|
Evaluate: $$ \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - (xy)^2}\, dx \, dy $$ Can someone provide a hint to tackle this problem? I only know how to do substitution and integration by parts for one variable. My difficulty so far is to express the term $ \dfrac{1}{1 - (xy)^2} $ in term of $ x $ treating $ y $ as a constant so that I can use substitution. |
|||||
|
|
$$\iint_{(0,1)^2}\frac{dx\,dy}{1-(xy)^2}=\iint_{(0,1)^2}\sum_{n\geq 0}x^{2n}y^{2n}\,dx\,dy =\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\color{red}{\frac{\pi^2}{8}}.$$ |
|||||||||||||||||
|
|
A much less elegant solution. Consider $$I=\int\frac{dx}{1 - (xy)^2} $$ and change variable $$xy=t\implies x=\frac ty\implies dx=\frac{dt}{y}$$ This make $$I=\frac 1y\int\frac{dt}{1 - t^2}= \frac {\tanh ^{-1}(t)}y=\frac {\tanh ^{-1}(xy)}y\implies \int_0^1\frac{dx}{1 - (xy)^2}=\frac {\tanh ^{-1}(y)}y$$ Now, using $$\tanh ^{-1}(y)=\sum_{n=0}^\infty \frac {y^{2n+1}}{2n+1}\implies \frac{\tanh ^{-1}(y)}y=\sum_{n=0}^\infty \frac {y^{2n}}{2n+1}$$ which makes $$J=\int \frac{\tanh ^{-1}(y)}y \,dy=\sum_{n=0}^\infty \frac {y^{2n+1}}{(2n+1)^2}$$ $$\int_0^1 \frac{\tanh ^{-1}(y)}y \,dy=\sum_{n=0}^\infty \frac {1}{(2n+1)^2}=\frac{\pi ^2}{8}$$ |
|||
|
|
