I am trying to prove this integral tends towards zero (a variation of the Hankel contour problem). I am not sure if my approximation is valid here,
My integral round a circular pole at the origin is, $$ I_{2} = \int_{2 \pi}^0 \frac{u^{s-1}}{e^{u} -1} du $$ I make a substitution, $u = \epsilon e^{i\theta}$ to get, $$ I_{2}= \int_{2\pi}^{0} \frac {\left(\epsilon e^{i\theta}\right)^{s-1}} {e^{\epsilon e^{i\theta}} - 1} d(\epsilon e^{i\theta}) $$ Here is the bit where I think I'm being a bit dodgy: In the limit of $\epsilon \to 0$, I take a binomial expansion of the denominator to get, $$ I_{2}= \int_{2\pi}^{0} \left(\epsilon e^{i\theta}\right)^{s-1} \left(1 + e^{\epsilon e^{i\theta}}\right) d(\epsilon e^{i\theta}) $$ Then for $\text{Re}(s>1)$, $I_2 \to 0$ as $\epsilon \to 0$ since the integrand is proportional to $\epsilon^{s}$.
Solved: This is not valid because $e^{0}$ is not small.
The actual solution cam to me whilst thinking about it. By considering the integral of
Firstly consider, $$ e^{u} > e^{u} - 1 $$ so, $$ \int \frac{u^{s-1}}{e^{u}} du \gt \int \frac{u^{s-1}}{e^{u} - 1} du $$ Now expanding the LHS integrand in a power series, we get, $$ \int u^s \left( \frac{1}{u} - 1 +\mathcal{O}(u)\right) du $$ and $u \propto \epsilon$, so as long as $Re(s) > 1$ then $\text{LHS} \to 0$ as $\epsilon \to 0$ and therefore $I_2 \to 0$ because of the inequality relationship.