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Complex Integration Function with multiple poles at origin

  1. Mar 30, 2013 #1

    VVS

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    Hello,

    I hope somebody can help me with this one.

    1. The problem statement, all variables and given/known data
    I want to find the integral of 1/x^N*exp(ix) from -inf to inf.


    2. Relevant equations
    It is very likely that this can somehow be solved by using Cauchy's integral formula.


    3. The attempt at a solution
    I tried to integrate it by defining a countour as follows:
    1. From -R to -r
    2. semicircle from -r to +r around the origin from below
    3. from r to R
    4. Semicircle from +R to -R around the origin from above.

    I can show that 4 tends to 0 as R tends to infinity
    But I can't somehow evaluate 2:
    I get 1/(r*exp(i*theta)^N*exp(i*r*exp(i*theta))r*exp(i*theta)d(theta)
    Nothing cancels as nicely as in the case of a simple pole.

    Thank you
     
  2. jcsd
  3. Mar 31, 2013 #2

    VVS

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    sorry i didnt solve it
     
    Last edited: Mar 31, 2013
  4. Mar 31, 2013 #3

    Ray Vickson

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    The integral may not converge. Since you have an improper integral, it needs to be *defined*, typically in terms of some limiting operations such as
    [tex] \int_{-\infty}^{\infty} \frac{e^{ix}}{x^N} \, dx = \lim_{L,U \to \infty,\: a,b \to 0+}
    \left[ \int_{-L}^{-a} \frac{e^{ix}}{x^N} \, dx
    + \int_{b}^{U} \frac{e^{ix}}{x^N} \, dx \right].[/tex] Does this limit exist in your case?
     
  5. Mar 31, 2013 #4

    VVS

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    Hey,
    Thanks for your help but I had solved the problem now before your answer.
    I called In=1/x^n*exp(ix). And I used integration by parts. I chose 1/x^n as the function to be integrated and exp(ix) as the function to be differentiated. That means that I get something proportiional to 1/x^(n-1)exp(ix) which is In-1. So I keep expressing In in terms of In-1, In-2 and so forth till I get to 1/x*exp(ix) which is easily integrated to Pi*i. All other terms vanish because they are of the from 1/x^N-n and the limits are +inf and-inf.
    Thank you again for your help
     
  6. Mar 31, 2013 #5

    Ray Vickson

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    I tried to warn you but you refused to listen.
     
  7. Apr 1, 2013 #6

    VVS

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    Hey, The integrals converge with those limits. So I did listen to you.
     
  8. Apr 1, 2013 #7

    Ray Vickson

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    All you did was to perform a sequence of illegal operations to obtain a wrong answer. I will say it only one more time: you need to look at limits when dealing with improper integrals.
     
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