# Evaluate this (improper) integral

1. Oct 18, 2008

### kingwinner

1. The problem statement, all variables and given/known data
Evaluate

∫ x1/2 e-x dx
0
(This is the gamma function evaluated at 3/2)

2. Relevant equations

3. The attempt at a solution
I've tried integration by parts, but it doesn't seem to work.

Thanks for any help!

2. Oct 18, 2008

### jjou

I tried integration by parts too, and it was giving me trouble. So I did some research. The solution I found hopefully is not the only one, since it's a little complicated (in my opinion). Here are the basic steps:

1. Show that $$G(\frac{3}{2})=\frac{1}{2}G(\frac{1}{2})$$ where G is the gamma function.
(This step is not difficult. Use integration by parts and it follows immediately.)

2. Show that $$G(\frac{1}{2})=\sqrt{\pi}$$.
To do this, you change variables: let $$u=\sqrt{x}$$. After this, you'll see that you have exactly a Gaussian integral (which is where the $$\sqrt{\pi}$$ comes from).

Have you seen a proof / calculation of the Gaussian integral before? If not, there's a really nice proof of it on Wikipedia:
http://en.wikipedia.org/wiki/Gaussian_integral

If you haven't seen any of this before, I'm hoping someone else will get a simpler solution for this integral.

3. Oct 18, 2008

### HallsofIvy

Staff Emeritus
The reason the "Gamma" function is given a special name is that it CANNOT be written in terms of elementary functions. If you let u= x1/2, then du= (1/2)x-1/2dx so dx= 2x1/2du. Of course $e^{-x}= e^{-u^2}$ so
$$\int_0^\infty x^{1/2}e^x dx= 2\int_0^\infty e^{-u^2}du$$
The Gamma function and Gaussian function are pretty much interchangable.

Here is the standard way of finding the Gaussian integral, which you will find in most (multi-variable) Calculus books.
Let $$I= \int_0^\infty e^{-x^2} dx$$

(I see now, after all that work, that this is exactly what jjou's link gives!)
Then it is certainly also true that $$I= \int_0^\infty e^{-y^2}dy$$
Multiplying those:
$$I^2= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy$$
which we can interpret as the double integral
$$= \int_{x=0}^\infty \int_{y= 0}^\infty e^{-x^2-y^2}dydx[/itex] over the first quadrant of the plane. Switching to polar coordinates, that is [tex]\int_{\theta= 0}^{2\pi} \int_{r= 0}^\infty e^{-r^2} (r dr d\theta)$$
$$= 2\pi \int_{r=0}^\infty e^{-r^2} rdr$$

Now having that "r" in the integrand makes all the difference. Let u= r2 so that du= 2r dr or (1/2)du= r dr. The integral becomes
$$2\pi (1/2)\int_{u= 0}^\infty e^{-u}du= -\pi e^{-u}\right|_0^\infty= \pi$$
and, since that was I2,
$$I= \int_0^\infty e^{-x^2}dx= \sqrt{\pi}$$

Since
$$\int_0^\infty x^{1/2}e^{-x}dx= 2\int_0^\infty e^{-u^2}du$$

$$\int_0^\infty x^{1/2}e^{-x}dx= 2\sqrt{\pi}$$

(After all that work I see that this is exactly what the jjou's link gives! Oh, well, consider it confirmation.)