Determine if an integral expression is in $L^2(\mathbb{R})$. Question: Consider
$$
v(x)\equiv v_0(x)\int_x^\infty \frac{f(x')dx'}{(1-e^{-2x'})v_0(x')},\;\;\;\;v_0(x)\equiv{e^x}{(1-e^{-2x})^n},\;\;n>0.
$$
where $f\in L^2(\mathbb{R}^+)\cap C^1(\mathbb{R}^+)$. I want to prove that
$v\in L^2(\mathbb{R}^+)$
Context: I am computing the resolvent set of a differential operator and I computed the solution of the resolvent equation. Using variation of parameters, I end up with expressions similar to what is above. Among other things, I need to determine if those expressions give me functions that are in $L^2(\mathbb{R}^+)$.
Reasoning: When $x\rightarrow 0^+$, I am good because the integral behaves in the worst case as $x^{-n}$, a bad behavior canceled by the multiplying $v_0(x)$ outside the integral. For the behavior at $x=\infty$, it "seems" fine to me also since the integral will go to zero exponentially fast and fast enough to cancel the behavior of the function $v_0(x)$ outside and make $v$ be in $L^2(\mathbb{R})$. Now, obviously, this last sentence is a heuristic argument, which needs to be made formal, if true at all. That's where I need help really.
Bonus: If $v$ is indeed in $L^2(\mathbb{R})$, am I right to say that it is not in $L^2(\mathbb{R})$ in the case $n=0$?
 A: Just a beginning
Let me re-write this just for clarity. Given some $n\in\mathbb{N}$, we have this operator $T_n$ acting as
$$ T_n:f(x) \mapsto e^x(1-e^{-2x})^n \int_{x}^{+\infty}\frac{f(t)\,dt}{e^t(1-e^{-2t})^{n+1}} $$
on $(C^1\cap L^2)(\mathbb{R}^+)$. The linear operator $T_0$ brings the indicator function $\mathbb{1}_{(0,M)}(x)$ into
$$ e^x\left(\operatorname{arctanh}(e^{-x})-\operatorname{arctanh}(e^{-M})\right)\mathbb{1}_{(0,M)}(x)$$
and the indicator function $\mathbb{1}_{(a,b)}(x)$ into
$$ e^x\left(\operatorname{arctanh}(e^{-x})-\operatorname{arctanh}(e^{-b})\right)\mathbb{1}_{(0,b)}(x)-e^x\left(\operatorname{arctanh}(e^{-x})-\operatorname{arctanh}(e^{-a})\right)\mathbb{1}_{(0,a)}(x)$$
which equals
$$ e^x\left(\operatorname{arctanh}(e^{-a})-\operatorname{arctanh}(e^{-b})\right)\mathbb{1}_{(0,a)}(x)-e^x\left(\operatorname{arctanh}(e^{-x})-\operatorname{arctanh}(e^{-b})\right)\mathbb{1}_{(a,b)}(x).$$
The decreasing function $g(x)=e^x\operatorname{arctanh}(e^{-x})$ behaves like $-\frac{1}{2}\ln(x)$ in a right neighbourhood of the origin and converges to $1$ as $x\to +\infty$. $-\frac{1}{2}\ln(x)$ is unbounded, but nevertheless it belongs to $L^2(0,1)$, so the range of $T_n$ should belong to $L^2(\mathbb{R}^+)$ even in the case $n=0$.
