# 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.

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.
$$\left|\, \int\limits_{\frac{\pi}{4}+2\pi k}^{\frac{3\pi}{4}+2\pi k}\frac{ln^\alpha x}{x}\sin(x)\, dx \, \right| \geq \frac{1}{\sqrt{2}}\frac{\pi}{2}.$$