I want to show that :
$$ \forall x \in E, \forall \alpha \in \mathbb{R} , \exists \beta < \alpha +1 , C\in \mathbb{R} \mbox{ such that } $$ $$\int_x^1 \frac{ |\ln y | ^\alpha }{ y (y-x)^{3/4} } dy \leq C \frac{|\ln(x)|^\beta} {x^{3/4}} $$
nevertheless I don't understand how you can compute this integral because in $y = x$, the expression is not defined. That's why I wrote $E$ for the domain of $x$. Idk where this expression is defined. I have no clue. Holder for instance gives me nothing (but one friend told me that it gives the case where $\beta = \alpha + 1 $).
So I'd like to understand how this expression does exists, and then understand why $ \beta,C $ could exists to finally find a solution.
Let $\,0<x<1\,$ , $\,y:=tx\,$ , $\,t:=e^z\,$ . $\enspace$ It’s $\,e^z\geq 1+z\,$ for $\,z\in\mathbb{R}\,$ .
$\displaystyle \int\limits_x^1 \frac{|\ln y|^{\alpha}}{y(y-x)^{3/4}}dy = \frac{1}{x^{3/4}} \int\limits_1^{1/x} \frac{|\ln (tx)|^{\alpha}}{t(t-1)^{3/4}}dt = \frac{1}{x^{3/4}} \int\limits_0^{-\ln x} \frac{(-\ln x - z)^{\alpha}}{(e^z-1)^{3/4}}dz \leq$
$\displaystyle \leq \frac{(-\ln x)^{\alpha}}{x^{3/4}} \int\limits_0^{-\ln x} \frac{dz}{(e^z-1)^{3/4}} \leq \frac{(-\ln x)^{\alpha}}{x^{3/4}} \int\limits_0^{-\ln x} \frac{dz}{z^{3/4}} = 4 \frac{(-\ln x)^{\alpha+\frac{1}{4}}}{x^{3/4}} = C \frac{|\ln x|^{\beta}}{x^{3/4}}$
with $\,C:=4\,$ and $\,\displaystyle \beta:=\alpha+\frac{1}{4}\,$