# Complex Integral Issue

1. Mar 2, 2012

### Airsteve0

1. The problem statement, all variables and given/known data

Consider the integral of the function (1) around a large circle of radius R>>b which avoids the singularities of ($e^{z}$+1)$^{-1}$. Use this result to determine the sum (2) and (3).

2. Relevant equations

(1) - f(z) = $\frac{1}{(z^2-b^2)(e^z+1)}$
(2) - $\sum$$\frac{1}{(2n+1)^2+a^2}$ from 1 to ∞
(3) - $\sum$$\frac{1}{(2n+1)^2}$ from 1 to ∞

3. The attempt at a solution

I understand the basics of contour integration and using residues to evaluate them; however, with this particular question I am lost at where I should start. I think that the contour should be a circle obvioulsy but going from the contour to the summation has me confused.

2. Mar 2, 2012

### Dick

The idea is that your function has poles at z=(2n+1)*pi*i for all n and at z=b and z=(-b). For large circle which avoids those singularities the contour integral will equal the sum of the residues of those poles times (2*pi*i). Now if you can argue that as R->infinity the contour integral goes to zero, then the infinite sum of (2n+1)*pi*i residues will be equal to minus the sum of the residues from the b and -b poles. That's how you get a formula for an infinite sum.

3. Mar 2, 2012

### Airsteve0

why is it that the the sum of the residues of one set of poles is equal to minus the sum of the other?

4. Mar 2, 2012

### Dick

Because if the value of contour integral goes to zero as R->infinity then the sum of all of the residues must go to zero. If you split them into two sets then the sum of those two sets must be zero.

5. Mar 2, 2012

### Airsteve0

ah ok, thanks I will see what I can do

6. Mar 2, 2012

### Dick

Good luck! The residue part is actually not that hard. Do that first. Be careful with the argument that you can find contours whose integral goes to 0 as R goes to infinity. I thought that was the hardest part.