Complex analysis - partial fraction expansion

[yy205001]

Homework Statement

Show that:
Ʃ(-1)ⁿ/(n^2+a^2) (from n=0 to ∞) = pi/[asinh(pi*a)], a\neq in, n\in Z.

Homework Equations

f(z) = f(0) + Ʃb_n(1/(z-a_n)+1/a_n) (from n=1 to ∞) , where b_n is the residue of f(z) at a_n.

The Attempt at a Solution

The main problem is I don't how to pick the function f(z) to start with. I am thinking is pi*cot(pi*z) a good starting point? But in this case, we are dealing with the hyperbolic function, so coth will be a better choice??

Any help is appreciated!

 

[vanhees71]

First of all the sum is over n \in \mathbb{Z}, and the correct statement is
\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+a^2}=\frac{\pi}{a \sinh(a \pi)}.
Now find a function with residua (-1)^n at z=n for n \in \mathbb{Z}.

You are close with the choice of your function, but the one you gave has residua +1. Multiply it with an entire function which takes the values (-1)^n at z=n \in \mathbb{Z}. Then choose an appropriate contour to integrate over and finally deform it appropriately to use the residue theorem.

 

[yy205001]

From my notes, i found that Ʃ(-1)ⁿf(n) (from -∞ to ∞) = - Ʃ Res(pi*csc(pi*z), z_j) (poles z_j of f(z)).

So, should I let f(n) = 1/(n²+a²) in this case?

 

[brmath]

First of all the sum is over n \in \mathbb{Z}, and the correct statement is
\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+a^2}=\frac{\pi}{a \sinh(a \pi)}.
Now find a function with residua (-1)^n at z=n for n \in \mathbb{Z}.

You are close with the choice of your function, but the one you gave has residua +1. Multiply it with an entire function which takes the values (-1)^n at z=n \in \mathbb{Z}. Then choose an appropriate contour to integrate over and finally deform it appropriately to use the residue theorem.

I've been trying myself to sort out this Laurent series stuff, but I'm pretty sure I'm not following this. Would you mind answering some questions?

Are we trying to find a function g(z) to expand in a Laurent series which will match the summation at z = n?

I understand that any appropriate g would have simple poles at ±ai, so would have just the $a_{-1}$ term in its Laurent series. But why would this residue necessarily show up in the other terms of the Laurent series?

Does this Laurent series converge to g everywhere? Somehow I'm thinking that the series around ai is good only down to -ai -- that is converges in a radius of 2a around the point +ai.

Or am I completely on the wrong track here?

If you can explain, I would very much appreciate it.