Cauchy residue theorem

**zetafunction** · Jun8-09, 07:19 AM

my question is , let us have the following complex integral

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

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

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

**HallsofIvy** · Jun8-09, 07:25 AM

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 $LaTeX Code: (z-\\infty )f(z)$ with z tending to infinity", is the residue itself.

**zetafunction** · Jun8-09, 07:40 AM

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 ??

**Count Iblis** · Jun8-09, 09:51 AM

Originally Posted by 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 ??

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.