Test the uniform convergence integral $\int_1^\infty\frac{ln^\alpha x}{x}\sin(x)\, dx$ Test the uniform convergence integral 
$$ \int_{1}^{\infty} \frac{\ln^\alpha x}{x}\, \sin x \, dx, \quad \alpha\in[1,\infty). $$
As popular tests don't work, I suspect it convergence not uniform.
 A: Let's show that the Cauchy criterion fails. 
First, note that for any $x$ we have $\lim\limits_{\alpha\to\infty}\frac{\ln^\alpha x}{x} = \infty$ and for any $x$ there is $\alpha$ s. t. $\frac{\ln^\alpha x}{x} > 1$.
Second, $\sin(x)>1/\sqrt{2}$ for $\frac{\pi}{4}+2\pi k \leq x \leq \frac{3\pi}{4}+2\pi k$.
Finally, for any $b>0$ there is $k\in\mathbb N$ and sufficiently large $\alpha$ s. t.
$$
\left|\, \int\limits_{\frac{\pi}{4}+2\pi k}^{\frac{3\pi}{4}+2\pi k}\frac{ln^\alpha x}{x}\sin(x)\, dx \, \right| \geq \frac{1}{\sqrt{2}}\frac{\pi}{2}.
$$
So the integral doesn't satisfy the Cauchy criterion and there is no uniform convergence.
A: It's a partial proof, inconvenient to be put as a comment.
Using l'Hospital principle repetitively, we have $\lim_{x\to\infty} \frac{\ln^\alpha x}{x}=\lim_{x\to\infty} \frac{\alpha\ln^{\alpha-1} x}{x^2}=\lim_{x\to\infty} \frac{\alpha(\alpha-1)\ln^{\alpha-2} x}{2x^2}=\dots=\lim_{x\to\infty} \frac{\alpha!\ln^{0} x}{2^{\alpha-1}x^2}=\lim_{x\to\infty} \frac{K}{x^2},$
where $K(\alpha)$ is postive.
Therefore $\int_{1}^{\infty} \frac{\ln^\alpha x}{x}\, \sin x \, dx 
\approx K\int_{1}^{\infty} \frac{1}{x^2}\, \sin x \, dx
=K\int_{1}^{\infty} \frac{1}{x^{1/2}}\frac{\sin x}{x^{3/2}}\,  \, dx$. For $\frac{1}{x^{1/2}}$ is positive and decreasing, and $\int_{1}^{\infty} \frac{\sin x}{x^{3/2}}\,  \, dx$ is bounded, we have the integral is convergent.
But since as $\alpha\to\infty$, $K(\alpha)\to\infty$, the convergent is possibly not uniform.
