Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$

Question

I would like to evaluate this integral, $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}$$

An approach: 1. Using integration by parts

$u=\ln\left(\frac{a-x^2}{a+x^2}\right)$, $du=\frac{x^2+a}{x^2-a}\left[\frac{2x}{a+x^2}+\frac{2x(a-x^2)}{a+x^2}\right]dx$

$dv=\frac{dx}{x^2\sqrt{1-x^2}}$, $v=-\frac{\sqrt{1-x^2}}{x}$

$$I=2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2+a}dx-2\int_{0}^{1}\frac{\sqrt{1-x^2}}{x^2-a}dx=2I_1-2I_2$$

Integral $I_1$: Make a substitution of: $x=\sin(v)$, $v=\arcsin(x)$, $dx=\cos(v)dv$

$$I_1=\int_{0}^{\pi/2}\frac{\cos^2(v)}{a+\sin^2(v)}dv$$

Using trig to rewrite: $$I_1=\int_{0}^{\pi/2}\sec^2(v)\frac{dv}{[1+\tan^2(v)][(a+1)\tan^2(v)+a]}$$

Make another substitiution of: $v=\tan(y)$, $dy=\frac{1}{\sec^2(v)}dv$ and using partial fraction decomp:

$$I_1=(a+1)\int_{0}^{\infty}\frac{dy}{a+(a+1)y^2}-\frac{\pi}{2}=(a+1)I_3-\frac{\pi}{2}$$

With this substitution: $t=\frac{\sqrt{a+1}}{\sqrt{a}}y$, $dy=\frac{\sqrt{a}}{\sqrt{a+1}}dt$

$$I_3=\frac{1}{\sqrt{a(a+1)}}\cdot \frac{\pi}{2}$$

$$I_1=\frac{\pi}{2}\left(\frac{\sqrt{a+1}}{\sqrt{a}}-1\right)$$

In the same manner, $$I_2=\frac{\pi}{2}\left(\frac{\sqrt{a-1}}{\sqrt{a}}-1\right)$$

Finally, $$I=\pi\left(\frac{\sqrt{a-1}-\sqrt{a+1}}{\sqrt{a}}\right)$$

I am looking for another short approach of evaluating integral $I$

While in your question, your formatting is fine, please don't format in a way that makes the vertical distance of a title to large. — amWhy, Jun 28 '18 at 22:35
Congratulations for the patience you show ! Good job, indeed and $+1$ — Claude Leibovici, Jun 29 '18 at 06:16

Jack D'Aurizio · Answer 1 · 2018-06-28T23:11:45.013

4

Assuming $a>1$, the original integral equals $$ \frac{1}{2}\int_{0}^{1}\log\left(\frac{a-x}{a+x}\right)\frac{dx}{x\sqrt{x(1-x)}}\,dx=-\int_{0}^{1}\frac{\text{arctanh}(x/a)}{x\sqrt{x(1-x)}}\,dx \tag{1}$$ or $$ -\sum_{n\geq 0}\int_{0}^{1}\frac{x^{2n-1/2}}{(2n+1)a^{2n+1}\sqrt{1-x}}\,dx=-\sum_{n\geq 0}\frac{\pi\binom{4n}{2n}}{(2n+1)16^n a^{2n+1}}\tag{2}$$ hence it is enough to recall the generating function for Catalan numbers to recover the wanted identity.

It is worth noticing that the Maclaurin series of the squared arcsine allows to compute the similar integral $$ \int_{0}^{1}\log\left(\frac{a^2+x^2}{a^2-x^2}\right)\frac{dx}{\color{red}{x}\sqrt{1-x^2}} $$ or the hypergeometric function $\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{4},\tfrac{5}{4},\tfrac{3}{2};z\right)$ in terms of $\arcsin^2$ and $\text{arcsinh}^2$, in a very similar and efficient way.

edited Jun 28 '18 at 23:11

answered Jun 28 '18 at 22:57

Jack D'Aurizio

353,855

Nice answer +1. Do you think you could do the integral by treating $a$ as a parameter and differentiating under the integral sign? – Rumplestillskin Jun 28 '18 at 23:21
1

@Rumplestillskin: of course, that is exactly what the OP did, but known GFs shorten the path to the solution really much. – Jack D'Aurizio Jun 28 '18 at 23:23
How do you mean? The OP used integration by parts? – Rumplestillskin Jun 28 '18 at 23:30
1

@Rumplestillskin: all right, I was inaccurate. What I really meant to say is that using Feynman's trick works, indeed, but it just produces very similar integrals to the ones managed by the OP. – Jack D'Aurizio Jun 28 '18 at 23:51

score 2 · Answer 2 · answered Jun 29 '18 at 14:42

Under $x\to\sin x$, one has $$I=\int_{0}^{1}\ln\left(\frac{a-x^2}{a+x^2}\right)\cdot\frac{\mathrm dx}{x^2\sqrt{1-x^2}}=\int_{0}^{\pi/2}\ln\left(\frac{a-\sin^2x}{a+\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}.$$ Let $$I(k)=\int_{0}^{\pi/2}\ln\left(\frac{a-k\sin^2x}{a+k\sin^2x}\right)\cdot\frac{\mathrm dx}{\sin^2x}$$ and then \begin{eqnarray} I'(k)&=&\int_{0}^{\pi/2}\left(-\frac1{a-k\sin^2x}-\frac1{a+k\sin^2x}\right)\mathrm dx\\ &=&-\frac{\pi}{2}\left(\frac1{\sqrt{a(a+k)}}+\frac1{\sqrt{a(a-k)}}\right). \end{eqnarray} So $$ I=I(1)=-\frac{\pi}{2\sqrt a}\int_0^1\left(\frac1{\sqrt{a+k}}-\frac1{\sqrt{a-k}}\right)\mathrm dk=-\frac{\pi}{\sqrt a}(\sqrt{a+1}-\sqrt{a-1}).$$

Short way to evaluate $\int_{0}^{1}\ln(\frac{a-x^2}{a+x^2})\cdot\frac{dx}{x^2\sqrt{1-x^2}}$

2 Answers2