I was really hoping someone could help me in calculating this contour integral (Schlaefli): $$ B_{n}= \frac{n!}{2\pi i}\oint_{C} \frac{1 }{z^{n+1}} \frac{z}{e^{ z} – 1}\,dz$$
My textbook (Arfken) quickly goes over some details and later on requests to find residues: \begin{align*}
\mathrm{Res}_{z =\pm 2 \pi n i} \left ( \frac{1 }{z^{n}} \frac{1}{e^{ z} – 1}\ \right ) &= ?
\end{align*}
and I just get lost. In addition, it uses this contour below(which I don't understand why it uses clockwise rather than counterclockwise, and why it bridges along the positive real axis):
If I could get explanation on how to find the indicated residue, how the integrand behaves asymptotically as $$\frac{1 }{\left | z \right |^{n}}$$ and/or an explanation on the contour that will greatly help me. Thank you in advance for all/any help !
Best Answer
The integral along the big circle vanishes as radius $\to \infty$ ($1/(e^{z}-1)$ is uniformly bounded on that contour), and that of pink/green segment cancels each other. The blue circle, as its radius tends to $0$, is the same as residue at the origin, hence $$B_{n}= \frac{n!}{2\pi i}\oint_{C} \frac{1 }{z^{n+1}} \frac{z}{e^{ z} - 1}\,dz$$ The big circle is chosen to be clockwise so as to make the blue circle counterclockwise.
I think the writer's intention is to derive another result using the same contour: $$\mathrm{Res}_{z = 2 \pi m i} \left ( \frac{1 }{z^{n}} \frac{1}{e^{ z} - 1}\ \right ) = \lim_{z\to 2\pi m i} \left(\frac{1}{z^n} \frac{z-2\pi m i}{e^z - 1} \right)=\frac{1}{(2\pi m i)^n}\lim_{z\to 2\pi m i} \left(\frac{z-2\pi m i}{e^z - 1} \right)$$ Where we used the fact that $2\pi m i$ is a simple pole, yielding the residue is $1/(2\pi m i)^n$. Thus, $${B_n} = -n!\sum\limits_{m \in \mathbb{Z} - \{ 0\} } {\frac{1}{{{{(2\pi im)}^n}}}} $$ When $n$ is even, we obtain the famous: $${B_{2n}} = -\frac{{2(2n)!}}{{{{(2\pi i)}^{2n}}}}\zeta (2n)$$ When $n$ is odd, we obtain $B_{2n+1} = 0$ when $n\geq 1$.