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Complex analysis

  1. Jan 25, 2009 #1
    Hey guys.
    I need to calculate this integral so I was thinking about using the residue theorem.
    The thing is that the point 0 is not enclosed within the curve that I'm about to build, it's on it.
    Can I still use the theorem?

    Thanks a lot.
     

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  3. Jan 25, 2009 #2
    So use Cauchy Principal value to bypass the singularity at 0.
     
  4. Jan 25, 2009 #3

    lurflurf

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    You have a bigger problem then that, the residue at x=0 is zero (removeable singularity). This is a standard example and is often done as follows
    consider
    To fix the residue=0 problem add in an odd function the usual one is to use
    exp(x i)/x=cos(x)/x+i sin(x)/x
    the cos(x)/x diverges as an improper integral, but as lqg states cauch principle sense may be used
    now consider ther contour formed by theese pieces
    (-R,r) straight line
    (-r,r) semicircle arc about z=0
    (r, R) straight line
    (R,r) semicircle arc about x=0
     
  5. Jan 25, 2009 #4
    I didn't quite understand the contour you described there.
    What's R and What's r?

    Thanks a lot.
     
  6. Jan 25, 2009 #5

    lurflurf

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    Take limits R-> +infinity,r->+0
    If you draw a pictuis an upper semicircle that in the limit is large with a tiny semicircle at the origin then you get for the various integrals

    (-R,r) straight line
    -infinity+(Integral you want)/2
    (-r,r) semicircle arc about z=0
    {+,-}[+ if it was upper - if it was lower] pi*i (residue theorem)
    (r, R) straight line
    -infinity+(integral you want)/2
    (R,r) semicircle arc about x=0
    0
     
  7. Jan 25, 2009 #6
    Is it something like that?

    Thanks again.
     

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  8. Jan 26, 2009 #7

    lurflurf

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    That is it. I guess that is one reasone why some people do not like old math books with no pictures. Were you able to finish?
     
  9. Jan 26, 2009 #8
    I'll try, if I'll have some troubles, I'll be back :smile:.
    Thanks a lot.
     
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