1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You are correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integral transformation?
Loading...