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

  • Thread starter rock.freak667
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In summary, the integral \int ^{\infty} _{-\infty} e^{-x^2} dx=\sqrt{\pi} is known as the Gaussian integral and can be solved using various methods, such as expanding it as a power series or using double integrals and converting to polar coordinates. It is a useful integral in mathematics and has various applications.
  • #1
rock.freak667
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Consider

[tex]\int e^{-x^2} dx[/tex]if that can't be expressed in terms of elementary functions, how did they compute

[tex] \int ^{\infty} _{- \infty} e^{-x^2} dx =\sqrt{\pi}[/tex]
(I think I have the limits wrong, but I know it has [itex]\infty[/itex] as the upper or lower limit)
 
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  • #2
rock.freak667 said:
Consider

[tex]\int e^{-x^2} dx[/tex]


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

[tex] \int ^{\infty} _{\infty} e^{-x^2} dx =\sqrt{\pi}[/tex]
well, one way of doing so is i guess expanding [tex] e^{-x^{2}}[/tex] 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
[tex] I = \int_{-\infty}^{\infty}e^{-x^2}dx[/tex]
[tex] I^2 = \int_{-\infty}^{\infty}e^{-x^2} dx\times \int_{-\infty}^{\infty} e^{-y^2}dy[/tex]

[tex] I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy[/tex]
[tex] I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-(r^2)}rdrd\theta[/tex]
[tex] I^2 = \pi - \pi e^{-\infty}[/tex]
 
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  • #4
FrogPad said:
[tex] I = \int_{-\infty}^{\infty}e^{-x^2}dx[/tex]
[tex] I^2 = \int_{-\infty}^{\infty}e^{-x^2} dx\times \int_{-\infty}^{\infty} e^{-y^2}dy[/tex]

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

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
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).
 
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  • #6
sutupidmath said:
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 [tex] \int_0^1 x dx [/tex] we get a number right? What about when we integrate [tex] \int_0^u x dx [/tex]? Well the second case returns a function dependent on [itex] u [/itex].

So in the first case, [tex] \int_0^1 x dx [/tex], why not just call this a number, how about [itex] I [/itex]. 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.
 
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  • #7
There are other ways to solve the above as well. You will learn these if you take a course in complex variables.
 
  • #8
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.
 

What is an "Unintegrateable fn."[s]?

An "Unintegrateable fn."[s] is a mathematical function that cannot be integrated using traditional methods. This means that it cannot be expressed as a single, closed-form equation, and may require more advanced mathematical techniques to evaluate.

Why are some functions unintegrateable?

There are several reasons why a function may be unintegrateable. One common reason is that the function may be too complex or involve too many variables for traditional integration techniques to handle. Another reason is that the function may contain special functions, such as trigonometric or exponential functions, that do not have simple antiderivatives.

Can unintegrateable functions be evaluated or used in calculations?

Yes, unintegrateable functions can still be evaluated and used in calculations. While they may not have a closed-form integral, they can still be approximated or evaluated numerically using methods such as numerical integration or computer algorithms.

Are there any techniques that can be used to integrate unintegrateable functions?

Yes, there are advanced techniques that can be used to approximate or evaluate unintegrateable functions. These techniques include numerical integration, series expansions, and special functions such as the Gamma function.

Are there any real-world applications for unintegrateable functions?

Yes, unintegrateable functions have many real-world applications, especially in physics and engineering. For example, many physical phenomena, such as fluid dynamics and electromagnetic fields, can be described by unintegrateable functions. These functions are also commonly used in computer programming and data analysis.

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