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Complex integral

  1. Dec 28, 2008 #1
    How to evaluate the following integral using residue theorem:

    [tex]
    \int_1^2 (x+1) \sqrt[6]{\frac{x-1}{2-x}}dx
    [/tex]

    (The answer is [tex]\frac{31}{36}\pi [/tex])

    Thanks for any help
     
  2. jcsd
  3. Dec 28, 2008 #2

    tiny-tim

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    Welcome to PF!

    Welcome to PF! :smile:

    Show us how far you've got, applying the residue theorem, and where you're stuck, and then we'll know how to help. :wink:
     
  4. Dec 28, 2008 #3

    Dick

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    That's actually a pretty tough question, in the sense that there's a lot of work involved in the calculation, but I'll get you started. You should notice that (x-1)/(2-x) goes from 0 at x=1 to infinity at x=2. That should suggest to you that you want to try the substitution u^6=(x-1)/(2-x). With some work you can use that to change the integral into a rational (even!) function in u. Since it's even you can change it into an integral over the whole real line (provided you remember to divide by 2 later). Now you just have to locate the poles etc. It looks nasty. I stopped doing the details after that. I hope this is a 'super challenge' question.
     
    Last edited: Dec 28, 2008
  5. Dec 30, 2008 #4
    Just for habit, when dealing with complex numbers I prefer changing x's into z's. So you know you're dealing with complex stuff, such as integrals.

    The problem when using residue calculus is that your contour is from 1 to 2, and in this case you residue is on 2. Which makes it break down quickly, however, if you do some nice conversions or apply some thoughtful series and you can change this =).

    So, think about how you can change this algebraically to make it work for you. As a note z can be expressed in many ways.
     
    Last edited: Dec 30, 2008
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