It is possible to find a power series approximation to the following smooth function $h(x)$?

Question

After giving this answer to a question trying to find a smooth transition between $f(x\leq 0)=0$ and $f(x\geq 1)=x$ through the integral of a smooth transition function:

$$h(x)=\begin{cases} 0,\quad x\leq -1, \\ x,\quad x\geq 1, \\ \int\limits_{-1}^x \frac{1}{1+\exp\left(\frac{4u}{u^2-1}\right)}\ du,\ \text{otherwise} \end{cases}$$

which derivative makes a smooth transition from $f(x\leq 0)=0$ into $f(x\geq 1)=1$.

You could see this function on Desmos:

Even when the plot looks like a quite "innocent" function, given I have been unable to find a closed-form to the integral (if you now it please share it in the answer), I then tried to find a good polynomial approximation to $h(x)$ on the interval $x\in (-1,\ 1)$, but Wolfram-Alpha don't show any series expansion for this function (power series like Taylor's, Laurent's, Puiseux's, Padé approximants, etc).

By playing on Desmos the best simple fit I found on $x\in (-1,\ 1)$ is the function $\hat{h}(x) =\left|\frac{x+1}{2}\right|^{2.5}$ but is not really good neither: it is not smooth, as a polynomyal expansion should be.

so my question is:

It is possible to find a power series approximation to the function $h(x)=\int\limits_{-1}^x \frac{1}{1+\exp\left(\frac{4u}{u^2-1}\right)}\ du$ on $x\in(-1,\ 1)$? (centered at $x=0$ if possible)

Answer 1

Though not quite a standard power series, one can derive the series representation

$$g(x)=\int\limits_{-1}^x \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du= \frac{x+1}{2}+\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{a_n}{2 n} \left(x^{2 n}-1\right)\right),\quad -1<x<1\tag{1}$$

from the Maclaurin series

$$f(u)=\frac{1}{e^{\frac{4 u}{u^2-1}}+1}=\frac{1}{2}+\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N a_n u^{2 n-1}\right),\quad -1<u<1 \tag{2}$$

where $a_n=\frac{f^{(2n-1)}(0)}{(2 n-1)!}$ via term-wise integration, i.e. $\int\limits_{-1}^x \frac{1}{2} \, du=\frac{x+1}{2}$ and $\int\limits_{-1}^x a_n u^{2n-1} \, du=\frac{a_n}{2 n} \left(x^{2 n}-1\right)$.

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The first few terms of the series for $f(u)$ are

$$f(u)=\frac{1}{2}+u-\frac{u^3}{3}-\frac{13 u^5}{15}+\frac{67 u^7}{315}+\frac{3083 u^9}{2835}...\tag{3}$$

and the first few terms of the series for $g(x)$ are

$$g(x)=\frac{x+1}{2}+\frac{x^2-1}{2}-\frac{x^4-1}{12}-\frac{13 \left(x^6-1\right)}{90}+\frac{67 \left(x^8-1\right)}{2520}+\frac{3083 \left(x^{10}-1\right)}{28350}..\tag{4}.$$

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Figure (1) below illustrates $f(u)=\frac{1}{e^{\frac{4 u}{u^2-1}}+1}$ in blue and the Maclaurin Series for $f(x)$ in orange where the series defined in formula (2) above is evaluated at $N=100$.

Figure (1): Illustration of $f(u)=\frac{1}{e^{\frac{4 u}{u^2-1}}+1}$ (blue) and associated Maclaurin series (orange)

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Figure (2) below illustrates $g(x)=\int\limits_{-1}^x \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du$ in blue and the series for $g(x)$ in orange where the integral in formula (1) above is evaluated using numerical integration (via Mathematica's NIntegrate function) and the series defined in formula (1) above is evaluated at $N=100$.

Figure (2): Illustration of $g(x)=\int\limits_{-1}^x \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du$ (blue) and associated series (orange)

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One can also derive the Maclaurin series

$$g(x)=\int\limits_{-1}^x \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du=b+\frac{x}{2}+\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{a_n}{2 n}\, x^{2 n}\right),\quad -1<x<1\tag{5}$$

where

$$b=\int\limits_{-1}^0 \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du\approx 0.137775\tag{6}$$

and $a_n$ is the same as in formulas (1) and (2) above.

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The first few terms of the Maclaurin series for $g(x)$ defined in formula (5) above are

$$g(x)=b+\frac{x}{2}+\frac{x^2}{2}-\frac{x^4}{12}-\frac{13 x^6}{90}+\frac{67 x^8}{2520}+\frac{3083 x^{10}}{28350}...\tag{7}$$

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Figure (3) below illustrates $g(x)=\int\limits_{-1}^x \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du$ in blue and the Maclaurin series for $g(x)$ defined in formula (5) above in orange where the integral in formula (5) above is evaluated using numerical integration and the Maclaurin series defined in formula (5) above is evaluated at $N=100$.

Figure (3): Illustration of $g(x)=\int\limits_{-1}^x \frac{1}{e^{\frac{4 u}{u^2-1}}+1} \, du$ (blue) and formula (5) Maclaurin series (orange)

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Answer 2

A good smooth transition is obtained with the function

$$ t(x,\lambda,x_0) = \frac{\ln (\cosh (\lambda (x-x_0)))}{2 \lambda }-\frac{\lambda x_0-\ln (2)}{2 \lambda }+\frac{x}{2} $$

This function is obtained integrating

$$ \frac{1}{2} (\tanh (\lambda (x-x_0))+1) $$

as

$$ \int_x \frac{1}{2} (\tanh (\lambda (x-x_0))+1)dx=\frac{\ln (\cosh (\lambda (x-x_0)))}{2 \lambda }+\frac{x}{2}+c_0 $$

and then calculating $c_0$ as

$$ \lim_{x\to-\infty}\frac{\ln (\cosh (\lambda (x-x_0)))}{2 \lambda }+\frac{x}{2}+c_0=0 $$

NOTE

With $x_0$ the transition is located at a precise point. If $x_0=0$ and $\lambda=2$ we have

$$ t(x) = \frac{\ln (\cosh (2 x))}{4 }+\frac{\ln (2)}{4 }+\frac{x}{2} $$