The function $$ f(x)\ :=\ e^{1-\frac{1}{\sqrt{1-x^2}}} $$ has the properties that

  • $\frac{d^nf}{dx^n}(\pm 1)=0$,
  • $\frac{d^n}{dx^n}\left(\sqrt{1-x^2}-f(x)\right) = 0$ at $x=0$,

and its integral $F(x) := \int_{-1}^{x}f(t)\,\mathrm{d}t$ provides $C^\infty$ transition between the two non-identical horizontal lines $y(x)=0$ and $y(x)\approx1.48756$.


  • can the integral $$ F(x) := \int_{-1}^{x}e^{1-\frac{1}{\sqrt{1-t^2}}}\,\mathrm{d}t $$ be expressed by standard functions or known special functions?
  • can at least $F(1)$ be expressed via already known functions?
  • do $f(x)$ or $F(x)$ already appear as the solution to a mathematical or physical problem?

Remark: I "discovered" $f(x)$ when trying to find a $C^\infty$ connection between two parallel lines.

  • 1
    Mathematica fails with both $F(x)$ and $F(1)$. – user64494 Oct 11 at 16:51
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    Sorry for my ignorance, but what does the $\ast$ operation denote? – M.G. Oct 11 at 17:34
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    What is $d^n\over x^n$? Is that meant to be $d^n\over dx^n$? – Gerry Myerson Oct 11 at 22:02
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    I tried cleaning up the notation, and the typo Gerry pointed out. Please edit if I messed anything up. – David Roberts Oct 11 at 23:16
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    In fact, if you just want a closed form smooth step function, it may be easier to look at things like $\tanh(\tan(x))$. – Willie Wong Oct 12 at 3:24

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