Evaluate $\int_0^\infty\frac{dl}{(r^2+l^2)^{\frac32}}$

Question

How to evaluate the following integral $$\int_0^\infty\frac{dl}{(r^2+l^2)^{\large\frac32}}$$

The solution is supposed to look like this, unfortunately I can't derive it. $$ \left[\frac{l}{r^2\sqrt{r^2+l^2}}\right]_{l=0}^\infty $$

Answer 1

Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes $$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$

Answer 2

$$ \int\frac{dl}{\left(r^2+l^2\right)^{3/2}} = -\frac{1}{r}\frac{d}{dr}\int\frac{1}{\left(r^2+l^2\right)^{1/2}}dl $$ change of variables $v=l/r$ we find

$$ -\frac{1}{r}\frac{d}{dr}\frac{1}{r}\int^{\infty}_{0}\frac{1}{\left(1+v^2\right)^{1/2}}rdv = -\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0} $$ $$ -\frac{1}{r}\frac{d}{dr}\left[\sinh^{-1}v\right]^{\infty}_{0} = -\frac{1}{r}\left[\frac{1}{\sqrt{1+v^2}}\frac{dv}{dr}\right]^{\infty}_{0} = -\frac{1}{r}\left[\frac{1}{\sqrt{1+\left(\frac{l}{r}\right)^2}}\left(\frac{-l}{r^2}\right)\right]^{\infty}_{0} $$ which leads to your result.

Answer 3

All integrals of the form $\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx$ can be expressed in terms of trigonometric functions

as follows: first, let $x=at$, then $u=\dfrac1{t^n+1}$. Now recognize the expression of the beta function in

the new integral, and finally use Euler's reflection formula for the $\Gamma$ function in order to arrive at

the desired result.