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Homework Help: Integral transformation?

  1. Jul 16, 2014 #1
    Change of variables for integral

    1. The problem statement, all variables and given/known data

    Determine whether the following equality holds:

    [itex]\displaystyle\int_0^{\infty} \frac{e^{-\dfrac{(x^2+4z^2)}{4z}}}{2z} \, \mathrm{d}z = \displaystyle\int_0^{\infty} \frac{e^{-\dfrac{(1+z^2)x}{2z}}}{2z} \, \mathrm{d}z, \forall x,z \in \mathbb R_+[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I obtained the first integral after fiddling around with the integral

    [itex]K_0(x) := \displaystyle\int_0^{\infty} e^{-x\cdot\cosh{z}} \, \mathrm{d}z[/itex]

    using the substitution [itex]\tau := \dfrac{xe^z}{2}[/itex]. Note that [itex]K_0(x)[/itex] is a modified Bessel function of the second kind of order 0. The second integral was something I came across after playing around with Rohatgi's "product convolution" integral for independent exponential random variables for a while. Mathematica claims that the second integral is also [itex]K_0(x)[/itex], and numerical integration seems to show that the original equality holds. The integrals look similar enough that some simple transformation seems relevant, but I haven't managed to find one yet.

    Obviously, determining how Mathematica concluded that the second integral is [itex]K_0(x)[/itex] would be helpful, too.

    Last edited: Jul 16, 2014
  2. jcsd
  3. Jul 17, 2014 #2


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    Try substituting a simple multiple of z.
  4. Jul 17, 2014 #3
    Substituting [itex]u=2z[/itex] in the first integral yields

    [itex]\displaystyle\int_0^{\infty} \frac{e^{-\dfrac{(x^2+4z^2)}{4z}}}{2z} \, \mathrm{d}z = \displaystyle\int_0^{\infty} \frac{e^{-\dfrac{(x^2+u^2)}{2u}}}{2u} \, \mathrm{d}u[/itex].

    Not bad! I'll try out some more elementary transformations tomorrow. Thanks!
  5. Jul 17, 2014 #4


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    If u = 2z, the lower denominator of the transformed integral (originally 2z) should be u rather than 2u.
  6. Jul 17, 2014 #5
    Well, maybe I'm too tired, but isn't [itex]\mathrm{d}z=\dfrac{\mathrm{d}u}{2}[/itex]?
  7. Jul 17, 2014 #6


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    You are correct.
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