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Tricky semi-infinite integral

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to integrate

    [tex]\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp[/tex].

    2. Relevant equations
    I know that

    [tex]\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} dp= \frac{\pi}{2b}e^{tb^2}erfc(\sqrt{a}x)[/tex]

    3. The attempt at a solution
    I rewrote the problem in terms of the complex exponential and reduced the integrand to a form similar to

    [tex]\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp= Re(\int^\infty_0 \frac{e^{-t(p-s)^2}}{{p^2+b^2}} dp)[/tex]

    where s is complex but I was stuck here. Any suggestions would be much appreciated.
  2. jcsd
  3. Feb 9, 2009 #2

    Tom Mattson

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    The integrand is an even function of p, so I would divide it by 2 and integrate over [itex](-\infty ,\infty )[/itex]. Then I would try to use the Residue Theorem. Are you familiar with that result?
  4. Feb 9, 2009 #3
    Yes I am familiar with it however not in this form. Are you suggesting forming a curve on the complex plane which encloses the points p=+/-ib. The integral over this curve would equal 2*pi*i* Sum of residues?
  5. Feb 9, 2009 #4

    Tom Mattson

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    Not quite. If the real axis is part of the contour, then you would close it with a semicircular arc in the upper half plane (or the lower half, your choice). In either case the contour only encloses one of the poles, not both.
  6. Feb 9, 2009 #5
    Cool. Thanks for the help! It was a very nice step.

    Here is the solution for anyone interested:

    \int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp = \frac{\pi e^{tb^2} cos(ibu)}{2|b|}}[/tex]
    Last edited: Feb 9, 2009
  7. Feb 9, 2009 #6
    Hi Tom. As my b is a real number, for the case b=0 this doesn't work. I want to find out

    the limit as b-> 0 of b * integral. I know that it is equal to zero but I am having proving this. All I really have to do is show that the integral with b=0 is finite.

    Any suggestions?
  8. Feb 10, 2009 #7

    Tom Mattson

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    How about setting b=0 in the original integral and calculating the residue of the pole of order 2 that results? You'll need a different contour of course.
  9. Feb 10, 2009 #8
    I ended up just using L'Hopital's rule and it came out quite easily. Thanks Tom you were a lot of help! Very much appreciated.
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