I've solved a problem, I presume that my calculations were rights using the dominated convergence theorem (I say this Wikipedia):
$$\lim_{n\to\infty}\int_0^1\frac{x^{\Gamma(nx)}}{1-\log(x)}dx=0,$$ where (for integers $n\geq 1$) we denote with $\Gamma(s)$ the Gamma function. My details are that my integrand is positive and $$\frac{x^{\Gamma(nx)}}{1-\log x}\leq \frac{1}{1-\log x},$$ having $$\int_0^1 \frac{dx}{1-\log x}=-e\operatorname{Ei}(-1)<\infty.$$ My sequence of real-valued functions $x^{\Gamma(nx)}/(1-\log x)\to 0$ as $n\to\infty$ for each fixed $x\in(0,1)$ because $\Gamma(nx)\to\infty$ as $n\to\infty$, and the terms os such sequence are measuble since are composition of measurable functions on $(0,1)$ with respect the Lebesgue meausure $dx$.
Exercise 1 Please if there are some mistake or detail that I need to add to such proof tell me.
After I tried a different exercise, was also a creation:
Exercise 2. Provide me hints to study the monotony of $$\int_0^1\frac{x\Gamma(\lambda x)}{1-\log x}dx,$$ where $0<\lambda$ is a real parameter. Thanks in advance.
Using Wolfram Alpha online calculator, I've that our integrals equals to $\approx 1.08165$ when $\lambda=1/2$; equals to $\approx 0.551193$ when $\lambda=1$; equals to $\approx 0.373987$ when $\lambda=2$ or equals to $\approx 10.2654$ when $\lambda=6$. You can calculate these with this code
int x Gamma(0.5x)/(1-log(x))dx, from x=0 to 1
I presume that I need to take the derivate inside the integral sign, and to combine with Stirling's asymptotic.