# Contour integration around a complex pole

• Remixex
In summary, the integral $\int_{-\infty}^{\infty} \frac{e^{-i \alpha x}}{(x-a)^2+b^2}dx$ can be solved using contour integration. The contour chosen depends on the location of the poles, which are at $z=a\pm ib$. By taking the limit as the contour approaches infinity, the integral over the semicircle tends to zero and the integral over the real line is equivalent to the desired integral. This approach is useful for solving inverse Fourier transforms involving exponentials in the numerator.
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!

## 1. What is contour integration around a complex pole?

Contour integration around a complex pole is a technique used in complex analysis to evaluate integrals involving functions with poles, which are points where the function becomes infinite. It involves integrating along a closed curve in the complex plane that encircles the pole.

## 2. Why is contour integration around a complex pole useful?

Contour integration around a complex pole is useful because it allows us to evaluate integrals that would otherwise be difficult or impossible to solve using traditional methods. It also provides a more efficient way to calculate certain types of integrals.

## 3. How do you choose the contour for contour integration around a complex pole?

The contour for contour integration around a complex pole is chosen based on the location and type of pole, as well as the function being integrated. In general, the contour should encircle the pole in a way that avoids other singularities and simplifies the integral.

## 4. What are the key steps in performing contour integration around a complex pole?

The key steps in performing contour integration around a complex pole include identifying the pole and its type, choosing an appropriate contour, parameterizing the contour, evaluating the integral using the residue theorem or Cauchy's integral formula, and simplifying the result if necessary.

## 5. Can contour integration around a complex pole be used for any type of function?

No, contour integration around a complex pole is only applicable to functions that have poles. It is also important to consider the location and type of pole, as well as the behavior of the function near the pole, in order to choose an appropriate contour and successfully evaluate the integral.

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