Exact values of error function

Question

$$\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} \, dt.$$

$$\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}.$$

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 00:58
@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 01:00
so you are saying that if a closed form can be found for some subset of the whole real line. — Seyhmus Güngören, Oct 11 '14 at 23:55
@user153012: you may be interested in this question as well. — Mårten W, Oct 11 '17 at 09:01
obviously there exists numbers a and b such that $$\int_a^b e^{-x^2} dx=0.1$$ (but a and b might not have closed forms) — Lucenaposition, Jun 18 '24 at 07:11
This question is similar to: Non-trivial values of error function $\operatorname{erf}(x)$?. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Maxime Jaccon, Jan 30 '25 at 20:44

score 2 · Answer 1 · answered Apr 29 '25 at 18:17

Something you might find interesting is the Mellin transform of error function, which can possibly open the door for investigation into values. First we can take the Mellin transform of the error function's derivative: $$2\pi^{-\frac{1}{2}}\int\limits_{0}^{\infty}e^{-x^2}x^{s-1}\,dx =\pi^{-\frac{1}{2}}\int\limits_{0}^{\infty}e^{-x}x^{\frac{s}{2}-1}\,dx=\pi^{-\frac{1}{2}}\Gamma\left(\frac{s}{2}\right)$$ There's a Mellin transform identity which gives us: $$\int_{0}^{x}f\left(y\right)\,dy \leadsto -s^{-1}\int\limits_{0}^{\infty} f\left(x\right)x^{s}\,dx$$ So the mellin transform of $\operatorname{erf}\left(x\right)$ gives us: $$-s^{-1}\pi^{-\frac{1}{2}}\Gamma\left(\frac{s+1}{2}\right)$$ And to recover the error function: $$\operatorname{erf}\left(x\right) = -\frac{1}{2\pi^{\frac{3}{2}}i}\int_\limits{c-i\infty}^{c+i\infty}\frac{\Gamma\left(\frac{s+1}{2}\right)}{s}x^{-s}\,ds$$

Exact values of error function

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