# Exact values of error function

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 $a<b$.

• for the error function there is no known closed form. Only very good approximations. But if you are talking about some specific points, it doesnt make sense because it will definitely pass through some $\pi$-like known values or some rational numbers. – Seyhmus Güngören Oct 11 '14 at 0:58
• @SeyhmusGüngören Yes I know that the error function has not closed-form. I'm talking about a closed-form at some $z$ beside $0$ and $\infty$. – user153012 Oct 11 '14 at 1:00
• See my answer to your other question. – Lucian Oct 11 '14 at 2:33
• The third case is $z=-\infty$. – GEdgar Oct 12 '14 at 0:54
• @user153012: you may be interested in this question as well. – Mårten W Oct 11 '17 at 9:01