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could anyone help me with finding the inverse function to the function $$y(x) = x + \frac{1}{1+\mathrm{e}^{-x}}?$$

The inverse functions to individual (separate) functions $x$ and $\dfrac{1}{1+\mathrm{e}^{-x}}$ are easy to determine. However, the determination of the inverse function to their sum seems to be more difficult. Does anyone have some idea?

Thanks to any advice.

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    Functions involving $x$ and $\exp(x)$ have a tendency to include the Lambert W function. https://en.wikipedia.org/wiki/Lambert_W_function , $W(x) = inverse\{xe^{x}\}$ – 2017-02-16
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    @mathreadler Apparently not this one, though! – 2017-02-16
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    We do get $xe^{-x}$ in some places if we put the terms on the same division. – 2017-02-16

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There seems to be no closed-form solution. However, there is a series solution involving powers of $e^{1-y}$, which should be good for large $y$:

$$ x = y-1+{{\rm e}^{1-y}}-2\,{{\rm e}^{2-2\,y}}+11/2\,{{\rm e}^{3-3\,y}}-{ \frac {53\,{{\rm e}^{4-4\,y}}}{3}}+{\frac {1489\,{{\rm e}^{5-5\,y}}}{ 24}} + \ldots $$

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    Dear Robert, thank you for your answer. As in case of your colleague Jack, may I ask you for a more detailed explanation how you derived your approximation? – 2017-02-16
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    Write the equation, with $x = y + u$ and $t = e^{-y}$, as $$u +\frac{1}{1 + t e^{-u}} = 0$$ Then in Maple: series(RootOf(u + 1/(1+t*exp(-u)),u),t); – 2017-02-16
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    BTW: you might also look at [OEIS sequence A052885](https://oeis.org/A052885) and links there. – 2017-02-16
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For $x>0$ the function is bounded between $x+\frac{1}{2}$ and $x+1$, hence the inverse function is bounded between $y-\frac{1}{2}$ and $y-1$. With a step of Newton's method, a more accurate approximation of the inverse function is given by $$ y-1+\frac{e+e^y}{e+3e^y+e^{2y-1}}.$$ What is the purpose for computing an explicit inverse function?

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    Dear Jack, thank you for your answer. I do not need the explicit inverse function. In fact, the approximation you proposed seems to be accurate enough for my purpose. May I ask you for a more detailed explanation how you derived your approximation? – 2017-02-16
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    @LubomirKlimes: I applied one step of Newton's method for finding the root of $f(x)=x+\frac{1}{1+e^{-x}}-y$ with starting point $y-1$. The given expression is just $y-1-\frac{f(y-1)}{f'(y-1)}$. – 2017-02-16