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Is there a geometric way of seeing why the integral $\int\limits_{-\infty}^\infty (x^2+y^2+z^2)^{-{3\over 2}}dz={2\over x^2+y^2}$? Otherwise what is a good way of evaluating it algebraically?

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That the result is $c/(x^2+y^2)$ for some positive $c$ follows from the change of variable $z=u\sqrt{x^2+y^2}$. To evaluate $c$, a useful change of variable is $u=\tan t$. – Did Feb 11 '12 at 19:32
@DidierPiau: How on earth did you spot that? :) – vec Feb 11 '12 at 19:46
up vote 0 down vote accepted

Since we are integrating over $z$ let's set $a^2=x^2+y^2$ and search this indefinite integral :

$$\int (a^2+z^2)^{-{3\over 2}}dz=-\frac 1a\frac d{da}\left[\int \frac 1{\sqrt{a^2+z^2}} dz\right]$$ $$=-\frac 1a\frac d{da}\left[\arg\sinh\left(\frac za\right)\right]=\frac z{a^2\sqrt{z^2+a^2}}$$

The definite integral from $-\infty$ to $\infty$ will be $$\left[\frac z{a^2\sqrt{z^2+a^2}}\right]_{-\infty}^{\infty}=\frac2{a^2}=\frac2{x^2+y^2}$$

Not very geometrical, hoping it helped anyway,

EDIT: another derivation for the indefinite integral is provided by WolframAlpha (use 'Show steps') with a nice illustration.

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Thanks, Raymond! – vec Feb 11 '12 at 21:01
The show steps from WA is exactly what DidierPiau suggested doing in the comment to the original question. – Sasha Feb 11 '12 at 22:09

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