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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.

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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.

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