Applying Cauchy Integral Theorem to Compute Integrals over Circular Paths

moo5003
Messages
202
Reaction score
0

Homework Statement


For r=1,3,5 compute the following integral:

Integral over alpha (e^(x^2)/(x^2-6x)dx
Alpha(t) = 2+re^(it) from 0 to 2pi



Homework Equations


Cauchy Integral Formula:
f(z) = 1/(2ipi)Integral over Alpha(f(x)/(x-z)dx)



The Attempt at a Solution



For r = 1, the integral is simply 0 since f(x) is analytic over the domain (critical points being 0 and 6).

For r = 3, I'm a little unsure how to proceed. Obviously plugging in alpha and alpha prime doesn't seem like the correct method (simply because the algebra involved is alot) not to mention a nasty integral from 0 to 2pi at the end. I'm assuming I'm supposed to use the cauchy integral theorem (the chapter we are on) in solving the problem.

Questions: How can I apply the cauchy integral theorem to help me solve this?

At first I tried spliting up the domain into separate chunks so that some would sum to 0 and then I would be left with something that I already knew, though I couldn't really manage. The best application of the cauchy integral theorem that I can think of is taking a f(z) that I know and setting it equal to the integral relation and then somehow deriving the original integral from the chaos... Is this the correct way of going about it?
 
Physics news on Phys.org
Good news, I figured out my problem. You have to use partial fractions to separate the deonominator so that you can better apply cauchy formula.

e^(x^2)/(x^2-6) = e^36 / (6x-6) - 1/6x and then just go from there depends on r it changes since domain changes.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top