Quote by gabbagabbahey
The integral goes from 0 to infinity correct?...If so a simple substitution [itex]u=\beta\sqrt{x}[/itex] gives
[tex]\int_0^{\infty} x^{1/2}e^{\beta x} dx=\frac{2}{\beta} \int_0^{\infty} e^{\beta x} du[/tex]
For which the solution is well known and easily derivable.

yeah, that is true ;)
but you still have x, but you integrate over u, that is strange