Antiderivative of $e^{x^2}$: Correct or fallacy?

Question

Please check where is the mistake in this following process. I could not make out.

$$e^x=\sum_\limits{n=0}^\infty \frac{x^n}{n!}$$ $$\implies e^{x^2}=\sum_\limits{n=0}^\infty \frac{{(x^2)}^n}{n!}=\sum_\limits{n=0}^\infty \frac{x^{2n}}{n!}$$ $$\implies \int e^{x^2} dx=\int \sum_\limits{n=0}^\infty \frac{x^{2n}}{n!} dx$$ $$\implies \int e^{x^2} dx=\sum_\limits{n=0}^\infty \frac{x^{2n+1}}{n!(2n+1)} + c$$

But $e^{x^2}$ has no antiderivative as such. How is then thus possible?

Is this correct or just a fallacy?

Answer 1

But $e^{x^2}$ has no antiderivative as such.

This is false. $e^{x^2}$ certainly has an antiderivative, since all continuous functions have antiderivatives. It's just that the antiderivative cannot be written with our standard set of functions (i.e. polynomials, trigonometrics, and exponentials).

So yes, what you did is entirely correct (because all the sums are absolutely convergent, the operations are valid).

Answer 2

What you have done looks correct. $e^{x^2}$ does have an antiderivative, it is just not expressible as any combination of elementary functions.

Answer 3

You have found a series representation of the imaginary error function

\begin{equation} \mathrm{erfi}(x) = \frac{2}{\sqrt{\pi}} \int\limits_{0}^{x} \mathrm{e}^{z^2} dz = \frac{2}{\sqrt{\pi}} \sum_{n = 0}^{\infty} \frac{x^{2n+1}}{n!(2n+1)} \end{equation}