# A question about the residue calculus

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?

• Is this integral even convergent? – Mercy King Apr 14 '13 at 21:34
• @ Mercy. Surely it will converge, apart from some pretty agressive $f(x) > x^2 e^x$, right? – user27182 Apr 14 '13 at 21:37
• – user27182 Apr 14 '13 at 21:40
• Calculating the residues of $\dfrac{f(x)}{x^2(e^x-1)}$ themselves are not difficult, only the difficulties are that whether they really work on $\int_{-\infty}^\infty\dfrac{f(x)}{x^2(e^x-1)}dx$ . – doraemonpaul Apr 16 '13 at 6:22

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 $\varepsilon \to 0$.
• Thanks for the reply. I chose $k\geq 0$ since it is for these values that the poles lie within the contour. Hence, we would not need to consider the poles corresponding to $k<0$. – Pixel Apr 15 '13 at 7:19