Integrating Around a Circle: Proving an Infinite Series Equation

[romeo6]

Here is my problem:

I need to integrate:

(\frac{sin \alpha z}{\alpha z})^2\frac{\pi}{sin\pi z}

around a circle of large radii and prove:

\sum_{m=1}^\infty(-1)^{n-1} <br /> <br /> <br /> (\frac{sin m\alpha}{m\alpha})^2<br /> <br /> <br /> =\frac{1}{2}

I'm kind stumped.

I've been looking at books for a while now and the only useful things I've discovered are:

\frac{\pi}{sin \pi z}=\Gamma(z)\Gamma(1-z)

and also

\frac{sin z}{z}=\sum_{n=0}^\infty C_n z^{2n-1}

Can anyone help me out?

Thanks!

 

[HallsofIvy]

Well,
(\frac{sin \alpha z}{\alpha z})^2\frac{\pi}{sin\pi z}
has poles at every integer, since sine will be 0 there.

The integral will be an infinite sum of residues.

 

[romeo6]

Thanks Halls, I just got all pleased with myself then realized I've hit another snag:

Ok, so I just used the rule:

Res=\frac{g(z_0))}{h(z_0)&#039;}

With :

g(z)=(\frac{sin(\alpha z)}{\alpha z})^2

and

g(z)=\frac{1}{\pi}sin(\pi z)

and so

g&#039;(z)=cos(\pi z)

Clearly there are zeros for m=0,1,2...and so the when z\rightarrow z_0

we end up with the following residues:

\sum_{m=1}^\infty(-1)^{n-1} (\frac{sin m\alpha}{m\alpha})^2

Now, the integral is 2\pi i \sum Res

Can you help me figure out how to get:

\sum_{m=1}^\infty(-1)^{n-1} (\frac{sin m\alpha}{m\alpha})^2=\frac{1}{2}

Essentially I need to figure how to get rid of the 'i' also figure out what the integral goes to...

Thanks!