Suppose I have a convergent definite integral of the form $$\int_{-\infty}^\infty \frac{f(x)}{x^2(e^x-1)}\text{d}x,$$ where $f(x)$ has no poles, and I want to try to evaluate it using the residue calculus. I can choose the semi-circular contour $\Gamma$ with positive orientation consisting of paths $\gamma_1 = z$, for $-R\leq z\leq R$, and $\gamma_2=Re^{i\theta}$, for $0\leq\theta\leq\pi$.

Assuming the integral over $\gamma_2$ is zero, then am I right in thinking that the integral is equal to $2\pi i$ times the sum of the residues of the integrand, and that the poles occur at $x=2\pi i k$, where $k\geq 0$ is an integer?

It is this last part I am unsure of. Is this the correct way to consider the poles and calculate the residues?


Your contour causes problems: the integrand is not holomorphic on the real axis.

The poles are given by $z^2(\operatorname{e}^z-1) = 0$.

There is a double pole at $z=0$ and a simple pole when $\operatorname{e}^z-1=0$.

The solutions to $\operatorname{e}^z-1=0$ are $z \in \{2\pi i k : k \in \mathbb{Z}\}$. Notice that $k$ need not be positive. Any integer value of $k$ will be sufficient. For example, $\operatorname{e}^{-2\pi i} = \operatorname{e}^{2\pi i} = 1.$

In reality, you will need a so-called beehive contour where you go along the real axis from $-R$ to $-\varepsilon$, then around $\varepsilon \operatorname{e}^{-it}$ with $\pi \le t \le 2\pi$, then along the real axis from $\varepsilon$ to $R$, and then back around along $R\operatorname{e}^{i\theta}$ with $0 \le \theta \le \pi$. Then you look at the limit as $R \to \infty$ and $\varepsilon \to 0$.

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