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?

Answer 1

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.

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

Answer 2

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 $(-\infty ,\infty)$.