Prove uniform convergence $$\int \limits_2^{+\infty} \frac{1}{x\ln(x)^{\alpha}}dx$$ Where $ \alpha \in [\alpha_0, +\infty), \alpha_0 > 1 $
I have made
$$ \int \limits_2^{+\infty} \frac{1}{x\ln(x)^{\alpha}}dx = \int \limits_2^{+\infty}\frac{d\ln x}{\ln(x)^{\alpha}} $$ let $t = \ln x $ $$ \int \limits_{\ln2}^{+\infty}\frac{d t}{t^{\alpha}} =\int \limits_{\ln2}^{e} \frac{d t}{t^{\alpha}} + \int \limits_{e}^{+\infty} \frac{d t}{t^{\alpha}} $$
It is clear that second part has uniform convergence by Weierstrass, but I stuck with the first part. How to explore it?
Uniform convergence of I(y) = $\int \limits_a^{+\infty} f(x,y)dx$ on the segment $[c,d]$
$\forall \epsilon > 0 \exists A(\epsilon) \geq a : \forall R > A$ and $\forall y \in [c; d] \Rightarrow |\int \limits_R^{+\infty} f(x, y)dx| < \epsilon$
Using the following definition of uniformly convergent parametric integral $$\forall\varepsilon > 0\; \exists\,A(\varepsilon): \forall R>A, \forall y\in I, \left|\int_{R}^{+\infty}\,f(x,y)\,dx\right|<\varepsilon $$ it is enough to check that
$$ \int_{\exp(k\alpha_0)}^{+\infty}\frac{dx}{x\log(x)^{\alpha}} =\int_{k\alpha_0}^{+\infty}\frac{dz}{z^{\alpha}}=\frac{(k\alpha_0)^{1-\alpha}}{\alpha-1}\leq\frac{\alpha_0^{1-\alpha_0}}{\alpha_0-1}\,k^{1-\alpha}\leq C_0 k^{1-\alpha_0}$$ so a suitable choice of $k$ in terms of $C_0$ ensures uniformity $\forall \alpha\in I = [\alpha_0,+\infty)$.