How to evaluate integrals like $ \int_C \frac{z}{e^z-1} dz$ without Residue Theorem? I'm trying to evaluate two integrals:
$$ I_1 = \int_{C(0, 1)} \frac{z}{e^z-1} dz,\quad  I_2 = \int_{C(1, 3)} \frac{\sin(iz)}{e^{iz}-1} dz $$
$ C(a, r) $ is a circular contour with radius $r$ centered at $a$, oriented counterclockwise.
In both cases there's a singularity point $ z=0 $. How can I handle it? Can I use some parametrization to get rid of exponential function?
In general, what can I do to evaluate integrals of type:
$$ I_3 = \int_{\gamma} \frac{f(z)}{e^{g(z)}-1} dz$$
Where $f, g$ holomorphic on open, simply connected subset U, $\gamma(t) \in U$?
 A: This is really using the fact that the singularity is removable without explicitly mentioning it.
Somewhat more generally, consider an integral of the form
$$J = \oint_\Gamma \frac{f(z)}{g(z)}\; dz$$
where $f$ and $g$ are analytic in a simply connected region $U$ containing the piecewise smooth simple closed contour $\Gamma$, $f(p) = g(p) = 0$ for some point $p$ on $\Gamma$, and $g$ has no other zeros in or on $\Gamma$.  So except for the singularity at $p$ we would have $J=0$ by the Cauchy integral theorem.  
We might compare this to another integral
$$ \widetilde{J} = \oint_\widetilde{\Gamma} \frac{f(z)}{g(z)}\; dz = 0$$
where $\widetilde{\Gamma}$ is obtained by modifying $\Gamma$ near $p$ so that $p$ is outside $\widetilde{\Gamma}$.
The difference between them is the integral over a small contour $\widetilde{\widetilde \Gamma}$ near $p$.
Now suppose we know that $f(z)/g(z)$ is bounded for $z$ near $p$, say $|f(z)/g(z)| < B$ for $|z - p| < \epsilon$.  If $\widetilde{\widetilde \Gamma}$ is all in the region $|z-p|<\epsilon$ and has length at most  $k \epsilon$ for some constant $k$, then we can
estimate
$$ |J |= |J - \widetilde{J}| = \left|\oint_{\widetilde{\widetilde \Gamma}}  \frac{f(z)}{g(z)}\; dz \right| \le B \cdot \text{length}(\widetilde{\widetilde \Gamma}) \le B k \epsilon $$
and taking the limit as $\epsilon \to 0$ we conclude that $J = 0$.
