# Unintegrateable fn.[s]

1. Feb 26, 2008

### rock.freak667

Consider

$$\int e^{-x^2} dx$$

if that can't be expressed in terms of elementary functions, how did they compute

$$\int ^{\infty} _{- \infty} e^{-x^2} dx =\sqrt{\pi}$$

(I think I have the limits wrong, but I know it has $\infty$ as the upper or lower limit)

2. Feb 26, 2008

### sutupidmath

well, one way of doing so is i guess expanding $$e^{-x^{2}}$$ as a power series, using taylor series. But i also think one can compute it using double integrals. I have just heard about this though, since i have no idea how to deal with double integrals yet!

3. Feb 26, 2008

$$I = \int_{-\infty}^{\infty}e^{-x^2}dx$$
$$I^2 = \int_{-\infty}^{\infty}e^{-x^2} dx\times \int_{-\infty}^{\infty} e^{-y^2}dy$$

$$I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy$$
$$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-(r^2)}rdrd\theta$$
$$I^2 = \pi - \pi e^{-\infty}$$

Last edited: Feb 26, 2008
4. Feb 26, 2008

### sutupidmath

THis looks cute, although i do not understand a damn thing what u did! I mean i haven't yet dealt with double integrals!

5. Feb 26, 2008

### D H

Staff Emeritus
This is the Gaussian integral. The wikipedia article (link here) on this integral goes over the derivation of this integral and does a rigorous job (check out the "careful" proof of the identity).

6. Feb 26, 2008

EDIT: Note: Read the article D H posted, and not the gibberish below.

Honestly, I can't believe I remembered how to do it. I've seen it a couple of times in class. To me it is, a trick.

But here's the jist of it. That x^2 looks like a beast, and integrating from -infinity to infinity seems like a problem.

We know that if we multiply two exponentials e^u*e^y we get e^(u+y). So when we multiply the two integrals together and get x^2+y^2 this should be screaming, convert me into polar coordinates.

So we multiply the two integrals together, and convert the x^2+y^2 into r^2.

First though, why is that even possible? Well remember that when you integrate with "numbers" you get a number. What I mean by this is the following.

If we integrate $$\int_0^1 x dx$$ we get a number right? What about when we integrate $$\int_0^u x dx$$? Well the second case returns a function dependent on $u$.

So in the first case, $$\int_0^1 x dx$$, why not just call this a number, how about $I$. So this makes sense to be able to multiply two numbers together, eg. [itex] I\timesI = \int_0^1 x dx \times \int_0^1 x dx [/tex]. Think about why we can "push" them together.

I think the most interesting part about it, was changing to polar coordinates. The part where we change from sweeping out -infinity to infinity in the x and y direction in rectangular coordinates to sweeping out all values by rotating from 0 to 2pi and extending the "arm" from 0 to infinity.

Last edited: Feb 26, 2008
7. Feb 26, 2008

### John Creighto

There are other ways to solve the above as well. You will learn these if you take a course in complex variables.

8. Feb 27, 2008

### Big-T

You will also need Fubini's theorem in order to justify that you may in this case convert the product of two integrals into a double integral.