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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?

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  • $\begingroup$ Maxim is deforming $C$ to the vertical line $(2-i\infty,2+i\infty)$. The direct calculation with $g(z) = \frac{e^{tz}}{z-1}$ gives $\int_C f(z)dz = \lim_{a \to 0} \int_{-1+a}^{1-a} 2i \pi g(z)dz + \int_{|z-1|=a, arg(z) \in (-\pi,\pi)} f(z)dz+O(a)$ $=\lim_{a \to 0} \int_{-1+a}^{1-a} 2i \pi g(z)dz + 2i\pi \log(2) e^t- e^t \frac12\log^2 (z-1)|_{1+e^{-i\pi} a}^{1+e^{i\pi}a}$ where $\int g(z)dz$ is $Ei$ something $\endgroup$
    – reuns
    Apr 5 '19 at 23:49
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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 $$\operatorname*{Res}_{z = \infty} \frac {e^{t (z + 1)}} z \ln \left( 1 + \frac 2 z \right) = e^t \sum_{k \geq 1} \frac {(-2 t)^k} {k! \,k}.$$

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