Is there a proof that gives the evidence that there is no closed form for $\int e^{-x^2} \cdot dx$?
Or just because they were not able to find that elementary form for a long time of trying without any useful results then they told us that such a form does not exist?
Error function is solving the integral problem but it is not elementary function.
Applying the integral is simple by using Taylor series.
$$e^{-x^2}=\sum_{n=0}^{\infty} \frac {(-1)^n x^{2n}}{n!} =1-x^2+ \frac {x^4}{2!}-\frac {x^6}{3!}+\frac {x^8}{4!}-\cdots$$
$$\int e^{-x^2} \cdot dx =\sum_{n=0}^{\infty} \frac {(-1)^n x^{2n+1}}{n!(2n+1)} =x-\frac {x^3}{3}+ \frac {x^5}{2!5}-\frac {x^7}{3!7}+\frac {x^9}{4!9}-\cdots$$
And if there is no sufficient proof, I will be more confident that there is one and we should note that:
$\displaystyle\int_{-\infty}^{\infty} e^{-x^2} \cdot dx = \sqrt {\pi}$ which means the formula contains inverse trigonometric function/s.
$\displaystyle\arctan(x) =\sum_{n=0}^{\infty} \frac {(-1)^n x^{2n+1}}{(2n+1)} =x-\frac {x^3}{3}+ \frac {x^5}{5}-\frac {x^7}{7}+\frac {x^9}{9}-\cdots$ it is similar somehow to $\displaystyle\int e^{-x^2} \cdot dx $ expansion series.
$\arctan(x)$ curve looks closely like error function curve.
Laplace proof (1812) ends with $\displaystyle I^2=2[\arctan(s)]_{0}^{\infty} =\pi$ , where $\displaystyle I=\int_{-\infty}^{\infty} e^{-x^2} \cdot dx$
And finally, the above 4 notes do not mean that $\arctan(x)$ can give a solution, but with some intelligent handling using possible inverse trigonometric functions and other tools may lead to some useful results and that is what I believe if and only if there is no proof tells that no elementary function exists for $\int e^{-x^2} \cdot dx $
