# (The area under the curve of the bell of Gauss)

$$Problem:$$ (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.

• It can be calculated. it is $\frac{\sqrt\pi}2\textrm{erf}(x_0)$ Dec 13 '19 at 15:00
• @Andrei the definition of $erf(x_0)$ makes that reasoning circular. You can't calculate it from that expression. Dec 13 '19 at 15:01
• @CyclotomicField correct, but the error function can be quickly approximated using some simple polynomials Dec 13 '19 at 15:08