Complex Integral

**canberra1454** · December 14th 2009, 03:33 PM

Show that

$\frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{1}{1-2r\cos \theta +r^2} d \theta = 1, 0 \leq r < 1$ .

Right now, I don't see how to show this. I do see that the integrand is an even function if that helps at all. Also, since this is not from $-\infty$ to $\infty$ , I do not know the method to proceed. I need a few hints on how to start.

**shawsend** · December 14th 2009, 05:12 PM

Hi. If I use the $z=e^{it}$ substitution, then I get:

$\int_{-\pi}^{\pi}\frac{1}{1-2 r\cos(t)+r^2}dt=-i\oint \frac{dz}{(r-z)(rz-1)}=\frac{2\pi}{1-r^2},\quad |r|<1$

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