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.

  • $\begingroup$ @Alex I guess I know it's a gamma function, as I mention in the question. The question is to find an easy way to evaluate this. $\endgroup$ – Cheeku Jan 25 '14 at 3:36

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.

  • $\begingroup$ That was helpful, thanks! $\endgroup$ – Cheeku Jan 25 '14 at 3:37
  • $\begingroup$ @Cheeku I'm glad. I'm not sure if there's any intuition, other than to know that there's a few none integrals involving $e$ and squares. $\endgroup$ – user61527 Jan 25 '14 at 3:38
  • $\begingroup$ I knew it was a gamma function, but didn't realize that a substitution would work if expanded. Note to self : Give it a shot before discarding a method. $\endgroup$ – Cheeku Jan 25 '14 at 3:39

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