I have to evaluate the following expression : $$\int^{\infty}_{0} x^{\frac{3}{2}}e^{-x}$$
Wolfram|Alpha evaluates to $\frac{3\sqrt{\pi}}{4}$.
I don't see how we got there.
A hint would be helpful.
My attempts were to use the "By Parts" rule, when I realized that this is the famous Gamma function. There are several sources on internet which give a way to evaluate this, but I want a prood, as intuitive you can make it.
The key point is the Gaussian integral:
$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$
This can be found in many sources (including this website) so I won't reproduce the proof here.
To get to the point where this is useful, integrate by parts once to find
$$\int_0^{\infty} x^{3/2} e^{-x} dx = \frac 3 2 \int_0^{\infty} x^{1/2} e^{-x} dx$$
Do it again to find that this is equal to
$$\frac{3}{4} \int_0^{\infty} x^{-1/2} e^{-x} dx$$
Now substitute $u = x^{1/2}$ so that $du = \frac 1 2 x^{-1/2} dx$; this changes the integral to
$$\int_{0}^{\infty} x^{-1/2} e^{-x} dx = 2 \int_0^{\infty} e^{-u^2} du = \int_{-\infty}^{\infty} e^{-u^2} du$$
Using the above, we're done.