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## Main Question or Discussion Point

integral of e^(-x^2)dx?

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integral of e^(-x^2)dx?

AKG

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Usually done as a double integral with polar coordinates. I think its Fubini's Theorem.

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Here's a nice Wikipedia link to show you the double integral. It's a page on the Gaussian Integral, which is the same integral as yours, but a definite one. It convertswhozum said:Usually done as a double integral with polar coordinates. I think its Fubini's Theorem.

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

to polar coordinates to show how it is evaluated.

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LeonhardEuler

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AKG

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Does there exist a proof that says that it is impossible to produce a closed anti derivative of f(x)=e^-x^2? I would like to see it if it does exist. If it is impossible to generate an anti derivative through the use of elementary functions could some one please produce an anti derivative with non-elementary functions?

LeonhardEuler

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I have not seen it myself, but I heard that it is quite long (a dozen pages, I heard)Watts said:Does there exist a proof that says that it is impossible to produce a closed anti derivative of f(x)=e^-x^2? I would like to see it if it does exist.

Well, there's an answer to that question, but it's kind of disapointing. By definition:Watts said:If it is impossible to generate an anti derivative through the use of elementary functions could some one please produce an anti derivative with non-elementary functions?

[tex]erf(z) \equiv \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{t^2}dt[/tex]

So then:

[tex]\int e^{x^2}dx=\frac{\sqrt{\pi}}{2}erf(z) +C[/tex]

Kind of disapointing because it just is basically defined as the answer to the question. But it is not so much different from other functions. It doesn't give exact answers, but niether do most functions. For example look at sin(x). It is defined by an integral, but in fact, in some books the real natural logarithm is defined by:

[tex]\ln{x} \equiv \int_{1}^{x}\frac{1}{t}dt[/tex]

The erf function could as well be defined by a taylor series as some other functions are, by integrating e^(x^2)'s Taylor series.

shmoe

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Yes, it is impossible. This is important enough that it's in just about every treatment I've seen on "integration in finite terms" (<-buzzwords for google), so there will be many options, such asWatts said:Does there exist a proof that says that it is impossible to produce a closed anti derivative of f(x)=e^-x^2? I would like to see it if it does exist.

http://www.claymath.org/programs/outreach/academy/LectureNotes05/Conrad.pdf [Broken]

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