Evaluate the integral: $$\int_0^{\pi} \frac{\cos 2\theta}{1 -2a\cos \theta +a^2}d\theta$$

The way I approach this problem is:

since, $\cos \theta = \frac{e^{it} + e^{-it}}{2}$; and $cos 2\theta = Re(z^2)$. Then, the integral will be written as follow: $$\frac{1}{2}\int_0^{2 \pi} \frac{Re(z^2)}{1 - 2a \left( \frac{z + z^{-1}}{2}\right) +a^2}dz$$ $$ = \frac{1}{2}\int_0^{2\pi} \frac{-Re(z^2)}{(z-z_1)(z-z_2)}dz$$ where: $$z_1 = \frac{(2a^2 - 2) + \sqrt{4(a^4+1)}}{4a}; \; z_2 = \frac{(2a^2 - 2) - \sqrt{4(a^4+1)}}{4a}$$ until here, I don't know how to use the Residue theorem to evaluate the integral. Can someone show me ?

  • 1
    $\begingroup$ First: you should pull the Re part outside the first integral. Second: do a case analysis on $a$ to determine where exactly the roots are. Third: your integration is no longer over $[0,2\pi]$ but over the unit circle with your change of variables to $z$. With those three things you can apply the residue theorem quite simply. $\endgroup$ – Cameron Williams Aug 9 '15 at 3:12
  • $\begingroup$ how can I pull out the real part outside the integral ? $\endgroup$ – Alexander Aug 9 '15 at 3:23

Notice, the following expression $$\frac{\cos 2\theta}{1-2a\cos \theta+a^2}=\frac{2\cos^2 \theta-1}{1-2a\cos \theta+a^2}=\frac{A\cos \theta(1-2a\cos \theta+a^2)+B(1-2a\cos \theta+a^2)+C}{1-2a\cos \theta+a^2}$$$$=A\cos \theta+B+\frac{C}{1-2a\cos \theta+a^2}$$

On solving for $A, B, C$, we get $$A=-\frac{1}{a}, \ B=-\frac{a^2+1}{2a^2}, \ C=\frac{a^4+1}{2a^2}$$ Now.we have $$\int_{0}^{\pi}\frac{\cos 2\theta}{1-2a\cos \theta+a^2}d\theta$$ $$=\int_{0}^{\pi}\frac{2\cos^2 \theta-1}{1-2a\cos \theta+a^2}d\theta$$

$$=\int_{0}^{\pi}\left(-\frac{1}{a}\cos\theta-\frac{a^2+1}{2a^2}+\frac{a^4+1}{2a^2}\frac{1}{1-2a\cos \theta+a^2}\right)d\theta$$ $$=-\frac{1}{a}\int_{0}^{\pi}\cos\theta d\theta-\frac{a^2+1}{2a^2}\int_{0}^{\pi} d\theta+\frac{a^4+1}{2a^2}\int_{0}^{\pi}\frac{1}{1-2a\cos \theta+a^2}d\theta$$

$$=0-\frac{(a^2+1)\pi}{2a^2}+\frac{a^4+1}{2a^2}\int_{0}^{\pi}\frac{1}{(a^2+1)-2a\cos \theta}d\theta$$

$$=-\frac{(a^2+1)\pi}{2a^2}+\frac{a^4+1}{4a^3}\int_{0}^{\pi}\frac{1}{\frac{(a^2+1)}{2a}-\cos \theta}d\theta$$

Now, set $\cos\theta=\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$

I hope you can take it from here.

  • $\begingroup$ Wow, it makes the problem easier to use residue theorem. $\endgroup$ – Alexander Aug 9 '15 at 5:44
  • $\begingroup$ Yes, you are right. $\endgroup$ – Harish Chandra Rajpoot Aug 9 '15 at 5:48

Another way is to use the following Fourier expansion: for $|a| < 1$,

$$ \frac{1-a^2}{1 - 2a\cos\theta + a^2} = \sum_{n=-\infty}^{\infty} a^{|n|} e^{in\theta} = 1 + 2 \sum_{n=1}^{\infty} a^n \cos(n\theta). $$

This series immediately gives us

$$ \int_{0}^{\pi} \frac{\cos n\theta}{1 - 2a\cos\theta + a^2} \, dx = \frac{\pi a^n}{1-a^2}, \quad n = 1, 2, \cdots. $$

  • $\begingroup$ I prefer using Residue theorem to evaluate the integral. $\endgroup$ – Alexander Aug 9 '15 at 5:42

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