# Evaluate Cauchy Integral of Contour Integral Homework

• kreil
In summary, the conversation discussed the evaluation of a complex contour integral using Cauchy's Integral Theorem. The arc term was evaluated using Jordan's Lemma, and the final result for the integral was determined to be (2π/e^k)(e^-k - 1).
kreil
Gold Member

## Homework Statement

I have to evaluate the following integral by means of the Cauchy Integral Theorem:

$$\int_{- \infty}^{\infty}\frac{e^{-ikx}}{(x+i)(x+2i)}dx$$

## Homework Equations

$$f(z_0)=\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-z_0}dz$$

## The Attempt at a Solution

The idea I had was to consider the complex contour integral,

$$\oint_C\frac{e^{-ikz}}{(z+i)(z+2i)}dz= \oint_C f(z)dz$$

If the contour C is a semi circular contour in the lower plane with radius > 2 (so that both poles are enclosed), we can break this integral up into two parts:

$$\oint_C f(z)dz = \int_a^{-a}f(z)dz + \int_{arc}f(z)dz$$

$$\implies \int_{-a}^a f(z)dz = \int_{arc}f(z)dz - \oint_C f(z)dz$$

At this point I can evaluate the contour integral by splitting into two integrals using partial fractions and then directly applying Cauchy's Integral Theorem. My problem is with this arc integral, which I think goes to zero in the limit $a,R \rightarrow \infty$? I know Jordan's Lemma States, for a semi circular contour in the upper plane:

$$\lim_{R \rightarrow \infty} \int_{C_R} e^{i \alpha z} f(z) dz = 0$$

but this formula is for alpha greater than or equal to zero. In this case, alpha is less than zero and the contour is in the lower plane.

So, does this arc term vanish or need to be evaluated, and in either case how can I proceed?

kreil said:
I know Jordan's Lemma States, for a semi circular contour in the upper plane:

$$\lim_{R \rightarrow \infty} \int_{C_R} e^{i \alpha z} f(z) dz$$

but this formula is for alpha greater than or equal to zero. In this case, alpha is less than zero and the contour is in the lower plane.
I do not believe $\alpha$ can be less than zero. Assuming that's correct, then carefully go through the derivation of Jordan's Lemma as you have written it above, then re-do the derivation with a contour in the lower half-plane ($-C_R$) with the expression:

$$\lim_{R \rightarrow \infty} \int_{-C_R} e^{-i \alpha z} f(z) dz$$

I guess my wording was a little awkward. I'm aware that alpha can't be less than zero in Jordan's lemma, and was trying to say that's why I can't apply the lemma.

Is it OK to just move the poles into the upper plane by taking z = -z? Then the contour would be in the upper plane and the form of the integral would be suitable for the lemma. This integral would differ from the one I have by only a sign (I think..), so if it is equal to zero by jordan's lemma then the other would be as well.

Last edited:
kreil said:
Is it OK to just move the poles into the upper plane by taking z = -z?

Why would it not be?

excellent. then the arc term goes to zero and i get, letting a and R diverge:

$$\int_{- \infty}^{\infty}\frac{e^{-ikz}}{(z+i)(z+2i)}dz=- \oint_C \frac{e^{-ikz}}{(z+i)(z+2i)}dz = i \oint \frac{e^{-ikz}}{z+i}dz - i \oint \frac{e^{-ikz}}{z+2i}dz = 2 \pi i^2 \left ( e^{-k} - e^{-2k} \right ) = \frac{2 \pi}{e^k} \left ( e^{-k}-1 \right )$$

$$\implies \bar f (k) = \int_{- \infty}^{\infty} \frac{e^{-ikx}}{(x+i)(x+2i)}dx = \frac{2 \pi}{e^k} \left ( e^{-k}-1 \right )$$

I was wrong with my last post and removed it. I noticed you incorporated the -1 into the expression. Sorry.

## 1. What is a Cauchy integral?

A Cauchy integral is a type of contour integral that is used to evaluate complex integrals along a closed curve in the complex plane. It was introduced by the French mathematician Augustin-Louis Cauchy in the early 19th century and is an important tool in complex analysis.

## 2. What is a contour integral?

A contour integral is a type of integral that is evaluated along a specific path or curve in the complex plane. It is also known as a line integral and is used to calculate the area under a curve in the complex plane. Contour integrals are widely used in the field of complex analysis.

## 3. How do you evaluate a Cauchy integral?

To evaluate a Cauchy integral, you first need to parameterize the curve or contour over which the integral is being evaluated. Then, you can use various techniques such as the Cauchy integral formula or the residue theorem to calculate the value of the integral. It is important to choose a suitable parameterization and use the correct method to ensure an accurate evaluation.

## 4. What is the significance of Cauchy integrals in mathematics?

Cauchy integrals are significant in mathematics because they allow us to solve complex integrals that would otherwise be difficult or impossible to evaluate using traditional methods. They are also important in the study of complex functions and have applications in many areas of mathematics, including physics and engineering.

## 5. Are there any real-world applications of Cauchy integrals?

Yes, there are many real-world applications of Cauchy integrals. For example, they are used in electromagnetism to calculate the electric field around a charged object. They are also used in fluid mechanics to calculate the flow of a fluid around a curved surface. Additionally, Cauchy integrals have applications in signal processing, probability theory, and many other fields.

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