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