# Evaluate $\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$

Evaluate by complex methods

$$\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$$

Sis.

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–  Mhenni Benghorbal Feb 26 '13 at 22:15
@ Mhenni Benghorbal: thanks! Your links are always helpful! :-) –  Chris's sis Feb 26 '13 at 22:36

This integral is $1/2$ the integral over $[0,2 \pi)$. Let $z=e^{i \theta}$, $d\theta = dz/(i z)$; the result is

$$\frac{1}{2 i}\oint_{|z|=1} \frac{dz}{z} \frac{-\frac{1}{4} (z^2-1)^2}{(a z^2-(1+a^2)z+a)(b z^2-(1+b^2)z+b)}$$

which can be rewritten as

$$\frac{i}{8} \oint_{|z|=1} \frac{dz}{z} \frac{(z^2-1)^2}{(a z-1)(z-a)(b z-1)(z-b)}$$

There are 5 poles, although because $0<a<b<1$, only 3 of them fall within the contour. This integral is then $i 2 \pi$ times the sum of the residues of these poles. The residues of these poles are actually straightforward:

$$\mathrm{Res}_{z=0}\frac{i}{8} \frac{(z^2-1)^2}{z(a z-1)(z-a)(b z-1)(z-b)} = \frac{i}{8 a b}$$ $$\mathrm{Res}_{z=a}\frac{i}{8} \frac{(z^2-1)^2}{z(a z-1)(z-a)(b z-1)(z-b)} = \frac{i}{8 a} \frac{a^2-1}{(a b-1)(a-b)}$$ $$\mathrm{Res}_{z=b}\frac{i}{8} \frac{(z^2-1)^2}{z(a z-1)(z-a)(b z-1)(z-b)} = -\frac{i}{8 b} \frac{b^2-1}{(a b-1)(a-b)}$$

There is vast simplification from adding these pieces together, which I leave to the reader. The result is

$$\int_0^{\pi} d\theta \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)} = \frac{\pi}{2} \frac{1}{1-a b}$$

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Thank you! Clearly explained. (+1) Actually, I did some integrals by going this way, but they were much simpler. Now, I realize that I should have made that substitution here. –  Chris's sis Feb 13 '13 at 18:00
@Chris'ssisterandpals: any rational function of sines and cosines can be done this way, in principle. It just so happened that the denominator was easily factorizable. –  Ron Gordon Feb 13 '13 at 18:22
Actually, not unexpected if you did the simpler one before, with just $a$. You can actually keep going with $c$, $d$, as many as you like; the factorization is just as easy, although adding up the residues may get tiresome. And, yes, complex analysis is the most beautiful branch of math IMO. –  Ron Gordon Feb 13 '13 at 18:27
Yes, I begin to think it is since things might be simplified a lot. –  Chris's sis Feb 13 '13 at 18:39

It can be done easily without complex, if we note that $$\frac{\sin x}{1-2a\cos x+a^2}=\sum_{n=0}^{+\infty}a^n\sin[(n+1)x]$$

Just saying.

EDIT: for proving this formula, we actually use complex method

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Interesting formula. (+1) –  Chris's sis Feb 13 '13 at 22:01
Did you mean $\sin{[(n+1) x]}$? –  Ron Gordon Feb 14 '13 at 1:12
@rlgordonma Yes. Thanks. –  Cortizol Feb 14 '13 at 9:23
@Cortizol: Yes, I see where this comes from and how you would accomplish the integration here, and it checks with my answer. (+1) But...can this be easily extended to the cases I described in the comments above, with 3 such products in the denominator, or , or...? –  Ron Gordon Feb 14 '13 at 10:14
@rlgordonma $$\sum_{n=0}^{+\infty}a^n\sin[(n+1)x]=\sum_{n=0}^{+\infty} \Im\left(a^n e^{i(n+1)x}\right)=\Im \left(\frac{e^{ix}}{1-ae^{ix}}\right),$$ and $$\Im(z) = \frac{1}{2i} (z-\bar{z}),$$ so $$\Im \left(\frac{e^{ix}}{1-ae^{ix}}\right)=\frac{1}{2i}\left(\frac{e^{ix}}{1-ae^{ix}}‌​-\frac{e^{-ix}}{1-ae^{-ix}}\right)=\frac{\sin x}{1-2a\cos x+a^2}$$ –  Cortizol Feb 15 '13 at 16:42

Using the series presented by Cortizol, the integral can be written as: $$\int_0^{\pi} dx\left(\sum_{n=1}^{\infty} a^{n-1}\sin (nx)\right)\left(\sum_{m=1}^{\infty} b^{m-1}\sin (mx)\right)=\frac{1}{ab}\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a^nb^m\int_0^{\pi} \sin(nx)\sin(mx)\,dx$$ Notice that for $n\ne m$, the integral is always zero, hence we look at only those cases where $n=m$ i.e $$\frac{1}{ab}\sum_{n=1}^{\infty} (ab)^n\int_0^{\pi} \sin^2(nx)\,dx=\frac{\pi}{2ab}\sum_{n=1}^{\infty} (ab)^n=\boxed{\dfrac{\pi}{2}\dfrac{1}{1-ab}}$$ $\blacksquare$

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I see you're quick to apply what you learn!(+1) And good job tracking down an example already on the site. I'm sure there are proper references somewhere. –  David H Jun 6 at 7:35
@DavidH: Thanks for the upvote. :) I am looking for references, this series must be somewhere available in Gradshteyn and Ryzhik Tables. –  Pranav Arora Jun 6 at 7:38