looking for a slowly varing function with given properties I am looking for a slowly varying function $L(.)$ such that the function $f(s)=s+(1-s)^2L\left( \frac{1}{1-s}\right)$ is a probability generating function with $f'(1)=1$ and $f''(1)=\infty$.
I tryed $L(x)=\ln(x)$ but in this case $f(s)$ was not probability generating function.
 A: Consider any convenient probability generating function $f(s)$ satisfying $f'(1)=1$ and $f''(1)=\infty$.
Define the function : $$\boxed{L(x)=x^2f\left(\frac{x-1}{x}\right)+x(1-x)}$$
This function satisfies $\quad f(s)=s+(1-s)^2L\left( \frac{1}{1-s}\right)\quad$ because :
with $\quad x=\frac{1}{1-s} \quad\implies\quad \frac{x-1}{x}=s$
$L\left(\frac{1}{1-s}\right)=\left(\frac{1}{1-s}\right)^2f\left(s\right)+\left(\frac{1}{1-s}\right)\left(1-\left(\frac{1}{1-s}\right)\right)$
$(1-s)^2L\left(\frac{1}{1-s}\right)=f\left(s\right)-s$
$f(s)=s+(1-s)^2L\left( \frac{1}{1-s}\right)$
With this method you can find as many functions $L(x)$ as you want.
COMMENT about the expected "slowly varying $L(x)$ function".
$L(x)$ and $f(s)$ are related functions through the specified equation $f(s)=s+(1-s)^2L\left( \frac{1}{1-s}\right)$ or inversely $L(x)=x^2f\left(\frac{x-1}{x}\right)+x(1-x)$ .
Insofar $f(s)$ is a given function there is no choice for $L(x)$ to exactly satisfy $f(s)=s+(1-s)^2L\left( \frac{1}{1-s}\right)$. This defines the function $L(x)$ and you can't expect that $L(x)$ be smooth or not. You have to accept the function $L(x)$ as this function is given by  $L(x)=x^2f\left(\frac{x-1}{x}\right)+x(1-x)$.
May be the wording of question has to be differently written in order to obtain what you are expecting. Of course if the question becomes different the answer would be different.
