3
$\begingroup$

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?

  • 1
    @DidierPiau: How on earth did you spot that? :)2012-02-11

1 Answers 1

0

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.

  • 0
    The show steps from WA is exactly what DidierPiau suggested doing in the comment to the original question.2012-02-11