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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 $\alpha=\frac xr$

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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.

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  • $\begingroup$ $r'^3$ is not a function of $\theta$ right? So this is just the integral of a constant + cosine. $\endgroup$ – Shashi Dec 26 '16 at 11:06
  • $\begingroup$ @Shashi Are you sure $r'$ is not a function of $\theta$? Look again at the picture $\endgroup$ – polfosol Dec 26 '16 at 11:07
  • $\begingroup$ Sorry my bad I misunderstood it $\endgroup$ – Shashi Dec 26 '16 at 11:12
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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,

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

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  • $\begingroup$ What expansion have you got? $\endgroup$ – Olivier Oloa Dec 26 '16 at 10:20
  • $\begingroup$ $$(1-\alpha\cos\theta)\left(\sum_{n=0}^\infty P_n(\cos\theta)\alpha^n\right)^3$$ $\endgroup$ – polfosol Dec 26 '16 at 10:29
  • $\begingroup$ @polfosol See my edit. $\endgroup$ – Olivier Oloa Dec 26 '16 at 10:54
  • $\begingroup$ No problem. Thanks for trying $\endgroup$ – polfosol Dec 26 '16 at 10:59
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    $\begingroup$ This is great :) But I didn't get the part where $\int$ turned into $\sum$. Also the link to Wallis integral does not work $\endgroup$ – polfosol Dec 26 '16 at 14:43

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