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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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    e^x^2 doesnt have a closed form but i think it could have a seires – 2017-02-24
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    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. – 2017-02-24
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    Congrats for figuring out the integral of $e^{x^2}$ !!! – 2017-02-24
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    you can evaluate first few terms and then reduce the rest as the error estimate. – 2017-02-24
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    If you can read french, here is a famous exam from the École Normale Supérieure on elementary functions and why $x\mapsto e^{x^2}$ has no elementary antiderivative. http://pomux.free.fr/corriges-1995/pdf/m95lm1ea.pdf – 2017-02-24
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    @PhilippeMalot Thanks but I cant read French. – 2017-02-24

3 Answers 3

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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).

3

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

1

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}