# How do I evaluate the Complex Integral $(z^{-n})/(e^z−1)$ using residue theory?

Over here we had a similar looking question:

How do I evaluate the Complex Integral $(z^n)/(e^z - 1)$ using residue theory?

and the answer was that (rather unremarkably) we could remove the singularity made by $1/(e^z-1)$ for $z=0$ and conclude Res$((z^n)/(e^z-1))=0$ for all $n>0$ and Res$(1/(e^z-1))=1$.

I think this residue calculation becomes rather more interesting when we put a $-$ in front of the $n$. Then we have an (n+1)th order pole at $z=0$. Can we calculate the residue other than with this limit formula?

-
Whats wrong with this limit formula? – Alexander Thumm Oct 12 '11 at 11:04
I quote the wikipedia link: "For higher order poles, the calculations can become unmanageable, and series expansion is usually easier.". For large $n$ you will have to take high order derivatives of $z/(e^z-1)$ - and you can imagine that this has more and more terms because of the quotient rule... – Peter Sheldrick Oct 12 '11 at 11:12

The coefficients of the power series $z/(e^z-1)$ are by definition the Bernoulli numbers (divided by a factorial), so they are the residues of your function.
Well for example in the book Basic Complex Analysis by Marsden i saw some formulas for the residue of a function $g/f$ where both $g$ and $f$ were holomorphic and $g/f$ had a pole of degree (2) - i could apply this formula, degree (3) formula failed for me or degree (n) but that formula was very complex, involving the determinant of a big matrix - so i couldnt apply that. Those formulas where derived from expanding $g$ and $f$ as laurent series. Shouldn't it be easier in this specific case? – Peter Sheldrick Oct 12 '11 at 11:45