As the title said, I want to find two-sided asymptotic expansion of functions in $C^0((0,1))$, but my target is a little different from usual. More specificly, given some $f\in C^0((0,1))$, but $f\not\in L^1([0,1])$, find an expansion $E(f)$ such that $f-E(f)\in L^1([0,1])$.
Let's say functions in $C^0((0,1))$ form ordered asymptotic scale $\phi_s(x)$ for $x\to0^+$ , the calculated asymptotics of $f(x)$ is: $$f(x)\sim\sum_{s\in S_0} a_s\phi_s(x)\ ;\ f(1-x)\sim\sum_{s\in S_1}b_s\phi_s(x)$$ If I'm not wrong, the expansion $E(f)$ should take the form: $$E(f)=\sum_{u,v}c_{(u,v)}\phi_u(x)\phi_v(1-x)\ $$
My question is: Can we find $E(f)$ , given expansion of $f(x)$ at $0^+$ and $1^-$ ? If yes, is there a systematic way?
For example: $f(x)=\frac{\log(1-x)}{x^2\log(x)}$, we have: $$\begin{cases}f(x)\sim\frac{-1}{x\log x}+\frac{\frac{-1}{2}}{\log x}+O\left(\frac{x}{\log x}\right)\\f(1-x)\sim-\frac{\log x}{x}+\frac{-3}{2}\log x+O(x\log x)\end{cases}$$ After trial and error, I found $E(f)=-\frac{1}{x\log x}-\frac{\log(1-x)}{1-x}-\frac{1}{1-x}$
Some notes
-I know some functions like $\sin(1/x)$ might cause troubles so let's stick with functions that well-behaved enough, my main interests are functions composed by $x^s,\log^n x,\log(-\log(x))$ and its $x\mapsto1-x$ version.
-$E(f)$ is not unique, but one non-trivial solution is enough. I need to deal with integrals with parameters in practice, so trivial solutions like $E(f)=f$ won't help me.
-Here is an answer where I used this expansion $E(f)$ to solve integral.
