I have this integral:
$$\int_0^\infty \sqrt{x} e^{-\frac{x}{a}}dx$$
I am not sure how to solve this. I think it may involve the erf function (courtesy of wolfram) but I am not sure how to appropriately use this. The answer should be:
$$0.5(\sqrt\pi)(a^{3.2})$$
I would appreciate any help
$$ I = \int_0^{\infty} \sqrt{x} e^{-\frac{x}{a}} \ dx $$
$$ x = au \Rightarrow dx = a \ du $$
$$ \Rightarrow I = a\sqrt{a} \int_0^{\infty} x^{\frac{1}{2}}e^{-x} \ dx $$
$$ \Gamma(t) = (t-1)! = \int_0^{\infty} x^{t-1} e^{-x} \ dx $$
$$ \Rightarrow I = a\sqrt{a} \ \Gamma \left(\frac{3}{2} \right) = a^{\frac{3}{2}}\frac{\sqrt{\pi}}{2} $$
Hint: (I suppose you mean $a>0$ otherwise the integral diverges)
Make the substitution $t=\sqrt{x/a}$, this will transform the integral into something you should be able to handle.
