I have an inverse function of this guy :
$$ y(x)=x^{-1}(y(x))$$ $$ x(y)= \frac{1}{4}\Big(-\log[1 - ( \sqrt{\frac{2}{\epsilon^2}} y + C_1)] + \log[1 + ( \sqrt{\frac{2}{\epsilon^2}} y + C_1)] - \frac{2 ( \sqrt{\frac{2}{\epsilon^2}} y + C_1)}{(\sqrt{\frac{2}{\epsilon^2}} y + C_1)^2-1}\Big) $$
Which can be written : $$ x(y)= \frac{1}{4}\Big(\operatorname{arctanh }( \sqrt{\frac{2}{\epsilon^2}} y + C_1) - \frac{2 ( \sqrt{\frac{2}{\epsilon^2}} y + C_1)}{(\sqrt{\frac{2}{\epsilon^2}} y + C_1)^2-1}\Big) $$
I'd like very much to have an expression for $y(x)$. I assume the only way is to calculate a serie that fits. But I have no clue how to start the calculation. I was thinking about a Fourier series or something with $\frac{e^{k_n x}}{x^p} \; ; \; k_n\in \mathbb{C} \; ;\; n,p\in \mathbb{N}$
Would you have an idea about this ?
Here's a possible start.
$\begin{array}\\ x(y) &=\frac{1}{4}\Big(\operatorname{arctanh }( \sqrt{\frac{2}{\epsilon^2}} y + C_1) - \frac{2 ( \sqrt{\frac{2}{\epsilon^2}} y + C_1)}{(\sqrt{\frac{2}{\epsilon^2}} y + C_1)^2-1}\Big)\\ &=\frac{1}{4}\Big(atanh ( a y + c) - \frac{2 ( ay + c)}{(ay + c)^2-1}\Big) \quad a=\sqrt{\frac{2}{\epsilon^2}}, c=C_1\\ &=\frac{1}{4}\Big(atanh(z) - \frac{2z}{z^2-1}\Big) \quad z = ay+c\\ &=\frac{1}{4}\Big(atanh(z) - atanh(w)\Big) \quad w = \tanh(\frac{2z}{z^2-1})\\ &=\frac{1}{4}atanh\Big(\dfrac{z-w}{1-zw}\Big)\\ \text{so}\\ \tanh(4x(y)) &=\dfrac{z-w}{1-zw}\ \end{array} $
Not sure where to go from here so I'll stop.