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A Why can't I use contour integration for this integral?

  1. Dec 16, 2016 #1

    ShayanJ

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    Consider the integral ##\displaystyle \int_{-\infty}^\infty \frac{e^{-|x|}}{1+x^2}dx ##. I should be able to use contour integration to solve it because it vanishes faster than ## \frac 1 x ## in the limit ## x \to \infty ## in the upper half plane. It has two poles at i and -i. If I use a semicircle in the upper half of the plane, I should only consider the residue at i which is:

    ## \displaystyle \lim_{z\to i} \frac{e^{-|z|}}{i+z}=\frac{e^{-|i|}}{2i}=\frac{1}{2ie} ##

    So the integral is equal to ## \frac{2\pi i}{2ie}=\frac \pi e \approx 1.15 ##. But wolframalpha.com gives an expression involving Ci(x) and Si(x) that approximates to 1.24. What is wrong here?

    Thanks
     
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  3. Dec 16, 2016 #2

    BvU

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    Looks to me Wolfram is doing some numerical integration. When I do a primitive numerical integration with excel I get about the same answers they do.
    Your answer is correct

    (How did you get this Ci and Si answer from wolfie ?)
     
  4. Dec 16, 2016 #3

    stevendaryl

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    I don't know off the bat how to derive the right answer, but contour integration by using residues only works for analytic functions, and the use of an absolute value, [itex]|x|[/itex], makes your integrand non-analytic.
     
  5. Dec 16, 2016 #4

    stevendaryl

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    That link is for the integrand [itex]\dfrac{e^{itz}}{1+z^2}[/itex], not [itex]\dfrac{e^{-|z|}}{1+z^2}[/itex]
     
  6. Dec 16, 2016 #5

    BvU

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    Yeah, I missed that completely. o:) o:) o:) though to be smart and use ##t=i##

    (For the numerics looked at ##[0,\infty>##)

    Changed your avatar, didn't you ? Less colorful but more cultural !
     
  7. Dec 16, 2016 #6

    stevendaryl

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    Last edited: Dec 16, 2016
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