I've been stuck with calculating the intgral of the following problem. Can you help me?
$$\int e^{x^2} \mathrm{d}x$$
I know that, by Using the definition of gamma and beta functions, we have $\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x = \sqrt{\pi}$.
Thanks in advance.
Among the best things we can do about this one is to expand the function into the Taylor series and formally integrate term by term: $$\int e^{x^2}dx= \sum_{k=0}^\infty \int\frac{x^{2k}}{k!}dx= \sum_{k=0}^\infty\frac{x^{2k+1}}{(2k+1)k!}+C$$
AD's point is right, however it can be written down in terms of non-elementary functions so, we make new function as its evaluation.
$\operatorname{erf} (x) $ is the error function, defined as $\operatorname{erf}(z)= \frac {2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt$
If $I(x)=\int e^{-x^2}dx$, $$\operatorname{erf}(x)= \frac {2}{\sqrt{\pi}} (I(x)-I(0)) \\ I(x)=\int e^{-x^2}dx=\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+I(0) = \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+ C$$