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$Problem:$ (The area under the curve of the bell of Gauss)

It is known that $ I =\int_0 ^ \infty e ^ {- x ^ 2} dx$ is possible to calculate, because ...

$$I^2=\left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right)$$ $$=\int_0^\infty e^{-(x^2+y^2)}dxdy=\int_0^\infty \int_0^{\frac{\pi}{2}} e^{-r^2}rdrd\theta$$ $$=-\frac{\pi}{4}\int_0^\infty e^{-r^2}(-2r)dr=\frac{\pi}{4}$$ Hence $I=\frac{\sqrt{\pi}}{2}$, My question is: why for $ x_0> 0 $ it is not possible to calculate $\int_0 ^ {x_0} e ^ {- x ^ 2} dx $? Is there any proof of this fact? Thanks for your attention.

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  • $\begingroup$ It can be calculated. it is $\frac{\sqrt\pi}2\textrm{erf}(x_0)$ $\endgroup$
    – Andrei
    Dec 13 '19 at 15:00
  • $\begingroup$ @Andrei the definition of $erf(x_0)$ makes that reasoning circular. You can't calculate it from that expression. $\endgroup$ Dec 13 '19 at 15:01
  • $\begingroup$ @CyclotomicField correct, but the error function can be quickly approximated using some simple polynomials $\endgroup$
    – Andrei
    Dec 13 '19 at 15:08
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It is proven that there are no known anti-derivative in terms of known functions. You can find such a proof here in the paper "Impossibility theorems for elementary integration" by Brian Conrad.

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