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I know that $\int{\frac{1}{x}}dx$ is simply $\ln{(x)}+c$ (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach $e^{x{^2}}$. Perhaps a substitution?

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    I don't know enough to give a detailed answer, but to save you the trouble in the meantime: you don't. It can't be done.2012-10-16
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    taylor expansion?2012-10-16
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    ah. Taylor expansion will work, yes. I assumed you wanted a closed-form formula for the answer.2012-10-16
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    While $e^{x^2}$ has no closed-form formula, it is integrable on some intervals (such as $(-\infty, \infty)$, by switching to polar coordinates).2012-10-16
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    @LevDub, since the exponent is $+x^2$, this integral will diverge on any interval that has $\pm \infty$ as an endpoint2012-10-16
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    Oh, crap, you're right. I was pretending that the exponent was $-x^2$.2012-10-16
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    @LevDub This is in fact integrable on any finite interval. Be careful with your terminology.2012-10-16
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    http://math.arizona.edu/~mleslie/files/integrationtalk.pdf2012-10-16

2 Answers 2

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Actually, neither the antiderivative of $e^{x^2}$ nor $e^{-x^2}$ can be expressed in terms of 'elementary functions', so we simply define a new function called the error function by

$$\textrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{-\infty}^x e^{-t^2} dt.$$

We can also define a related function, the imaginary error function, by

$$\textrm{erfi}(z)=\frac{\textrm{erf}(iz)}{i}$$

(where $z\in\mathbb{C}$).

Then of course the map $z\mapsto\frac{\sqrt{\pi}}{2}\textrm{erfi}(z)$ is an antiderivative of $z\mapsto e^{z^2}$.

As is alluded to in the comments, the situation is more tractable for (improper) definite integrals of this form, e.g.

$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}.$$

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    How can you be sure it cannot be expressed in terms of 'elementary functions'? Is it a proved theorem that it cannot be expressed so? For me it looks like a theorem that cannot be proved with our time methods2012-10-16
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    Yes, this has been proven. Have a look at Liouville's theorem (from differential algebra, there are other "Liouville's theorems"), for example http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28differential_algebra%292012-10-16
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    wolfram alpha says something slightly different. Can you clarify?2012-10-16
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    What exactly is the difference you found? Oh wait, is it that they define the error function by integrating from zero instead from negative infinity? Because if that's the case, it doesn't matter, since the two definitions only differ by a constant.2012-10-16
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    @porton: See also http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral2012-10-17
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For the sake of accuracy and regarding some of the comments made:

1) $\int e^xdx=e^x+C$ and not $\ln(x)$. 2) The indefinite integral $\int e^{x^2}dx$ exists on any finite interval simply because the integrand is continuous. However, a primitive function can't be expresses as a combination of elementary functions (it is not a trivial proof that that is the case). 3) Using the Taylor expansion of $e^{x^2}$ one can integrate term by term to obtain a power series expansion for a primitive function and to obtain approximations of it. 4) The function $e^{x^2}$ is not integrable on $(-\infty ,\infty)$.