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Feb8-09, 02:30 PM
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Quote Quote by gabbagabbahey View Post
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