The error function is define like this: $$\operatorname{erf}(x):=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$

If i take the derivative i get $$\operatorname{erf'}(x)=\frac{2}{\sqrt\pi}e^{-x^2}$$because say $$\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt=\frac{2}{\sqrt{\pi}}\left(G(x)-G(0)\right)$$and $G(0)$ is constant.

Because $\forall x,\frac{2}{\sqrt\pi}e^{-x^2}>0$ we know that $\operatorname{erf}(x)$ is strictly increasing, so it suppose to have an inverse function.

I can understand few of things about $\operatorname{erf^{-1}}(x)$:

  • the domain is $\operatorname{erf}((-\infty,\infty))=(-1,1)$

  • the codomain is $(-\infty,\infty)$

  • 3 points:

    • $\lim\limits_{x\to\infty}\operatorname{erf}(x)=1\implies\lim\limits_{x\to1}\operatorname{erf^{-1}}(x)=\infty$
    • $\operatorname{erf}(0)=0\implies \operatorname{erf^{-1}}(0)=0$
    • $\lim\limits_{x\to-\infty}\operatorname{erf}(x)=-1\implies\lim\limits_{x\to-1}\operatorname{erf^{-1}}(x)=-\infty$
  • it is strictly increasing

But apart from those 4 things I can't think about any other information.

So my question is the following: Is there a way to define inverse error function using combination of known functions and transforms? And if not, what more can we know about the inverse function?

  • $\begingroup$ You might look at L. Carlitz, The inverse of the error function, Pacific J. Math. Volume 13, Number 2 (1963), 459-470, for some properties of the Maclaurin series $$\text{erf}^{-1}(x) = \frac{\sqrt{\pi} x}{2} + \frac{\pi^{3/2} x^3}{24} + \frac{7 \pi^{5/2} x^5}{960} + \ldots$$ $\endgroup$ – Robert Israel Dec 20 '17 at 22:44

Recall that $$\frac{d}{dx}f^{-1}(x)=\frac{1}{(f'\circ f^{-1})(x)}$$ and so we have $$\frac{d}{dx}\text{erf}^{-1}(x)=\frac{1}{(\text{erf}'\circ \text{erf}^{-1} )(x)}$$ or $$\frac{d}{dx}\text{erf}^{-1}(x)=\frac{\sqrt \pi}{2}e^{(\text{erf}^{-1}(x))^2}$$ and so the inverse error function can be defined by the differential equation $$y'=\frac{\sqrt\pi}{2}\exp (y^2)$$ with initial condition $y(0)=0$.


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