How do I prove that $\int_0^1 \int_0^1 {{2+x^2-y^2}\over {2-x^2-y^2}} \, dx \, dy=2G?$

Question

Let $G$ the Catalan's constant, prove that

$$\int_0^1 \int_0^1 {{2+x^2-y^2}\over {2-x^2-y^2}} \, dx \, dy=2G$$

I not sure what do, anyway I try to integrate with respect to x first

$$\int_0^1 {2-y^2+x^2\over 2-y^2-x^2}\,dx$$

Make a substitution $x^2=2+y^2$

$$-\int_0^1 {2\over xy} \, dy$$

$$-\int_0^1 {2\over y\sqrt{2+y^2}} \, dy$$

I can't find a standard integral for this, some help please.

Answer 1

First we can observe, by symmetry, that $$\int_{0}^{1}\int_{0}^{1}\frac{x^{2}-y^{2}}{2-x^{2}-y^{2}}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{x^{2}}{2-x^{2}-y^{2}}dxdy-\int_{0}^{1}\int_{0}^{1}\frac{y^{2}}{2-x^{2}-y^{2}}dxdy=0 $$ hence $$I=\int_{0}^{1}\int_{0}^{1}\frac{2+x^{2}-y^{2}}{2-x^{2}-y^{2}}dxdy=2\int_{0}^{1}\int_{0}^{1}\frac{1}{2-x^{2}-y^{2}}dxdy. $$ Now observe, again by symmetry, that the contributions with $x\leq y $ and $y\leq x $ are the same, so $$I=4\int_{0}^{1}\int_{0}^{x}\frac{1}{2-x^{2}-y^{2}}dydx. $$ Now taking the polar coordinates we have $$I=2\int_{0}^{\pi/4}\int_{0}^{\sec\left(\theta\right)}\frac{2\rho}{2-\rho^{2}}d\rho d\theta $$ $$=2\int_{0}^{\pi/4}\log\left(\frac{2}{2-\sec^{2}\left(\theta\right)}\right)d\theta $$ but $$\frac{2}{2-\sec^{2}\left(\theta\right)}=\frac{\tan\left(2\theta\right)}{\tan\left(\theta\right)} $$ hence $$I=2\int_{0}^{\pi/4}\log\left(\tan\left(2\theta\right)\right)d\theta-2\int_{0}^{\pi/4}\log\left(\tan\left(\theta\right)\right)d\theta $$ and now it is sufficient to note that $$2\int_{0}^{\pi/4}\log\left(\tan\left(2\theta\right)\right)d\theta=\int_{0}^{\pi/2}\log\left(\tan\left(u\right)\right)du=0 $$ since $$\int_{0}^{\pi/2}\log\left(\sin\left(u\right)\right)du=\int_{0}^{\pi/2}\log\left(\sin\left(\frac{\pi}{2}-u\right)\right)du=\int_{0}^{\pi/2}\log\left(\cos\left(u\right)\right)du $$ (and, for personal culture, $\int_{0}^{\pi/2}\log\left(\sin\left(u\right)\right)du=-\frac{\pi}{2}\log\left(2\right) $) so we have done $$I=-2\int_{0}^{\pi/4}\log\left(\tan\left(\theta\right)\right)d\theta=\color{red}{2G}$$ as wanted.

Answer 2

Noticing that $$ \begin{aligned} \quad \int_0^1 \int_0^1 \frac{x^2}{2-x^2-y^2} d x d y=\int_0^1 \int_0^1 \frac{y^2}{2-x^2-y^2} d x d y ,\\ \end{aligned} $$ we have \begin{aligned}I&=\int_0^1 \int_0^1 \frac{2+x^2-y^2}{2-x^2-y^2} d x d y \\&=2 \int_0^1 \int_0^1 \frac{1}{2-x^2-y^2} d x d y\\&= -2 \int_0^1 \int_0^1 \frac{1}{x^2+y^2-2} d xd y \\&=-4 \int_0^1 \int_0^y \frac{1}{x^2+y^2-2} d xd y\end{aligned}

Letting $x=ty$ transforms the integral into $$ \begin{aligned} I&= -4 \int_0^1 \int_0^1 \frac{y}{(1+t^2) y^2-2} d td y \\ &= -2\int_0^1 \int_0^1 \frac{2y}{(1+t^2) y^2-2} d yd t \\ & =-2 \int_0^1\left[\frac{\ln \left|\left(1+t^2\right) y^2-2\right|}{1+t^2}\right]_0^1 d t \\ & =-2 \int_0^1 \frac{\ln \left|t^2-1\right|-\ln 2}{1+t^2} d t \\ & =-2\left[ \underbrace{\int_0^1 \frac{\ln \left(1-t^2\right)}{1+t^2} d t}_{J}-\ln 2 \underbrace{ \int_0^1 \frac{d t}{1+t^2}}_{=\frac\pi4} \right] \end{aligned} $$ For the integral $J$, let $t=\tan \theta$, then $$ \begin{aligned} J & =\int_0^{\frac{\pi}{4}} \ln \left(1-\tan ^2 \theta\right) d \theta \\ & =\int_0^{\frac{\pi}{4}} \ln (\cos 2 \theta) d \theta-2 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta \\ & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \ln (\cos \theta) d \theta-2\left(\frac{G}{2}-\frac{\pi}{4} \ln 2\right) \\ & =-\frac{\pi}{4} \ln 2-G+\frac{\pi}{2} \ln 2 \\ & =\frac{\pi}{4} \ln 2-G \end{aligned} $$ Now we can conclude that $$ \begin{aligned} I & =-2\left(\frac{\pi}{4} \ln 2-G-\frac{\pi}{4} \ln 2\right) =2 G \end{aligned} $$ where $G$ is the Catalan’s constant.