# Contour integration around a complex pole

• Remixex

#### Remixex

Homework Statement
Prove that $$\int_{-\infty}^{\infty} \frac{e^{-i \alpha x}}{(x-a)^2+b^2 }dx=(pi/b) e^{-i \alpha a}e^{-b |a|}$$ via contour integration
Relevant Equations
Cauchy's integral $$\int_{C} \frac{f(z)}{z-z_0} dz= 2 \pi i f(z_0)$$ given C a closed curve and f(z) analytic over this curve.
Residue theorem might be useful.
$$\int_{-\infty}^{\infty} \frac{e^{-i \alpha x}}{(x-a)^2+b^2}dx=(\pi/b) e^{-i \alpha a}e^{-b |a|}$$
So...this problem is important in wave propagation physics, I'm reading a book about it and it caught me by surprise.
The generalized complex integral would be
$$\int_{C} \frac{e^{-i \alpha z}}{(z-a)^2+b^2}dz$$
I am having trouble with several things, one of them is how to define the actual curve, I imagined a conterclockwise semicircle with a keyhole on the complex pole a+ib (from -R to -epsi, upwards from -epsi to ib, semicircle from -epsi + ib to epsi +ib, downwards from epsi+ib to epsi, epsi to R, and another semicircle connecting R and -R) , I haven't been able to find much info on curves around complex poles.
Second problem I am having is that even for the first segment, I cannot for the life of me solve this integral, I tried some help with a computer and the expression involves more integrals, I do not know (and I don't think) it has an antiderivative.

Maybe there's some manipulation that I can do to take advantage of the Cauchy-Goursat theorem? maybe a variable change t=z-a might make this possible. (But given the exponential maybe one of the residues will not be zero...)

I am not looking for a full answer, but rather a good direction to take this problem to...I've been out of tune with these techniques for a while.

Contour integration is the correct approach.

The poles are at $z = a \pm ib$.

If $b \neq 0$ then the poles are not on the real axis, so one part of the contour is $$C_1 = \{z = x : x \in [-R,R]\}$$ traversed from $-R$ to $R$. The other is a semicircle $$C_2 = \{z = Re^{i\theta} \}$$ traversed from $R$ to $-R$. Whether this contour in the upper ($\theta \in [0, \pi]$) or lower ($\theta \in [-\pi,0]$) half-plane depends on the sign of $\alpha$: You need $$\sup_{\theta} \left|\frac{e^{i\alpha Re^{i\theta}}}{(Re^{i\theta} - a)^2 + b^2}iRe^{i\theta}\right| \to 0$$ as $R \to \infty$.

Thus the contour only ever encloses one of the poles. You then have that $$\int_{C_1} f(z)\,dz + \int_{C_2} f(z)\,dz = \pm 2\pi i \mathrm{Res}(z_0)$$ where $z_0$ is the pole lying inside the contour. If the contour is closed in the upper half plane then the plus sign is taken as the contour is traversed anticlockwise; otherwise the minus sign is taken as the contour is traversed clockwise. Taking the limit $R \to \infty$ and using the bound $$\left| \int_C f(z)\,dz \right| \leq L(C) \sup \{ |f(z)| : z \in C\}$$
where $L(C)$ is the length of the contour $C$ you should find that the integral over $C_2$ tends to zero whilst the integral over $C_1$ is the integral you want.

Last edited:
• Remixex
Perfect, thank you, I had never dealt with exponentials in the numerator doing residue theorem, this was very helpful, especially for inverse Fourier transforms!