Integral behavior with small o notation I would like to prove this simple result that seems true to me for an eigenvalue problem I am solving. This result would help me analyze the solutions to the resolvent equation.
Let $f$ be a continuous function such that $f(0)=0$ and let $k>1$ be real. Consider the integral
$$
I\equiv \int_x^1 \frac{f(x')}{x'^k}dx'\text{ for }x>0.
$$
I would like to prove that
$$
f\sim \mathcal{o}\!\left(x^{1-k}\right)\text{ as }x\rightarrow 0^+,\text{ i.e. }\lim_{x\to 0^+}x^{k-1}I(x)=0.
$$
I think it is true. I just don't know what to use exactly to prove it. Of course, I can be wrong, in which case a counter example is most welcome.
 A: Your result is true, at least so long as $\lim_{x\to0^+}I(x)$ exists (I believe this is always the case since $f$ is continuous at $0$, but I don't have a proof off the top of my head).
To see this, it is easiest to split into two cases:
If $\lim_{x\to0^+}I(x)=a$ for some $a\in\mathbb{R}$, we find clearly that $\lim_{x\to0^+}x^{k-1}I(x)=0\cdot a=0$.
If on the other hand $\lim_{x\to0^+}I(x)=\pm\infty$ we may use L'Hôpital's rule.
Indeed, suppose $\lim_{x\to0^+}I(x)=\infty$ (the other case is achieved by multiplying by $-1$) and note that $I'(x)=-\frac{f(x)}{x^k}$.
Thus we get that
$$
 \lim_{x\to0^+}x^{k-1}I(x)
 =\lim_{x\to0^+}\frac{I(x)}{x^{1-k}}
 =\lim_{x\to0^+}\frac{I'(x)}{\frac{\mathrm{d}}{\mathrm{d}x}[x^{1-k}]}
 =\lim_{x\to0^+}\frac{-f(x)}{x^k(1-k)x^{-k}}
 =0.
$$
EDIT: It is, in fact not always the case that $\lim_{x\to0^+}I(x)$ exists just because $f$ is continuous,
take e.g. the function $f(x)=x\sin(1/x)$ for $x\neq0$ and $f(0)=0$.
However the result above is true even if $I(x)$ is divergent at $0$ - simply let $f^+$ and $f^-$ denote the positive and negative parts of $f$, respectively and write $I(x)=I^+(x)-I^-(x)$, where $I^+(x)=\int_x^1\frac{f^+(x')}{x'^k}\mathrm{d}x'$ and similarly for $I^-$.
These will have a limit as $x\to0^+$ as they are monotone, and hence the above argument works.
