How do I expand $\int e^{x^2} \, dx$? I know this cannot be finitely expanded but how can I expand it using a Taylor or a Power series? $$\int e^{x^2} \, dx=?$$
 A: $$e^x = 1 + x + \frac{x^2}{2!} + \cdots$$
$$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \cdots$$
$$\int e^{x^2} \, dx = x+\frac{1}{3 \cdot 1!}x^3 +\frac{x^5}{5 \cdot 2!}+\cdots+C$$
A: That it can't be finitely expanded depends on which building blocks you're accepting. The Gaussian is that common that it's primitive function has a name of it's own:
$$\operatorname{erf}(x) = {2\over\sqrt\pi}\int_0^x e^{-t^2}dt$$
Since it's the antiderivate of $2e^{-x^2}/\sqrt\pi$ it's analytical with the McLaurin expansion
$$\operatorname{erf}(z) = {2z\over\sqrt\pi}\sum{(-1)^nz^{2n}\over n!(2n+1)}$$ 
On the other hand the antiderivate of $e^{x^2}$ is also analytical with the McLaurin expansion 
$$\int e^{x^2}dx = x\sum{x^{2n}\over n!(2n+1)}$$ 
Now consider
$$\operatorname{erf}(ix) = 
{2ix\over\sqrt\pi}\sum{(-1)^n(ix)^{2n}\over n!(2n+1)} = 
{2ix\over\sqrt\pi}\sum{(-1)^n(-1)^n(x)^{2n}\over n!(2n+1)} = 
{2ix\over\sqrt\pi}\sum{(x)^{2n}\over n!(2n+1)}$$
And compare that of the antiderivate of $e^{x^2}$ we would have:
$$\int e^{x^2}dx = {\sqrt\pi\over2i}\operatorname{erf}(ix)$$
