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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?

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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

1 Answer 1

up vote 3 down vote accepted

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.

Now the question remains what you mean by "calculate", i.e. what kind of answer do you seek?

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+1. See here. –  Did Oct 12 '11 at 11:37
    
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
    
@Peter You misunderstand my point. There is no simple closed formula for the Bernoulli numbers, so you have to say what you want. Asymptotics? Recurrence? Double sum? You find all this in the wikipedia article on Bernoulli numbers. –  Phira Oct 12 '11 at 11:48
    
@Peter What is "the usual" way to calculate Bernoulli numbers? –  Phira Oct 12 '11 at 11:51

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