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Definite integral of gaussian

  1. Mar 22, 2013 #1
    Is there a closed form expression for the following definite integral?

    [itex]\int_{-∞}^{∞} exp(\frac{-|z|^2}{2{\sigma}^2}-\alpha |\mu + z|)dz[/itex]

    where [itex]z[/itex] is complex, and [itex]\alpha, \sigma, \mu[/itex] are real constants.

    I couldn't obtain an expression similar to Gaussian integral, so I couldn't take the integral. So, how can we obtain a closed form expression for such an integral?
     
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  3. Mar 22, 2013 #2

    HallsofIvy

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    If z is complex, what is the meaning of the limits of integration- what path are you taking from [itex]-\infty[/itex] to [itex]\infty[/itex]?
     
  4. Mar 22, 2013 #3
    Thank you for your reply, and I'm sorry for the incorrect statement of my question. Let me try to explain as follows:

    Let z = x + jy. Then, the limit of x is from -∞ to ∞, and the limit of y is also from -∞ to ∞. In this case, I'm not sure about the limit of z. However, we can now restate the integral as follows:


    [itex]\int_{-∞}^{∞}\int_{-∞}^{∞} exp(\frac{-(x^2+y^2)}{2{\sigma}^2}-\alpha ((x+\mu)^2 + y^2)^{1/2})dxdy[/itex]
     
    Last edited: Mar 22, 2013
  5. Mar 22, 2013 #4

    Ray Vickson

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    There is something very wrong here. Integrals with respect to a complex z are defined, but they are 1-dimensional, like "line integrals" in the 2-dimensional plane. Your first message suggested that is what you want, but your later message implies that is NOT the case, but that, instead, you want a double integral over the plane. Which is it? What do you *really* want?
     
  6. Mar 22, 2013 #5
    Actually, the second integral is what I want to ask.

    However, if I'm not wrong, the second integral can be expressed as a contour integral in the complex plane, which I have tried to write in my first post. If I'm wrong please correct ( or what should be the path of z? )
     
    Last edited: Mar 22, 2013
  7. Mar 23, 2013 #6

    Ray Vickson

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    I don't think you can do the integral exactly, but you can reduce it to a one-dimensional integration involving a Bessel function, and that can be done numerically. In your function, change variables to
    [tex] x = - \mu + r \cos(\theta), \: y = r \sin(\theta). [/tex]
    Your integrand f(x,y) becomes
    [tex]f = \exp \left(-\frac{r^2 + \mu^2 + 2 \alpha \sigma^2 r}{2 \sigma^2} \right)
    e^{b \cos(\theta)}, \;\; b = \mu r \sigma^2. [/tex]
    The integral you want is
    [tex] \int_0^{\infty} \int_0^{2 \pi} r f \, dr \, d\theta.[/tex]
    We can do the ##\theta## integration first, to get
    [tex] \text{Answer} =
    \int_0^{\infty} r \exp \left(-\frac{r^2 + \mu^2 + 2 \alpha \sigma^2 r}{2 \sigma^2} \right)
    \text{BessellI}(0,\mu r/\sigma^2),[/tex]
    where ##\text{BesselI}(0,v)## is a modified Bessel function of the first kind. This Bessel function is a solution of the Bessel differential equation
    [tex] x^2 y'' + xy' +x^2 y = 0, [/tex] and with intitial terms of its series given by
    [tex] y = 1 + \frac{1}{4} x^2 + \frac{1}{64} x^4 + \cdots .[/tex] (Knowing a few terms of this series helps to relate this definition of BesselI(0,v) to possible other definitions you may know or encounter.)
     
    Last edited: Mar 23, 2013
  8. Mar 25, 2013 #7
    Ray, thank you very much for your detailed answer.
     
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