Help on an integral I can't figure out how to compute the following integral:
$$\int_{-\infty}^{+\infty}\frac{\mathrm{d}z}{\sqrt{(x^2+y^2+z^2)^3}}$$
I think I should do some substitution, but I didn't figure it out. Can you please give me a hint?
According to Mathematica the result should be
$$\left[\frac{z}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}\right]_{-\infty}^{+\infty}=\frac{2}{x^2+y^2}$$
Thank you very much for your effort.
 A: Subbing $z=\sqrt{x^2+y^2} \tan{t}$, the integral becomes
$$\frac1{x^2+y^2}\int_{-\pi/2}^{\pi/2} dt \, \frac{\sec^2{t}}{\sec^3{t}} = \frac1{x^2+y^2}\int_{-\pi/2}^{\pi/2} dt \,\cos{t} = \frac{2}{x^2+y^2}$$
A: Let $z=\sqrt{x^2+y^2}\tan\theta$, then $dz=\sqrt{x^2+y^2}\sec^2\theta\ d\theta$.
$$
\begin{align}
\int_{-\infty}^{+\infty}\frac{dz}{\sqrt{(x^2+y^2+z^2)^3}}&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta\ d\theta}{\sqrt{((x^2+y^2)+(x^2+y^2)\tan^2\theta)^3}}\\
&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta}{\sqrt{(x^2+y^2)^3}\sqrt{(1+\tan^2\theta)^3}}d\theta\\
&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sec^2\theta\ }{\sqrt{(x^2+y^2)^2}\sec^3\theta}d\theta\\
&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\cos\theta\ }{x^2+y^2}d\theta\\
&=\left.\frac{1}{x^2+y^2}\sin\theta\right|_{\Large-\frac\pi2}^{\Large\frac\pi2}\\
&=\frac{2}{x^2+y^2}.
\end{align}
$$
Since $\tan\theta=\dfrac{z}{\sqrt{x^2+y^2}}$, then $\sin\theta=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$. Hence
$$
\left.\frac{1}{x^2+y^2}\sin\theta\right|_{\Large-\frac\pi2}^{\Large\frac\pi2}=\left.\frac{1}{x^2+y^2}\cdot\dfrac{z}{\sqrt{x^2+y^2+z^2}}\right|_{-\infty}^{\infty}=\frac{2}{x^2+y^2}.
$$
