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Complex Gaussian Integral - Cauchy Integral Theorem

  1. Mar 4, 2015 #1

    VVS

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    1. The problem statement, all variables and given/known data
    I have to prove that [itex]I(a,b)=\int_{-\infty}^{+\infty} exp(-ax^2+bx)dx=\sqrt{\frac{\pi}{a}}exp(b^2/4a)[/itex] where [itex]a,b\in\mathbb{C}[/itex].
    I have already shown that [itex]I(a,0)=\sqrt{\frac{\pi}{a}}[/itex].
    Now I am supposed to find a relation between [itex]I(a,0)[/itex] and [itex]\int_{-\infty}^{+\infty} exp(-a(x-c)^2)dx[/itex] where [itex]c\in\mathbb{C}[/itex] using the Cauchy integral theorem and prove using this the result above.

    2. Relevant equations
    The Cauchy integral theorem states that [itex]\oint_\gamma f(z) dz=2\pi i Res(f(z))[/itex].

    3. The attempt at a solution
    This is what I got, but I am pretty sure it doesn't lead me anywhere.
    Now I am not sure whether [itex]\int_{-\infty}^{+\infty} exp(-a(x-c)^2)dx[/itex] is analytic in c. But I think what I can do is integrate over a closed contour [itex]\gamma[/itex] and use Cauchy's integral theorem, change the order of integration and thus equate it to the Residue.
    [itex]\int_{-\infty}^{+\infty} \oint_\gamma exp(-a(x-c)^2)dcdx=2\pi i Res(f(z))[/itex]
    Then I can expand to yield:
    [itex]\int_{-\infty}^{+\infty} exp(-ax^2) \oint_\gamma exp(-ac^2+2axc)dcdx=2\pi i Res(f(z))[/itex]
    I can sort of see the relation between the integrals now but I am kind of stuck.
     
  2. jcsd
  3. Mar 5, 2015 #2

    Svein

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    Science Advisor

    I think you are complicating this a bit too much. After all, [itex] -ax^{2}+bx=-a(x^{2}-\frac{b}{a}x)=-a(x-\frac{b}{2a})^{2}+\frac{b^{2}}{4a}[/itex].
     
  4. Mar 20, 2015 #3

    Ray Vickson

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    Are you being forced to use Cauchy's integral theorem here? I really don't see how it helps or why anyone would think it is needed. To me, it just gets in the way. A simple change of variables does the trick.
     
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