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A Little Problem with a Simple Integral

  1. Dec 8, 2004 #1
    If we have

    [tex]P(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2} x^2},[/tex]


    [tex]I=\int^{+\infty}_{-\infty}dx \, P(x) \theta(x)
    = \int^{+\infty}_0 dx \, P(x)=\frac{1}{2},[/tex]

    where [tex]\theta[/tex] is the step function. Now, using its integral representation, we have

    I=\int^{+\infty}_{-\infty}dx \, P(x) \frac{1}{2\pi i} \int^{+\infty}_{-\infty} \frac{dk}{k} \, e^{ikx} = \frac{1}{2\pi i} \int^{+\infty}_{-\infty} \frac{dk}{k} \int^{+\infty}_{-\infty}dx \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2+ikx}= \frac{1}{2\pi i}\int^{+\infty}_{-\infty} \frac{dk}{k} e^{-\frac{1}{2} k^2} .

    But the integral over [tex]k[/tex] is calculated using residues and gives [tex]2\pi i[/tex] times the residue at [tex]k=0[/tex], what gives simply [tex]2\pi i[/tex] for the integral over [tex]k[/tex] and the wrong result 1 for the integral [tex]I[/tex]. I cannot see the catching. What is wrong with this calculation?
  2. jcsd
  3. Dec 8, 2004 #2


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    Homework Helper

    Could you explain how you got from the second to last to the last expression? Also, I don't think you can close that last integral, because k is squared. So it will only go to 0 as R goes to infinity if -(x^2-y^2)<0, or |x|>|y|, where k=x+iy, and this is not true over the entire contour.
    Last edited: Dec 8, 2004
  4. Dec 9, 2004 #3
    The passage you're referring to is a gaussian integration. I'm not so sure (maybe all the problem is here), but I think that I can consider the integrated function as [tex]e^{-|k|^2/2}[\tex], what gives k squared for the reals and has not the problem you mentioned.
  5. Dec 9, 2004 #4


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    I don't think you can do that. |k|^2 does not equal k^2 for complex numbers. In particular, the first is real while the second isn't.
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