¿Cómo puedo probar que$\int_0^1 \int_0^1 {{2+x^2-y^2}\over {2-x^2-y^2}} \, dx \, dy=2G?$

Question

Vamos a$G$ la constante catalana, demostrar que

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No sé qué hacer, de todos modos trato de integrar con respecto a x primero

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

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No puedo encontrar una integral estándar para esto, alguna ayuda por favor.

Answer 1

Primero que podemos observar, por simetría, que $$\int_{0}^{1}\int_{0}^{1}\frac{x^{2}-y^{2}}{2-x^{2}-y^{2}}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{x^{2}}{2-x^{2}-y^{2}}dxdy-\int_{0}^{1}\int_{0}^{1}\frac{y^{2}}{2-x^{2}-y^{2}}dxdy=0 $$ 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 $s\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 $$I=2\int_{0}^{\pi/4}\int_{0}^{\sec\left(\theta\right)}\frac{2\rho}{2-\rho^{2}}d\rho d\theta $$ $$=2\int_{0}^{\pi/4}\log\left(\frac{2}{2-\s^{2}\left(\theta\right)}\right)d\theta $$ 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 $$I=-2\int_{0}^{\pi/4}\log\left(\tan\left(\theta\right)\right)d\theta=\color{red}{2G}$$ como quería.