Explore BrainMass

Contour integration problems

This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!

Using the method of complex variables and contour rotations, evaluate:

Please see the attachment for the questions.

Please show the complete steps.

© BrainMass Inc. brainmass.com October 25, 2018, 6:25 am ad1c9bdddf


Solution Preview


The integral from zero to infinity is half the integral from minus infinity to infinity, because the integrand is an even function of x. Consider the contour C that consists of a line from -R to R, and then a semicircle in the upper half plane of radius R and center z = 0, from R back to -R. We then consider the integral of dz/(1+z^2) along C. We have:

Integral over C of dz/(1+z^2) = Integral from -R to R of dx/(1+x^2) + Integral along semicircle of dz/(1+z^2)

The fintegrand 1/(1+z^2) has poles at z = i and z = -i, the former lies within the contour (when R > 1). The residue there is:

Limit z to i of (z-i)/(1+z^2) = 1/(2 i)

The integral over the contour C is thus 2 pi i times the residue which is pi. This means that:

Integral from -R to R of dx/(1+x^2) + Integral along semicircle of dz/(1+z^2) = pi

Taking the limit of R to infinity here and using that in this limit the integral along the semicircle goes to zero, yields:

Integral from -infinity to infinity of dx/(1+x^2) = pi ----->

Integral from 0 to infinity of dx/(1+x^2) = pi/2


b) ...

Solution Summary

We show how the compute the integrals, step by step. The methods of complex variables and contour rotations are evaluated.

See Also This Related BrainMass Solution

Alternative way to compute the integral of sin(x)/x from -∞ to ∞ using contour integration

The standard textbook way to compute the integral of sin(x)/x from minus infinity to plus infinity is to replace this integral by the principal value which then allows one to replace sin(x) by Im[exp(ix)], and we can take the imaginary part operator out of the integral. One then proceeds by completing the contour in the upper half plane and by plugging the hole that arises when taking the principal part using a small half circle in either the upper or lower half plane that makes the contour avoid the singularity at the origin. The contribution from the half circle must then be subtracted from the entire contour integral, and taking the appropriate limits will then yield the final result.

One problem with this approach is the complexity of attacking the problem this way. Many of the steps taken here are not obvious to beginning students, they are needed to avoid problems later due to choices one needs to make later, and when these avoidance steps come first this makes the entire approach quite dense. One can then ask if there exists a more direct, straightforward, way to compute the integral using contour integration.

View Full Posting Details