How to evaluate $\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{r'^3}d\theta$

Question

Let $P$ be a fixed point inside a circle with radius $r$ and $Q$ a non-fixed point on its perimeter as shown below, and let $\alpha=\frac xr$

Is there any closed-form solution for the following integral? $$\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{{r'}^{3}}d\theta$$ I tried Wolfram-Alpha without any luck. It doesn't look similar to any form of the integrals I have ever seen before. I appreciate if anybody gives it a try.

Answer 1

Hint. One may recall that $$ r'=\sqrt{r^2+x^2-2rx\cos \theta} $$ we are then looking for $$ \begin{align} &\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{{r'}^{3}}d\theta \\&=\frac1{r^3}\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{\left(1+\alpha^2-2\alpha\cos \theta\right)^{3/2}}d\theta \\&=\frac1{r^3}\frac1{(1+\alpha^2)^{3/2}}\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{\left(1-\frac{2\alpha}{1+\alpha^2}\cos \theta\right)^{3/2}}d\theta \\&=\frac1{r^3}\frac1{(1+\alpha^2)^{3/2}}\sum_{n=0}^\infty \frac{2\:\Gamma\big(n+\frac32\big)}{\Gamma\big(n+1\big)\Gamma\big(\frac12\big)}\cdot\int_0^{2\pi}\left(\cos^n{\theta}-\alpha\cos^{n+1}{\theta}\right)d\theta\cdot\left(\frac{2\alpha}{1+\alpha^2} \right)^n \end{align} $$ setting $a=\dfrac{2\alpha}{1+\alpha^2}$, $0<a<1$, one may then use Wallis' integral $$ \int_0^{2\pi}\cos^n{\theta}\:d\theta= 2\cdot\frac{\Gamma\big(\frac12\big)\Gamma(\tfrac{n+1}{2})}{\Gamma\big(\tfrac{n}{2}+1\big)} $$ to obtain $$ \begin{align} &\sum_{n=0}^\infty \frac{2\:\Gamma\big(n+\frac32\big)}{\Gamma\big(n+1\big)\Gamma\big(\frac12\big)}\cdot\int_0^{2\pi}\left(\cos^n{\theta}-\alpha\cos^{n+1}{\theta}\right)d\theta\cdot a^n \\\\&=\sum_{n=0}^\infty \frac{4\:\Gamma\big(n+\frac32\big)\Gamma(\tfrac{n+1}{2})}{\Gamma\big(n+1\big)\Gamma\big(\tfrac{n}{2}+1\big)} \cdot a^n-\sum_{n=0}^\infty \frac{4\:\Gamma\big(n+\frac52\big)\Gamma(\tfrac{n}{2}+1)}{\Gamma\big(n+2\big)\Gamma\big(\tfrac{n}{2}+\tfrac32\big)} \cdot a^n \\\\&=\frac{4\mathrm{E}\left(\frac{2 a}{1+a}\right)}{(1-a)\sqrt{1+a}}-\frac{4\alpha \mathrm{K}\left(\frac{2 a}{1+a}\right)}{a\sqrt{1+a}}+\frac{4\alpha\mathrm{E}\left(\frac{2 a}{1+a}\right)}{a(1-a)\sqrt{1+a}} \end{align} $$ where we have used elliptic integrals $\mathrm{E}(r):=\mathrm{E}(\tfrac\pi2,r)$ and $\mathrm{K}(r):=\mathrm{K}(\tfrac\pi2,r)$.

Finally,

$$ r^3(1+\alpha^2)^{3/2}\int_0^{2\pi}\frac{1-\alpha\cos{\theta}}{{r'}^{3}}d\theta=\frac{4\mathrm{E}\left(\frac{2 a}{1+a}\right)}{(1-a)\sqrt{1+a}}-\frac{4\alpha \mathrm{K}\left(\frac{2 a}{1+a}\right)}{a\sqrt{1+a}}+\frac{4\alpha\mathrm{E}\left(\frac{2 a}{1+a}\right)}{a(1-a)\sqrt{1+a}}. $$