I want to compute this contour integral:
\begin{equation} \int\limits_C \! \mathrm{d}z \; \log(\frac{z+1}{z-1}) \frac{e^{tz}}{z-1} \end{equation}
Where $C$ is a path going around the branch cut. My perplexities arise because the branching point $z=1$ presents also a "pole-like" behaviour, so that if we naively choose the contour to pass on the upper and lower parts of the cut we get a non integrable singularity. I thought of avoiding the singularity with a little circle around it, but then you cannot apply the residue theorem because of the branch cut. Is there a standard technique for these pathological cases?
Let $C$ be any closed contour around the branch cut $[-1, 1]$ and $I(t)$ be the integral over $C$. $I(t)$ does not depend on $C$ and is entire. Next, we have $$\mathcal L^{-1} {\left[ \ln \frac {z + 1} {z - 1} \right]} = \frac {2 \sinh t} t, \\ f(t) = \mathcal L^{-1} {\left[ \frac 1 {z - 1} \ln \frac {z + 1} {z - 1} \right]} = \frac {2 \sinh t} t * e^t = e^t (\Gamma(0, 2 t) + \ln(2 t) + \gamma).$$ By deforming the contour, we get $I(t) = 2 \pi i f(t)$ for $t > 0$. Extending $f$ to an entire function, we get $I(t) = 2 \pi i f(t)$ for any $t \in \mathbb C$.
Another way is to find $I(t)$ as $-2 \pi i$ times the residue of the integrand at infinity. Multiplying the series expansions gives $$\operatorname*{Res}_{z = \infty} \frac {e^{t (z + 1)}} z \ln \left( 1 + \frac 2 z \right) = e^t \sum_{k \geq 1} \frac {(-2 t)^k} {k! \,k}.$$