Series expansion of a function defined by a definite integral I am working on understanding the resolvent set of a certain eigenvalue problem. The solutions of the resolvent equation involve integrals similar to what I present below. Proving or contradicting the result claimed below would be useful.
Consider the expression below defined for $x\geq 0$ as:
$$
g(x)=x^p\int_1^x\left(\frac{f(x')}{x'^{p+1}}\right)dx'\text{ where }p>0.
$$
If $f$ is analytic, then it easy to prove (by expanding $f$ and integrating term-by-term) that
$$
g(x)=-\frac{f(x)}{p}+\mathcal{O}(x)+\mathcal{O}(x^p)\text{ as }x\to 0^+.\tag{1}
$$
On the RHS of (1), I really mean "$f(x)$", not "$f(0)$", because, below, I will allow the case where $f(0)$ is not necessarily defined.
I really would like to extend (1) to cases where $f$ is "less nice", that is:
(1) If $f$ is absolutely continuous everywhere.
(2) If $f$ is continuous and bounded on $(0,\infty)$. If that helps, also in $H^1(a,\infty)$ for any $a>0$.
Using L'Hopital's rule, it is not difficult to prove the following for the cases (1) and (2) above:
$$
g(x)=-\frac{f(x)}{p}+\mathcal{o}(1).\tag{2}
$$
Actually, if, in (2), I could replace "$\mathcal{o}(1)$" by "$\mathcal{O}(x^\epsilon)$" for some $\epsilon>0$, I would be happy, even if (1) would not be extended.
 A: Taking in account formulas for the derivative of production, one can get:
$$\text I(x)=\left(\dfrac {x^{p+1}}{p+1}\right)'_x\ \cdot\left(\int\limits_1^x \dfrac1{t^{p+1}}\,f(t)\,\text dt\right)$$
$$=\left(\dfrac {x^{p+1}}{p+1} \cdot \int\limits_1^x \dfrac1{t^{p+1}}\,f(t)\,\text dt\right)'_x
-\left(\dfrac {x^{p+1}}{p+1}\right)\,\left(\int\limits_1^x \dfrac1{t^{p+1}}\,f(t)\,\text dt\right)' $$
$$=\left(\dfrac{x}{p+1}\,\text I(x)\right)'
-\left(\dfrac {x^{p+1}}{p+1}\right)\,\dfrac1{x^{p+1}}\,f(x)
=\left(\dfrac{x}{p+1}\text I(x)\right)'-\dfrac {f(x)}{p+1},$$
$$(p+1)\,\text I(x) =x\,\text I'(x)+\text I(x) -f(x),$$
$$x\,\text I'(x)-p\,\text I(x)=f(x),\quad I(1)=0.\tag1$$
Easily to see that the equation $(1)$ in the form of
$$x^{p+1}\left(x^{-p}\text I\right)' =f(x)$$
corresponds to the given task.
Let
$$f(x)=\sum\limits_{j=0}^\infty a_j\,(x-1)^j,\quad \text I(x)=\sum\limits_{k=1}^\infty b_j\,(x-1)^j,\,\tag2$$
then
$$\sum\limits_{j=0}^\infty a_j\,(x-1)^j 
=\sum\limits_{j=1}^\infty jb_j\,(x-1)^j
+\sum\limits_{j=1}^\infty b_j\,(x-1)^{j-1}
-p\sum\limits_{j=1}^\infty b_j\,(x-1)^j$$
$$=b_1+\sum\limits_{j=1}^\infty \big((j-p)b_j+b_{j+1}\big)\,(x-1)^j,$$
\begin{cases}
b_1=a_0\\
b_2=a_1+(p-1)b_1\\
b_3=a_2+(p-2)b_2\dots\\
b_{j+1}=a_{j}+(p-j)b_j\dots\tag3
\end{cases}
Recurrence relations $(3)$ leads to the fast increment of $b_j.$ However, convergence can be achieved for the parameters $\;p=5, a_j=\dfrac1{(5j)!};$
