Let $\mathrm{erf}(x) := \frac{2}{\sqrt{\pi}} \int_{-\infty}^x \exp(-t^2) \, dt$ be the error function $\mathrm{erf}: \mathbb{R} \to (-1,1)$. It is monotonously increasing and therefore has an inverse $\mathrm{erf}^{-1}: (-1,1) \to \mathbb{R}$.

Now I have two questions:

1) On https://en.wikipedia.org/wiki/Error_function#Inverse_functions it is said that $\mathrm{erf}^{-1}$ can be extended to the open unit disc. Unfortunately, there is no reference given for this. Anyone knows a reference for the derivation of the MacLaurin series of $\mathrm{erf}^{-1}$?

2) There are so-called Hardy spaces $H^p$ of functions analytic on the open disc which satisfy $$\sup_{0<r<1}\left(\frac{1}{2\pi} \int_0^{2\pi}\left|f \left (re^{i\theta}\right )\right|^p \; \mathrm{d}\theta\right)^\frac{1}{p}<\infty.$$ see https://en.wikipedia.org/wiki/Hardy_space#Hardy_spaces_for_the_unit_disk.

I wonder wether $\mathrm{erf}^{-1}$ belongs to some $H^p$ with $p<\infty$?

As it was pointed out on math.SE, where I asked this question before (https://math.stackexchange.com/questions/1546570/inverse-error-function-its-analytic-continuation-and-hardy-space), the MacLaurin coefficients seem to be in $\ell^2$ (which implies membership in $H^2$), but can this also be proven?


  • $\begingroup$ With your definition erf maps R onto $(0,2)$ not $(-1,1)$. $\endgroup$ Dec 1, 2015 at 14:18

1 Answer 1


To obtain a function that maps $R\to(-1,1)$ one has to define the error function as $f(x)=(2/\sqrt{\pi})\int_0^x\exp(-x^2)dx$, not as you defined.

With this definition it is an entire function which has no critical points (that is $f'(z)\neq 0,\; z\in C$) and two finite asymptotic values, $\pm 1$. From the general theory of singularities of the inverse of an analytic function follows that these two asymptotic values are logarithmic singularities of the inverse, and that the inverse analytic branches exist in any simply connected domain which does not contain $\pm1$. In particular there is an inverse branch in the unit disk satisfying $f^{-1}(0)=0$, and this branch is unique. An explicit expression for its Taylor coefficients is given in Wikipedia, but it is complicated. It is obtained from the Burmann-Lagrange formula.

Now this inverse branch has two logarithmic singularities at $(-1,1)$, therefore it behaves like $\sqrt{\log(z\pm1)}$ at these points and is analytic at all other points. It follows that the inverse branch belongs to all $H_p$ with any $p>0$, in particular it belongs to $H_2$ and thus its coefficients belong to $\ell^2$ by Parseval. For some general information on the singularities of inverse functions you may look in


Of course, for an elementary function $f$ you do not need any general theory explained in this preprint.


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