Uniform convergence of improper integral 6

Question

Does integral $I$ converges uniformly on sets $E_1, E_2$, where $$ I = \int\limits_0^{+\infty} \frac {\ln^{\alpha} x}{x} \sin x \, \text{d} x, \\ E_1 = [0,1], E_2 =[1, +\infty) \,? $$ And may you please share some ideas concerning uniform convergence of integrals? Feeling lack of knowledge there.

Answer 1

My wild guess is that the given integral has to be interpreted in the following way: $$ \int_{0}^{+\infty}\left(\log x\right)^{\alpha}\frac{\sin x}{x}\,dx = \left.\frac{d}{db^{\alpha}}\int_{0}^{+\infty}\frac{x^b \sin(x)}{x}\,dx\right|_{b=0^+} = \left.\frac{d}{db^\alpha}\Gamma(b)\sin(\pi b/2)\right|_{b=0^+}$$ where $\frac{d}{db^\alpha}$, if needed, has to be interpreted as a fractional derivative operator.
The radius of convergence of the Taylor series of $\Gamma(b)\sin(\pi b/2)$ at the origin is $1$, since the function $\Gamma(z)$ function has simple poles at $z=0$ and $z=1$, but the pole at the origin is compensated by the sine term. In particular the wanted integral behaves like $(\alpha+1)!$, that is a bounded function on $(0,1)$ and an unbounded function over $(1,+\infty)$.