# Non-trivial values of error function $\operatorname{erf}(x)$?

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

• What do you mean by "known"? Apr 8, 2013 at 22:19
• 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 Apr 8, 2013 at 22:25
• 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 $4 \int_0^1 \sqrt{1 - t^2} \, dt$. Apr 8, 2013 at 22:35
• @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? Apr 8, 2013 at 22:44
• @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). Apr 9, 2013 at 0:16

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.

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 $$\operatorname{erf}(x)$$.

• 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? Oct 26, 2023 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? Oct 27, 2023 at 2:08