Computing Series with Residues

[jweygna1]

Homework Statement

It is a not a particular problem. Just how does one handle the situation of computing sums with residues if the residue comes out purely imaginary, as ultimately the sum must be real?

Homework Equations

sum(f(k), negative infinity, positive infinity)=-sum(Res(pi*cot(pi*z)f(z);p), for p poles of f)

The Attempt at a Solution

I guess for purpose of illustration I will show an example. Say we want to compute the sum S of f(k)=1/(k^2+1) from negative infinity to infinity. Then f(z)=1/(z^2+1)=((z+i)(z-i))^(-1) which has poles at i and -i. We construct g(z)=pi*cot(pi*z)*f(z) so S=-Res(g;i)-Res(g;-i).
Now all we have to do is find these two residues. But isn't Res(g;i)=Res(g;-i)=pi*coth(pi)/2i? This gives S=pi*coth(pi)/i which appears to be imaginary which doesn't make sense. The answer given is mine without the i in the denominator! What happened to it??

 

[lippyka]

The answer lies in the fact that coth(pi) is an even function, so coth(pi)=-coth(-pi). Therefore, Res(g;i)=pi*coth(pi)/2i=-Res(g;-i), and thus S=-Res(g;i)-(-Res(g;-i))=pi*coth(pi)/2i+pi*coth(pi)/2i=pi*coth(pi)/i which is real.