Solving Improper Rational Trig Integral

Question

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.

Answer 1

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 $$\begin{align} \int_{r=0}^\infty r f_a(r) \, dr = 2. \end{align}$$

Answer 2

$$I(a)=\int_0^{\infty} rdr \int_0^{\pi} \frac{a\sin{(\theta)}^3 d\theta}{(r^2 + a^2 - 2ar\cos{(\theta)})^{3/2}}\overset{\cos\theta=x}{=}\int_0^{\infty} rdr \int_{-1}^1 \frac{a(1-x^2) dx}{(r^2 + a^2 - 2arx)^{3/2}}$$ $$=\int_0^ardr\int_{-1}^1 \frac{a(1-x^2) dx}{(r^2 + a^2 - 2arx)^{3/2}}+\int_a^\infty rdr\int_{-1}^1 \frac{a(1-x^2) dx}{(r^2 + a^2 - 2arx)^{3/2}}=I_{1{(r<a)}}+I_{2{(r>a)}}$$ $$I_1=\frac a{a^3}\int_0^ardr\int_{-1}^1\frac{(1-x^2)dx}{\left((\frac ra)^2+1-2\frac rax\right)^{3/2}}\overset{t=\frac ra}{=}\int_0^1dt\int_{-1}^1(1-x^2)\left(\frac d{dx}\frac1{\sqrt{t^2+1-2tx}}\right)dx$$ $$\overset{IBP}{=}\int_0^1dt\left(\frac{1-x^2}{\sqrt{t^2+1-2tx}}\,\bigg|_{x=-1}^1\right)+2\int_0^1dt\int_{-1}^1\frac{xdx}{\sqrt{t^2-2tx+t^2}}$$ But $\displaystyle \frac1{\sqrt{t^2+1-2tx}}$ is the generating function for Legendre polynomials: $$ \frac1{\sqrt{t^2+1-2tx}}=\sum_{n=0}^\infty P_n(x)t^n,\,\,|t|<1$$ Legendre polynomials are orthogonal, and $P_1(x)=x$. Hence, at $r<a$ $$I_1=2\int_0^1dt\int_{-1}^1P_1(x)\cdot P_1(x)\,t\,dx=2\int_0^1tdt\int_{-1}^1x^2dx=\frac23\tag{1}$$ In the same way, for the case $r>a$ $$I_2=a\int_a^\infty \frac{rdr}{r^3}\int_{-1}^1\frac{(1-x^2)dx}{\left((\frac ar)^2+1-2\frac arx\right)^{3/2}}\overset{t=\frac ar}{=}\int_0^1dt\int_{-1}^1\frac{(1-x^2)dx}{(t^2+1-2tx)^{3/2}}$$ $$=\int_0^1\frac{dt}t\int_{-1}^1(1-x^2)\left(\frac d{dx}\frac1{\sqrt{t^2+1-2tx}}\right)dx$$ $$\overset{IBP}{=}2\int_0^1\frac{dt}t\int_{-1}^1\frac{xdx}{\sqrt{t^2+1-2tx}}=2\int_0^1dt\int_{-1}^1x^2dx=\frac43\tag{2}$$ $$I(a)=I_1+I_2=2$$ the result obtained by @heropup .

Answer 3

$\def\th{\theta}$\begin{align*} \int_0^{\infty} rdr \int_0^{\pi} \frac{a\sin^3\th \,d\th}{(r^2 + a^2 - 2ar\cos\th)^{3/2}} &= \int_0^{\pi} a\sin^3\th \int_0^{\infty} \frac{r}{(r^2 + a^2 - 2ar\cos\th)^{3/2}} dr\,d\th & \textrm{can justify} \\ &= \int_0^{\pi} a\sin^3\th \int_0^{\infty} \frac{r}{(r-a\cos\th)^2 + a^2\sin^2\th)^{3/2}} dr\,d\th \\ &= \int_0^{\pi} a\sin^3\th \int_{-a\cos\th}^{\infty} \frac{u+a\cos\th}{(u^2 + a^2\sin^2\th)^{3/2}} du\,d\th & u = r-a\cos\th \\ &= \ldots & \textrm{standard trig sub} \\ &= \int_0^\pi \left.\left( \frac{\sin\th(u\cos\th-a\sin^2\th)}{\sqrt{u^2+a^2\sin^2\th}} \right)\right|_{-a\cos\th}^\infty d\th \\ &= \int_0^\pi (1+\cos\th)\sin\th\,d \th \\ &= 2 \end{align*}