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I Integrate translated gaussian function by change of variable

  1. Dec 10, 2017 #1
    I want to integrate this:[tex] \int_0^∞ re^{-\frac{1}{2σ^2} (r-iσ^2q)^2} \, dr. [/tex]
    If I change the variable r into t with this relation:[tex]r-iσ^2q=t,[/tex]
    then the integral becomes[tex]\int_{-iσ^2q}^∞ (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt[/tex]
    so it seems I cannot use the famous gaussian integral formula. But I got the correct result for the imaginary part of this problem with just ignoring the lower limit(actually I just forgot to rearrange upper and lower limits but anyway):[tex]iσ^2q\int_{0}^{∞} e^{-\frac{1}{2σ^2} t^2}\,dt= \frac{iσ^2q}{2} \int_{-∞}^∞ e^{-\frac{1}{2σ^2}t^2} \, dt =\frac{iσ^2q}{2}\sqrt{2σ^2π}[/tex] So why are they the same: the integration over the interval {##[-iσ^2q, 0]+[0, ∞]##} and ##[0, ∞]##?
     
  2. jcsd
  3. Dec 10, 2017 #2

    haruspex

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    No, it becomes [tex]\int_{-iσ^2q}^{∞-iσ^2q} (t+iσ^2q)e^{-\frac{1}{2σ^2} t^2} \, dt[/tex], where the integral is understood to be along the straight line ##t=x-iσ^2q##.
     
  4. Dec 11, 2017 #3
    Yes, that's what my mentor noted on my homework, but I don't get it..
    Could you tellme how you think about my explanation below?
    Maybe I could write it as

    [tex]\int_{-iσ^2q}^{∞-iσ^2q}e^{-\frac{1}{2σ^2} t^2} \, dt[/tex]
    [tex] =\frac{1}{2} \int_{-∞-iσ^2q}^{∞-iσ^2q} e^{-\frac{1}{2σ^2}t^2} \, dt [/tex]
    [tex]≅\frac{1}{2} \int_{-∞}^{∞} e^{-\frac{1}{2σ^2}t^2} \, dt.[/tex]

    Sorry I don't know what the heck is wrong with the second code...Hope you recognize the limits:(
     
  5. Dec 11, 2017 #4

    haruspex

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    I do not understand how you got that form.
    Perhaps you should consider integrals around different contours. What do you know about integrals around poles in the complex plane?
     
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