# Definite Integral of function

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

$$I = \int_0^{\infty} \sqrt{x} e^{-\frac{x}{a}} \ dx$$

$$x = au \Rightarrow dx = a \ du$$

$$\Rightarrow I = a\sqrt{a} \int_0^{\infty} x^{\frac{1}{2}}e^{-x} \ dx$$

$$\Gamma(t) = (t-1)! = \int_0^{\infty} x^{t-1} e^{-x} \ dx$$

$$\Rightarrow I = a\sqrt{a} \ \Gamma \left(\frac{3}{2} \right) = a^{\frac{3}{2}}\frac{\sqrt{\pi}}{2}$$

• Don't use \frac on exponents it makes it impossible to read. Use a/b instead. – Ali Caglayan Aug 28 '13 at 21:34

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.

• your hint gives a very simple answer . (+1) – what'sup Aug 28 '13 at 21:04