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I Complex integral path

  1. Jun 27, 2017 #1
    Hello! If I have a real integral between ##-\infty## and ##+\infty## and the function to be integrated is holomorphic in the whole complex plane except for a finite number of points on the real line does it matter how I make the path around the poles on the real line? I.e. if I integrate on the semicircle in the upper plane and I have a pole at 0 let's say, do I get the same result if I go around the pole above or below the real axis? Thank you!
     
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  3. Jun 27, 2017 #2

    FactChecker

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    The sign will change because one path is clockwise around the pole and the other is counter-clockwise.
     
  4. Jun 27, 2017 #3
    Do you mean the sign of the semicircle around the pole (whose radius t=you take to go to 0 in the end)? But you also have to consider that in one case the pole is inside the contour (so you must use Residue theorem) and in the other is outside (SO by Cauchy theorem the integral around everything would be 0), right?
     
  5. Jun 27, 2017 #4

    strangerep

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    You'll get the same result IF you do both integrals properly.

    In the first case, the pole is outside the contour, so no residue contribution. But there's potentially a contribution from the small semicircle over the pole, traversed clockwise.

    In the second case, the pole is inside the contour so there's a residue contribution. But there's also a contribution from the small semicircle under the pole, this time traversed anticlockwise.

    You should end up with the same result either way.
     
  6. Jun 28, 2017 #5

    FactChecker

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    Yes. Sorry. My post was too brief. @strangerep gave a better answer.
     
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