How do you integrate $e^{x^2}$?

Question

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?

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. — Robert Mastragostino, Oct 16 '12 at 21:06
ah. Taylor expansion will work, yes. I assumed you wanted a closed-form formula for the answer. — Robert Mastragostino, Oct 16 '12 at 21:09
While $e^{x^2}$ has no closed-form formula, it is integrable on some intervals (such as $(-\infty, \infty)$, by switching to polar coordinates). — GMB, Oct 16 '12 at 21:12
@LevDub, since the exponent is $+x^2$, this integral will diverge on any interval that has $\pm \infty$ as an endpoint — Ganesh, Oct 16 '12 at 21:21
Oh, crap, you're right. I was pretending that the exponent was $-x^2$. — GMB, Oct 16 '12 at 21:23
@LevDub This is in fact integrable on any finite interval. Be careful with your terminology. — Potato, Oct 16 '12 at 21:42

score 12 · Accepted Answer · answered Oct 16 '12 at 21:22

12

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.

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

answered Oct 16 '12 at 21:22

Aleksandar Bahat

214
3
4

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 methods – porton Oct 16 '12 at 21:25
5

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%29 – mrf Oct 16 '12 at 21:53
wolfram alpha says something slightly different. Can you clarify? – Magpie Oct 16 '12 at 22:22
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. – Aleksandar Bahat Oct 16 '12 at 23:58
@porton: See also http://math.stackexchange.com/questions/155/how-can-you-prove-that-a-function-has-no-closed-form-integral – Hans Lundmark Oct 17 '12 at 10:08

score 2 · Answer 2 · edited Oct 16 '12 at 22:31

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 $(-\infty ,\infty)$.

How do you integrate $e^{x^2}$?

2 Answers2

Linked

Related