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Poisson integral

  1. Dec 13, 2013 #1
    ## \int^{\infty}_{-\infty}dxe^{-ax^2}=\sqrt{\frac{\pi}{a}}##
    Is it correct also when ##a## is complex?
  2. jcsd
  3. Dec 13, 2013 #2


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    It's correct, if [itex]\mathrm{Re} \; a>0[/itex].
  4. Dec 13, 2013 #3
    For a complex ##a##, there are two different roots. Which root is the correct one?
  5. Dec 14, 2013 #4


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    That's in fact a very good question. I've never thought about this before. I'm not sure whether the following is mathematically rigorous.

    I'd start with the simple case [itex]a=1[/itex]. The standard way to evaluate the integral is by setting
    [tex]I=\int_{-\infty}^{\infty} \exp(-x^2)>0.[/tex]
    [tex]I^2=\int_{\mathbb{R}^2} \mathrm{d} x \mathrm{d} y exp(-x^2-y^2).[/tex]
    Then we use polar coordinates in the [itex]xy[/itex] plane to get
    [tex]I^2=2 \pi \int_0^{\infty} \mathrm{d} r r \exp(-r^2)=-\pi \exp(-r^2)|_{r=0}^{\infty}=\pi.[/tex]
    Since [itex]I>0[/itex] we uniquely get
    with the usual positive square root for a positve real number.

    For real [itex]a>0[/itex] then you get by substitution [itex]y=\sqrt{a} x[/itex]
    [tex]I_a=\int_{-\infty}^{\infty} \mathrm{d} x \exp(-a x^2) = \frac{1}{\sqrt{a}} \int_{-\infty}^{\infty} \exp(-y^2)=\frac{\pi}{\sqrt{a}},[/tex]
    where again, I used the usual positive square root of a positive real number. You could as well have substituted [itex]y=-\sqrt{a} x[/itex], but then the infinite boundaries change signs, and thus the integral also flips signs, so that you get the same positive result as it must be.

    Now, for [itex]a \in \mathbb{C}[/itex], for convergence you obviously should have [itex]\mathrm{Re} \; a > 0[/itex]. Again we can take both roots in the substitution above, but it's most convenient to write
    [tex]a=|a| \exp(\mathrm{i} \varphi),[/tex]
    where [itex]\varphi \in (-\pi/2,\pi/2)[/itex] (this is one of many possible choices of the argument for a complex number with positive real part) and then use the square root as
    [tex]\sqrt{a}=\sqrt{|a|} \exp(\mathrm{i} \varphi/2), \quad \sqrt{|a|}>0. \qquad (*)[/tex]
    Now consider the Integral [itex]I_{a}[/itex] as an integral in the complex [itex]x[/itex] plane along the real axis. Then again we substitute [itex]x=\sqrt{a} t[/itex] with the meaning of [itex]\sqrt{a}[/itex] given by (*).

    The real integration path in the complex [itex]x[/itex] plane then maps to an integration path in the complex [itex]t[/itex] plane, which is a straight line through the origina, running from the lower left quadrant into the upper right (for [itex]\varphi>0[/itex]) or from the upper left to the lower right quadrant (for [itex]\varphi<0[/itex]). In both cases, you can define a closed path by adding the real axis (run from right to left in both cases) and two vertical parts at infinity. If you integrate [itex]\exp(-t^2)[/itex] along that closed path you get 0 due to Cauchy's integral theorem, and this means that instead to integrate along the straight line in the [itex]t[/itex] plane you can as well integrate along the real axis (in the normal positive sense). Thus we find
    [tex]I_a=\frac{\sqrt{\pi}}{|a|} \exp(-\mathrm{i} \varphi/2),[/tex]
    where the angle [itex]\varphi \in (-\pi/2,\pi/2)[/itex].
  6. Dec 14, 2013 #5
    (*) selects one the roots arbitrarily. Would that not mean that subsequent derivation inherits the arbitrariness?
  7. Dec 15, 2013 #6


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    You can of course choose the other root, i.e.,
    [tex]\sqrt{a}=-\sqrt{|a|} \exp(\mathrm{i} \varphi/2), \quad \varphi \in (-\pi/2,\pi/2).[/tex]
    Then in substituting
    [tex]t=-\sqrt{|a|} \exp(\mathrm{i} \varphi/2)x[/tex]
    the integration paths in the [itex]t[/itex] plane are straight lines through the origin either running from the upper right to the lower left quadrant (for [itex]\varphi>0[/itex]) or from the lower right to the upper left quadrant (for [itex]\varphi<0[/itex]).

    Then closing the path with the line running along the real axis runs from [itex]-\infty[/itex] to [itex]+\infty[/itex]. Thus the different sign in [itex]-1/\sqrt{|a|} \exp(-\mathrm{i} \varphi/2)[/itex] is compensated by the sign of the path along the real axis in the integral along the closed loop, giving 0 due to Cauchy's integral theorem. This shows that the integral is uniquely defined, no matter which root of [itex]a[/itex] you choose in the above substitution.
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