Let us look at the effects of integrating the powers of along a line in the complex plane (note that we implicitely assume that the answer is independent of the position of the line, and only depends on beginning and end!)
(A.3) |
We know how to integrate powers, so apart from the case , we get
So the first of these integrals is indepdent of the path between begin and end. It thus goes to zero as we look at a closed path, but the second integral actually does depend on the path of integration:
(A.6) |
We have two options: the contour (the technical word for the closed path) either encloses the origin where the singularity resides or not, see Fig. A.1 .
As we move the around the curve, you notice
for the blue curve that the phase (the number plotted on the
line connecting to the point for which we calculate the phase)
gently oscillates up and down, and thus the answer of the
contour integral is zero; for the orange curve we see a jump at
the negative x-axis. This is due to a convention; in principle
we can put this jump along any half-line in the complex plane
ending at the origin.
If we move the value at the right of the
jump down by 2
(making the function continuous) we realise that the
begin and endpoint differ in phase by 2
. Thus for any contour
enclosing the origin the phase changes by
, and thus we expect that
(A.7) |
for a contour enclosing the origin. The sign is positive if we integrate around in the positive sense (anti-clockwise), and negative if we do the integral along a contour that encircles the origin in a clockwise fashion.
If a function behaves like near the point , we say that the function has a pole at .