The error function is defined as
$$\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt.$$
We know that the Gaussian integral is
$$\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}.$$
Because of the symmetry of the integrand it is also true, that
$$\int_{-\infty}^{0} e^{-x^2}\,dx=\int_{0}^{\infty} e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}.$$
From here and from the definition of definite integral we know two exact values of the $\operatorname{erf}$ function, these are $\operatorname{erf}(0)=0$ and $\operatorname{erf}(\infty)=1$.
Question. Is there an exact closed-form value of $\operatorname{erf}(z)$ for some $z\neq0,\infty\,$?
Because of $(3)$ and $(4)$ here, the question is equivalent to the following. Are there a closed-form expression for some $z\neq 0,\infty$ of the following confluent hypergeometric functions?
$${_1F_1}\left(\frac{1}{2};\frac{3}{2};-z^2\right),$$ $${_1F_1}\left(1;\frac{3}{2};z^2\right).$$
Or because of $(5)$ here with the lower incomplete gamma function, for some $z \neq 0,\infty$
$$ \gamma\left(\frac 12, z^2 \right).$$
More general, is there any $(a,b)$ nonzero and finite pair for that we have a closed-form of the following?
$$\int_a^b e^{-x^2} \, dx,$$
where $a<b$.
