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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 $\operatorname{erf}(\pm\infty)$?

Mårten W
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  • Interesting question, but I've never seen anything other than those three. – robjohn Apr 08 '13 at 22:15
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    What do you mean by "known"? – Qiaochu Yuan Apr 08 '13 at 22:19
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    Wolfram only knows additional "values" at erf($\pm i \infty$) (which are $\pm i\infty$), see http://functions.wolfram.com/GammaBetaErf/Erf/03/. It's not a proof of nonexistence but it's rather discouraging – Cocopuffs Apr 08 '13 at 22:25
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    @QiaochuYuan: I mean an actual example, as opposed to existence proofs (although that would also be interesting). – Mårten W Apr 08 '13 at 22:33
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    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 $4 \int_0^1 \sqrt{1 - t^2} , dt$. – Qiaochu Yuan Apr 08 '13 at 22:35
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    @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? – Mårten W Apr 08 '13 at 22:44
  • @MårtenW: See here. – Mhenni Benghorbal Apr 08 '13 at 23:17
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    @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 http://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). – Qiaochu Yuan Apr 09 '13 at 00:16
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    @robjohn There is another one. Kindly see my answer below. – Tito Piezas III Oct 26 '23 at 19:16
  • Values $z\infty,a i\pm \infty$ give closed forms. For example Wolfram Alpha implies $\text{erf}((1-i)\infty)=1$ and states $\text{erf}(i-\infty)=-1$ – Тyma Gaidash Oct 27 '23 at 00:22
  • Let's name the mathematical problem: We want to see where the values and the arguments of the error function both are elementary numbers or closed-form numbers. – IV_ Mar 02 '24 at 14:15

2 Answers2

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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.

HackerBoss
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I'm 10 years late for this 2013 party, but better late than never. Yes, there is another closed-form value of $\operatorname{erf}(x)$. This is,

$$\operatorname{erf}\Big(\sqrt{\tfrac12}\Big)=1-\frac{\Gamma\big(\tfrac12,\tfrac12\big)}{\Gamma\big(\tfrac12\big)}=0.68268949\dots$$

the decimal expansion of which is https://oeis.org/A178647. This was found using a beautiful identity by Ramanujan (which I was coincidentally investigating),

$$\sqrt{\frac{\pi\,e}{2}} =1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots\color{blue}+\,\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

Since $\Gamma\big(\tfrac12\big)=\sqrt{\pi}$ and the series and continued fraction have closed-forms, Ramanujan's identity can also be expressed in two ways,

\begin{align} \Gamma\big(\tfrac12\big)\sqrt{\frac{e}{2}} &= \,\sqrt{\frac{e}{2}} \times\,\Gamma\big(\tfrac12\big)\,\operatorname{erf}\Big(\sqrt{\tfrac 12}\Big) \;\color{blue}+\; \sqrt{\frac{e}{2}} \times\,\Gamma\big(\tfrac12\big)\left(1-\operatorname{erf}\Big(\sqrt{\tfrac 12}\Big)\right)\\ &= \sqrt{\frac{e}{2}}\times\Big(\Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)\Big) \color{blue}+\, \sqrt{\frac{e}{2}}\times\Big(\Gamma\big(\tfrac12,\tfrac12\big)\Big) \end{align}

where the first way is in this post, and the second in this post. Equating the first addends,

$$\Gamma\big(\tfrac12\big)\,\operatorname{erf}\Big(\sqrt{\tfrac 12}\Big) = \Gamma\big(\tfrac12\big)-\Gamma\big(\tfrac12,\tfrac12\big)$$

one can then easily solve for $\operatorname{erf}(x)$.

Tito Piezas III
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    This uses a more generalized function, $\Gamma(a,x)$, as a non-trivial value of the error function. One also may use $\text{erf}(x)=1-\frac{\Gamma(\frac12,x^2)}{\Gamma(\frac12)}$ to get a closed form for any positive value of the error function. Are you sure it cannot be “expressed as values of some elementary function” as said in a comment by the OP? – Тyma Gaidash Oct 26 '23 at 20:54
  • @ТymaGaidash I see. At least I found the relation via a rather unusual context. Regarding the OP's statement, I've been in fact trying to find expressions for $\Gamma\big(\tfrac12,\tfrac12\big)$ and $\Gamma\big(\tfrac13,\tfrac13\big)$ using other functions since they have nicer cfracs compared to others. Maybe the polygamma function? – Tito Piezas III Oct 27 '23 at 02:08