# Problem with Complex contour integration

• Joker93
In summary, the conversation discusses the computation of an integral with a complex variable in the integrand by converting it to one with a real variable. The path of integration and the resulting expression for the integral are shown. The steps taken are considered legitimate, but there may be a simpler form for the integral. The simplification of the denominator is also mentioned.
Joker93

## Homework Statement

I want to compute ##I=\int_C \dfrac{e^{i \pi z^2}}{sin(\pi z)}##, where C is the path in the attached figure (See below). I want to compute this by converting the integral to one whose integration variable is real.

## Homework Equations

There are not more relevant equations.

## The Attempt at a Solution

From the path of the integration, we have that ##y=x-\frac{1}{2}##. Thus ##z=x+iy=x(1+i)-\frac{i}{2}##. With this, we can write the integral as
##I=-(1+i)\int_{-\infty}^{+\infty} \dfrac{e^{i \pi (x(1+i)-\frac{i}{2})^2}}{sin(\pi (x(1+i)-\frac{i}{2}))}##.
where we pick up a minus sign due to reversing the direction of integration.
Is this correct? I am using this integral for a more general proof, but it does not seem to fit nicely with the proof. In particular, what makes things ugly is the fact that when we expand the sin function in the denominator of the integrand function, we get things like ##e^{\pm \pi/2}##, which makes me feel not-so-sure about the steps leading from ##z## to ##x##.
Thank you!

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Joker93 said:

## Homework Statement

I want to compute ##I=\int_C \dfrac{e^{i \pi z^2}}{sin(\pi z)}##, where C is the path in the attached figure (See below). I want to compute this by converting the integral to one whose integration variable is real.

## Homework Equations

There are not more relevant equations.

## The Attempt at a Solution

From the path of the integration, we have that ##y=x-\frac{1}{2}##. Thus ##z=x+iy=x(1+i)-\frac{i}{2}##. With this, we can write the integral as
##I=-(1+i)\int_{-\infty}^{+\infty} \dfrac{e^{i \pi (x(1+i)-\frac{i}{2})^2}}{sin(\pi (x(1+i)-\frac{i}{2}))}##.
where we pick up a minus sign due to reversing the direction of integration.
Is this correct? I am using this integral for a more general proof, but it does not seem to fit nicely with the proof. In particular, what makes things ugly is the fact that when we expand the sin function in the denominator of the integrand function, we get things like ##e^{\pm \pi/2}##, which makes me feel not-so-sure about the steps leading from ##z## to ##x##.
Thank you!

Everything you did is perfectly legitimate.

To go further, you could re-express ##\exp(i \pi (x(1+i) - \frac 1 2)^2) ## by expanding out the exponent in real and imaginary terms. That will turn out to give a nicer-looking result, although it will not make the integral any more "do-able".

As for expansion of the denominator: I did it using the computer algebra package "Maple" and obtained
$$\text{denominator} = \cos(\pi x) \cosh(\pi x) - i \: \sin(\pi x) \sinh(\pi x)$$
There are no "##\pm##" ambiguities when you carry out the simplifications all the way to the bitter end. (Getting the result above by manual computation would not be difficult, but I am tired of doing such stuff by hand---hence the use of a computer algebra package.)

Joker93

## What is complex contour integration?

Complex contour integration is a mathematical technique used to evaluate integrals of complex functions along a path in the complex plane. It involves breaking down a complex function into simpler components and integrating each component along a closed contour in the complex plane.

## What are some common problems encountered in complex contour integration?

Some common problems encountered in complex contour integration include identifying the correct contour, dealing with branch cuts and branch points, and ensuring that the contour encloses all singularities of the function being integrated.

## How is complex contour integration different from real contour integration?

Complex contour integration differs from real contour integration in that the path of integration is along a curve in the complex plane instead of the real line. This allows for the evaluation of integrals of complex functions that cannot be evaluated using traditional real integration techniques.

## What are the benefits of using complex contour integration?

Complex contour integration allows for the evaluation of integrals of complex functions that cannot be evaluated using traditional real integration techniques. It also helps to simplify the evaluation of complex integrals by breaking them down into simpler components.

## What are some applications of complex contour integration?

Complex contour integration has various applications in physics, engineering, and other fields where complex functions are encountered. It is used in the study of electromagnetic fields, fluid dynamics, quantum mechanics, signal processing, and many other areas.

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