 Quote by Hans de Vries
That's the point. It does not become zero. The contribution of a half-circle is always
[itex]i\pi[/itex] 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
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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.