# Cauchy residue theorem

1. Jun 8, 2009

### zetafunction

my question is , let us have the following complex integral

$$\oint f(z)dz$$ where f(z) has a simple pole at $$z=\infty$$

then by Residue theorem $$\oint f(z)dz =2\pi i Res(z,\infty,f(z)$$

or equal to the limit $$(z-\infty )f(z)$$ with 'z' tending to infinity

2. Jun 8, 2009

### HallsofIvy

I'm sorry, I don't see a question here! Are you asking if it is one or the other of those?

If so, it is the first. The second, the "limit $$(z-\infty )f(z)$$ with z tending to infinity", is the residue itself.

3. Jun 8, 2009

### zetafunction

sorry i am not from US or England so my english could be a little mistaken

the idea of the post is: can we use 'Cauchy's residue theorem' even in the case the function f(z) has a pole at infinity ??

4. Jun 8, 2009

### Count Iblis

Yes. This is useful if the function is not meromorphic in the interior of the contour. Take e.g. the real integral (x^2 - x^3)^(1/3)dx from
x = 0 to 1. You can evaluate this by considering the so-called "dogbone" contour that goes from zero to 1 just below the real axis, encircles the branch point at z = 1 and then goes to zero just above the real axis and then encicles the branch point at z = 0.

Clearly the function is not meromophic inside the contour, but it is outside the contour (if you choose the branch cuts so that they cancel out outside the interval from zero to 1).

If you then perform the conformal transformation z ---> 1/z, you see that what was outside the contour is now inside the contour and what was inside is now outside. You also see that the function is now meromorphic inside the the contour and has a pole at zero.