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Contour integration problems

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

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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.