Does Integral Transformation Apply in Equality of Modified Bessel Functions?

[Combinatus]

Change of variables for integral

Homework Statement

Determine whether the following equality holds:

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

Homework Equations

The Attempt at a Solution

I obtained the first integral after fiddling around with the integral

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

using the substitution $\tau := \dfrac{xe^z}{2}$. Note that $K_0(x)$ 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 $K_0(x)$, 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 $K_0(x)$ would be helpful, too.

Hints?

 

[haruspex]

Try substituting a simple multiple of z.

 

[Combinatus]

Substituting $u=2z$ in the first integral yields

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

Not bad! I'll try out some more elementary transformations tomorrow. Thanks!

 

[SteamKing]

If u = 2z, the lower denominator of the transformed integral (originally 2z) should be u rather than 2u.

 

[Combinatus]

If u = 2z, the lower denominator of the transformed integral (originally 2z) should be u rather than 2u.

Well, maybe I'm too tired, but isn't $\mathrm{d}z=\dfrac{\mathrm{d}u}{2}$?

 

[SteamKing]

Well, maybe I'm too tired, but isn't $\mathrm{d}z=\dfrac{\mathrm{d}u}{2}$?

You are correct.