My friend gave me this problem, and I'm not too sure where to start. I know single integrals relatively well, but I'm pretty new to double integrals. If someone could show a step-by-step solution, that would be great.
$$\int_{-\infty}^{\infty}\int_{-\infty}^\infty \frac 1 {\cosh(x)+\cosh(y)} \, dy \, dx$$
I've tried hyperbolic tangent half-angle substitution and normal substitution, but I have no idea if that applies to double integrals as well.
$$\int_{-\infty}^{+\infty}\frac{dx}{\cosh x}=2\int_{0}^{+\infty}\frac{2\,dx}{e^{x}+e^{-x}}=4\int_{1}^{+\infty}\frac{du}{1+u^2}=\pi\tag{A} $$ $$\int_{\mathbb{R}}\frac{dz}{\cosh(z\sqrt{2})}=\frac{\pi}{\sqrt{2}}\tag{B}$$ hence $$ \iint_{\mathbb{R}^2}\frac{dx\,dy}{\cosh(x)+\cosh(y)}=\iint_{\mathbb{R}^2}\frac{dx\,dy}{2\cosh\left(\frac{x+y}{2}\right)\cosh\left(\frac{x-y}{2}\right)} =\iint_{\mathbb{R}^2}\frac{du\,dv}{2\cosh(u\sqrt{2})\cosh(v\sqrt{2})}$$ simply equals $\color{red}{\large \pi^2}$.