I found myself stuck with an "elementary" claim in some article. A simplified version of the problem is:   

Let $p : [0,1]\to [0,1]$ be a continuous and non decreasing function such that $p(0)=0$ and
$$
\int\_0^1 \left(\log \frac{1}{u}\right)^{\!1/2} \, dp(u) < + \infty.
$$
Show that
$$
\lim\_{h\to 0} \frac{\displaystyle\int\_0^h \left(\log \frac{1}{u}\right)^{\!1/2} \, dp(u)}{p(h) \left(\log \dfrac{1}{h}\right)^{1/2}}
$$
exists and is finite.


I think that a good idea is to use integration by parts formula which gives
$$
\int\_0^h \left(\log \frac{1}{u}\right)^{\!1/2}\, dp(u) = p(h) \left(\log \frac{1}{h}\right)^{\!1/2} + \int\_0^h \frac{p(u)}{2u\left(\log \frac{1}{u}\right)^{\!1/2}} du,
$$
but I don't see how to conclude...



Asymptotic of an improper integral
I think you reproduced the "elementary" claim correctly (for the case $n=1$). However, as stated in my earlier comment, the claim is false in general, even if you assume that $p>0$ on $(0,1]$.


Indeed, the claim was that the limit
\begin{equation\*}
 \lim\_{h\downarrow0}r(h) \tag{1}
\end{equation\*}
exists and is finite, where
\begin{equation\*}
 r(h):=\frac{\text{num}(h)}{\text{den}(h)},
\end{equation\*}
\begin{equation\*}
 \text{num}(h):=\int\_0^h \ln^{1/2}\frac1u\;dp(u),\quad 
 \text{den}(h):=p(h)\ln^{1/2}\frac1h. 
\end{equation\*}
Note that
\begin{equation\*}
 \text{num}(h)\ge\int\_0^h \ln^{1/2}\frac1h\;dp(u)=\text{den}(h),
\end{equation\*}
so that $r(h)$ is always $\ge1$.


However, let us show that the limit in (1) can be $\infty$:


Indeed, let
\begin{equation\*}
 p(u):=\frac1{\sqrt{\ln\frac1u}\,\ln^2\ln\frac1u} 
\end{equation\*}
for $u\in(0,1/3)$, with $p(0):=0$ and $p(u):=\frac1{\sqrt{\ln3}\,\ln^2\ln3}$ for $u\in[1/3,1]$.
Then $p$ is a real-valued continuous nondecreasing function on $[0, 1]$. Also,
\begin{equation\*}
 p'(u)\sim\frac1{2u\ln^{3/2}\frac1u\,\ln^2\ln\frac1u}
\end{equation\*}
(as $u\downarrow0$) and hence for $h\downarrow0$
\begin{equation\*}
 \text{num}(h)
 \sim\int\_0^h \frac{du}{2u\ln\frac1u\,\ln^2\ln\frac1u}
 =\frac1{2\ln\ln\frac1h}, 
\end{equation\*}
whereas
\begin{equation\*}
 \text{den}(h)=\frac1{\ln^2\ln\frac1h},
\end{equation\*}
so that
\begin{equation\*}
 r(h)\sim\frac12\,\ln\ln\frac1h\to\infty. 
\end{equation\*}




---


On a somewhat positive note, suppose that $p$ is continuous on $[0,1]$ with $p(0)=0$, has a positive derivative $p'$ in a right neighborhood of $0$, and
\begin{equation}
 p(h)=(c+o(1)) hp'(h)\ln\frac1h \tag{2}
\end{equation}
as $h\downarrow0$ for some $c\in[0,2)$. This condition will hold e.g. if (i) $p(h)=Ch^a\ln^b\frac1h$ for some positive real $C$ and $a$, some real $b$, and all small enough $h>0$ or if (ii) $p(h)=C/\ln^t\frac1h$ for some positive real $C$, some real $t>1/2$, and all small enough $h>0$.


Given the additional condition (2), you will have $r(0+)=\frac1{1-c/2}\in(0,\infty)$, by l'Hospital's rule. Indeed, then for the "derivative ratio" we have
\begin{equation\*}
 \frac{\text{num}'(h)}{\text{den}'(h)}
 =\frac{p'(h)\ln^{1/2}\frac1h}{p'(h)\ln^{1/2}\frac1h-\frac{p(h)}{2h}\ln^{-1/2}\frac1h}
 =\frac1{1-\dfrac{p(h)}{2hp'(h)\ln\frac1h}}\to\frac1{1-c/2}
\end{equation\*}
as $h\downarrow0$.


