The so called error function $\operatorname{erf}(x)$ is defined as $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt,$$ and it is well known that $\operatorname{erf}(\infty)=1$.

Are there any other known closed-form values of $\operatorname{erf}(x)$, except for $\operatorname{erf}(0)$ and $\operatorname{erf}(\pm\infty)$?

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    $\begingroup$ What do you mean by "known"? $\endgroup$ Apr 8, 2013 at 22:19
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    $\begingroup$ Wolfram only knows additional "values" at erf($\pm i \infty$) (which are $\pm i\infty$), see functions.wolfram.com/GammaBetaErf/Erf/03. It's not a proof of nonexistence but it's rather discouraging $\endgroup$
    – Cocopuffs
    Apr 8, 2013 at 22:25
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    $\begingroup$ What do you mean by "actual example"? In what sense is, say, $\pi$ a known number but $\int_0^1 e^{-t^2} \, dt$ an unknown number? After all, $\pi$ is just $4 \int_0^1 \sqrt{1 - t^2} \, dt$. $\endgroup$ Apr 8, 2013 at 22:35
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    $\begingroup$ @QiaochuYuan: I believe I mean something like this: Does there exist any $x$ such that both $x$ and $\operatorname{erf}(x)$ can be expressed as values of some elementary function? $\endgroup$
    – Mårten W
    Apr 8, 2013 at 22:44
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    $\begingroup$ @Mårten: this is not what you want. Every real number is the value of some elementary function (namely the constant elementary function with value that real number). But see www-math.mit.edu/~tchow/closedform.pdf for another suggestion (with the definition in that paper I would be extremely surprised if the answer was yes, but I would also be extremely surprised if you could prove it). $\endgroup$ Apr 9, 2013 at 0:16

1 Answer 1


If the values aren't listed on the Wolfram function page, I would be surprised if you found them anywhere else. The only listed closed form values are for $0$, $\pm\infty$, and $\pm i\infty$. However, you can find various equivalent formulations, continued fractions, and the like on that page. A good reference, generally, for most well-known functions.


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