (without Numerical assistance). If you can provide a proof as to how to do that integral, I will mark as answer also. +1 anyways $\endgroup$ – Jeff Faraci May 16 '14 at 3:54

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