While trying to solve a problem on my own (Maxwell-Juttner distribution). I found the following integral:
$$I = \int_{0}^{\infty} p^2 e^{-βmc^2 \sqrt{1 + (\frac{p}{mc})^2}} dp$$
I know the answer is supposed to be in terms of the modified Bessel function of the second kind $K_2 (βmc^2)$.
My guess was that the obvious way to go was using the formula I found in Wikipedia:
$$K_α (x) = \int_{0}^{\infty} e^{-x \cosh(t)} \cosh(αt) dt$$
Which gives:
$$K_2 (βmc^2) = \int_{0}^{\infty} e^{-βmc^2 \cosh(t)} \cosh(2t) dt$$
My idea was first to use the variable substitution $\frac{p}{mc} = u$ such that
$$I = (mc)^3 \int_{0}^{\infty} e^{-βmc^2 \sqrt{1 + u^2}} u^2 du$$
The integrals look very similar, but I still haven't been able to write $I$ in terms of $K_2 (βmc^2)$. My attempt was to make the substitution $u = \sinh(t)$ because $\sqrt{1 + u^2} = \sqrt{1 + \sinh ^2 (t)} = \cosh(t)$. so:
$$\frac{I}{(mc)^3} = \int_{0}^{\infty} e^{-βmc^2 \sqrt{1 + u^2}} u^2 du = \int_{0}^{\infty} e^{-βmc^2 \cosh(t)} \sinh ^2 (t) d(\sinh(t)) = \int_{0}^{\infty} e^{-βmc^2 \cosh(t)} \sinh ^2 (t) \cosh(t) dt$$
This however is not quite the same as the $K_2 (βmc^2)$ integral and so far I've found no way of actually expressing $I$ purely in terms of $K_2 (βmc^2)$ at all. What would be a good approach? I believe I am very close to the solution.