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distler Administator 20 posts

edited 8 years ago

At some point, in Homework 3, you will encounter the integral

I= \int \frac{dk''}{2\pi} \frac{1}{k^2-{k''}^2+i\epsilon}

For those of you who took complex analysis, this should be easy to do, using the methods of contour integration.

If you haven’t taken complex analysis, the answer is

I= -\frac{i}{2|k|}

To see how the integral is done, read on…

First, let’s rewrite

I = \int_{-\infty}^\infty \frac{dk''}{2\pi} \frac{1}{(|k|-k''+i\epsilon)(|k|+k'' +i\epsilon)}

The $i\epsilon$ displaces the location of the poles of the integrand off the real axis. The are located at $|k| +i\epsilon$ and at $-|k|-i\epsilon$ . The contour can be closed either in the upper half-plane or in the lower half-plane. Closing in the upper half-plane, we pick up the pole at $|k| +i\epsilon$ :

-|k|-i\epsilon

and we get $I= -\frac{i}{2|k|}$ . If we closed the contour in the lower half-plane, we would pick up the pole at $-|k|-i\epsilon$ . But, that contour is in the clockwise (instead of counter-clockwise) direction, the two minus-signs cancel, and we get the same result for the integral.

A Contour Integral

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