I am trying to solve a problem that involves evaluating the integral
$$ \int_0^{\infty} rdr \int_0^{\pi} \frac{a\sin{(\theta)}^3 d\theta}{(r^2 + a^2 – 2ar\cos{(\theta)})^{3/2}}$$
Mathematica computes the answer to be 2, which reproduces the final answer in the textbook. I have tried evaluating the integral by using trigonometric identities and then substituting $u = \cos{\theta}$ and have also tried evaluating the corresponding contour integral but had no success.
Could any one explain how one goes about evaluating the integral?
(If it's necessary, you can assume $a>0$)
Thank you.
Best Answer
I will provide an outline; you will need to fill in the details yourself. The substitution $$x = r^2 + a^2 - 2ar \cos \theta, \\ dx = 2ar \sin \theta \, d\theta, $$ yields
$$\begin{align} f_a(r) &= \int_{\theta=0}^\pi \frac{a \sin^3 \theta \, d\theta}{(r^2 + a^2 - 2ar \cos \theta)^{3/2}} \, d\theta = \frac{4}{3} \begin{cases}a/r^3, & a < r \\ 1/a^2, & a \ge r. \end{cases} \end{align}$$ Then $$\begin{align} \int_{r=0}^\infty r f_a(r) \, dr = 2. \end{align}$$