# Hard problem

1. Oct 8, 2006

### 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:

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

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!

Last edited: Oct 8, 2006
2. Oct 9, 2006

### 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.

Last edited by a moderator: Oct 9, 2006
3. Oct 9, 2006

### romeo6

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

Ok, so I just used the rule:

$$Res=\frac{g(z_0))}{h(z_0)'}$$

With :

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

and

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

and so

$$g'(z)=cos(\pi z)$$

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!