Thread: Very simple QFT questions View Single Post
P: 145
 Quote by Hans de Vries That's the point. It does not become zero. The contribution of a half-circle is always $i\pi$ times the residue, independent of the radius. If the contribution would be zero then it would not make any difference if you close the half-circle in the upper or the lower half. Regards, Hans
Now we might come closer to settle this.

You can not choose, how you close the circle, when you want to get a zero contribution.
Usually in the integrand there is some function say exp(i*x*...). This one you want to close with an upper half circle to get a zero contribution, because only then the e function goes to zero (assume x > 0)

For exp(-i*x*...) or when x<0 it is the opposite of course.

The residue theorem does not make any statement on the contribution of a half circle
or any other part of the integration path. You only now from it, what the result for a full closed loop will be.