Notice that
so f does not have a zero at zero.
So to calculate the residue at 0 we have
So the residue is
1)I must calculate . I'm not sure if I should see it as the real part of .
Anyway the problem is obviously when .
I get an infinite residue: .
2)I'd like to get an example of a real integral where when I use complex analysis methods and I choose a contour of integration, I fall over a singularity on the real axis.
Thanks in advance.
I have noticed that you have been calculating a lot of residues this help.
Suppose you have a complex function that is holomorphic except for a finite number of poles. Now if is a pole of order k then has a Laurent expansion at a that looks like
Then
The residue is the coefficient at to recover this from we take take derivatives to get
This gives that
This gives that
Thanks chip, I'll try it.
Thanks TheEmptySet, that was very useful. However the function f(z) should be but it doens't change the fact that . Also, how did you know that g(z) had a pole of order 2? Is it because it gave me infinity when I thought it was of order 1?
Correct me if I'm wrong in the following: If a function f has a singularity at and if then is either a pole of order greater than 1 or an essential singularity. On the other hand, if then f has a removable singularity at . While if with and , then f has a pole of order 1 at .
For this problem we had a function of the form
the numerator does not have a zero at z=0 and is analytic so the constant term of its power series must be non zero and the denominator has a zero or order 2. The first term in its Laurent series must look like So it has a pole of order 2.