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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?

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  • $\begingroup$ e^x^2 doesnt have a closed form but i think it could have a seires $\endgroup$ – Saketh Malyala Feb 24 '17 at 8:11
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    $\begingroup$ Where is the problem? The power series $\sum\limits_{n=0}^\infty \frac{x^{2n+1}}{n!(2n+1)}$ has radius of convergence $R=\infty$ and you differentiate it term by term. It is a very different matter to show that $e^{x^2}$ has no "elementary" antiderivative. $\endgroup$ – Jochen Feb 24 '17 at 8:14
  • $\begingroup$ Congrats for figuring out the integral of $e^{x^2}$ !!! $\endgroup$ – user399078 Feb 24 '17 at 8:27
  • $\begingroup$ you can evaluate first few terms and then reduce the rest as the error estimate. $\endgroup$ – user29418 Feb 24 '17 at 10:19
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    $\begingroup$ @PhilippeMalot Thanks but I cant read French. $\endgroup$ – SchrodingersCat Feb 24 '17 at 11:42
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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).

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

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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}

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protected by SchrodingersCat Oct 17 '17 at 16:25

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