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I have to calculate real integral $\displaystyle \int_{-\infty}^\infty \frac{e^{cx}}{1+e^x} \, \text{d}x$ for $c \in (0,1)$. I have also hint, to integrate over the squere, which consists of $x$ axis, 2 lines paralel to $y$ axis, and line with $z=x+2\pi i$, for real $x$.

I think that idea is to use Residue theorem. I think, that for the function $\frac{e^{cz}}{1+e^z}$, all singularities are $z=\pi i+2k\pi i$, so the only one within the area is $2 \pi i$. But I am not sure how to calculate residue in this point, neither how to evaluate other $3$ integrals on rectangle. I have concluded, that both integrals on the vertical line cancel each other, becouse: $$ \lim_{A \to \infty} \int_0^{2\pi} \frac{e^{cA}e^{ix}}{1+e^Ae^{ix}} \, \text{d}x=0 $$ and $$ \lim_{A \to \infty} \int_{2\pi}^0 \frac{e^{-cA}e^{ix}}{1+e^{-A} e^{ix}} \, \text{d}x=0 $$ How to determine integral on the other horizontal line?

V.G
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  • A pole occurs where $1+e^z=0,$ which is where $e^z=-1,$ and that happens where $z=\pi i$ and at other points differing from that by integer multiples of $2\pi i.$ It seems you're integrating over a square, and only one of those points is in the interior of that square. – Michael Hardy Sep 05 '20 at 15:18
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    This problem has already a few duplicates, look 1, 2, 3 and probably many others. – zwim Sep 05 '20 at 15:25
  • $$\begin{align} & 1+e^z = 1 + e^{z-\pi i} e^{\pi i} \ {} \ = {} & 1 - e^{z-\pi i} = 1 -\left( 1 +(z-\pi i) + \frac {(z-\pi i)^2} 2 + \cdots \right) \ {} \ = {} & -(z-\pi i)\left( 1 - \frac{z-\pi i} 2 + \text{higher-degree terms} \right) \end{align}$$ And so: $$ \frac{e^{cz}}{1 + e^z} = \frac{-e^{cz}}{1 - \frac{z-\pi i} 2 + \text{higher-degree terms}} \times \frac 1 {z-\pi i} $$ The residue at $\pi i$ is the product of the value of the fraction to the left of $\text{“} {\times} \text{''}$ and the residue of the function of $z$ that is to the right of$\text{ “}{\times}\text{''.}$ – Michael Hardy Sep 05 '20 at 15:49

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