Is There a Way to Compute the Unintegrateable Function e^{-x^2}?

[rock.freak667]

Consider

\int e^{-x^2} dxif 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)

 

[sutupidmath]

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}

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!

 

[FrogPad]

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}

 

[sutupidmath]

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}

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

 

[D H]

This is the Gaussian integral. The wikipedia article (http://en.wikipedia.org/wiki/Gaussian_integral" ) on this integral goes over the derivation of this integral and does a rigorous job (check out the "careful" proof of the identity).

 

[FrogPad]

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

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. I\timesI = \int_0^1 x dx \times \int_0^1 x dx [/tex]. Think about why we can &quot;push&quot; them together. <br /> <br /> 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 &quot;arm&quot; from 0 to infinity.

 

[John Creighto]

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

 

[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.