TITLE: Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$
QUESTION [6 upvotes]: How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\rho^2=y^2+a^2\\
x=\rho\tan\theta\\
dx=\rho\sec^2\theta \, d\theta\\
x^2+\rho^2=\rho^2\sec^2\theta\\
\int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy=
\int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\
\int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy=
\int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\
\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$
Update:
$$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$

REPLY [3 votes]: Let $y=\sqrt{a^2+x^2} \tan \theta$,  $dy=\sqrt{a^2+x^2}~ \mbox{sec}^2 \theta~ d\theta$ Then
$$\int_{0}^{a} \int_{0}^{a} \frac{dx ~ dy} {(a^2+x^2+y^2)^{3/2}} =\int_{0}^{a}\frac{dx}{a^2+x^2}\int_{0}^{\tan^{-1}(a/\sqrt{a^2+x^2})} \cos \theta ~d\theta$$ $$ = a \int_{0}^{a} \frac{dx}{(x^2+a^2)\sqrt{(2a^2+x^2)}}.$$
$$ \Rightarrow I=\int_{0}^{a}\frac{1}{2i} \frac{dx}{\sqrt{2a^2+x^2}} \left [ \frac{1}{x-ia}-\frac{1}{x+ia}\right]=\Im \left( \int_{0}^{a}\frac{dx}{(x-ia)\sqrt{2a^2+x^2}}\right).$$
Use $(x-ia)=1/t$, then
$$I=-\Im \left (\int_{i/a}^{(1+i)/(2a)} \frac{dt}{\sqrt{a^2t^2+2iat+1}}\right) =-\Im  \int_{2i}^{(1+3i)/2} \left(\frac{dv}{a\sqrt{v^2+(\sqrt{2})^2}}\right).$$
Letting $v=\sqrt{2} \tan \phi$, we find that $$I=-\Im \left( a^{-1} \ln \left[ \frac{1+3i+\sqrt{6i}}{2i \sqrt{2}(1+\sqrt{2})}\right]\right)= \frac{\pi}{6a}.$$