Convergence of two integrals. 
*

*I am given a family of integrals: $I(\alpha)=\int_3^\infty (\ln^{\alpha}(x)/x^{5/4})dx$ and I want to show that this family of integrals converges uniformly with respect to the exhaustion $E_k = [3,k]$ for $\alpha\in [ 0,2]$.

*I am given $J(\alpha)=\int_0^\infty \frac{\sin(\alpha x)}{x}dx$, and I want to show that it converges conditionally with respect to the exhaustion $E_k = [0,k]$ in the interval $\alpha \in [b,\infty)$ for each $b>0$.
I think I need to use here some inequalities for both cases, I am just not sure what; in the second integral I need to prove that $\lim_{k\to \infty}\int_0^k \sin(\alpha x)/x dx$ that there's some finite limit, but for the integral with the absolute value it doesn't converge.
I don't see how to show this?
 A: For (1): 
Uniform convergence follows from the Weierstrass M-test.  For $x \geqslant 3$ and $0 \leqslant \alpha \leqslant 2$, we have $\ln x > 1$ and $(\ln x)^\alpha \leqslant (\ln x)^2$.  For any $c > 0$ we have $\ln x = \frac{1}{c} \ln x^c < \frac{x^c}{c}$ and
$$\frac{(\ln x)^\alpha}{x^{5/4}} \leqslant \frac{(\ln x)^2}{x^{5/4}} = \frac{(\ln x^c)^2}{c^2 x^{5/4}}< \frac{x^{2c}}{c^2 x^{5/4}}$$
See if you can finish by choosing an appropriate value for $c$ such that the RHS is integrable over $[3,\infty)$.
For (2):
Changing variables with $u = \alpha x$, we  have $J(\alpha) = \int_0^\infty \frac{\sin u}{u} \, du$ and the conditional convergence of this improper integral has been established many times on this site.  There are a number of approaches -- for example, use integration by parts and a comparison with $x^{-2}$ to prove convergence of the integral over $[1,\infty)$.  Divergence of $\int_0^\infty \frac{|\sin x|}{x} \, dx$ is given in a multitude of answers here as well as in commonly used textbooks.
To establish uniform convergence for $\alpha \in [b,\infty)$, note that by the second mean value theorem for integrals, we have for some $\xi \in (c_1,c_2)$
$$\left|\int_{c_1}^{c_2} \frac{\sin \alpha x}{x} \, dx \right| = \left|\frac{1}{c_1}\int_{c_1}^{\xi} \sin \alpha x \, dx \right| = \frac{|\cos \alpha c_1 - \cos \alpha \xi|}{ \alpha c_1} \leqslant \frac{2}{b c_1}$$
The RHS can be made smaller than any $\epsilon > 0$ by choosing $c_1$ sufficiently large (independently of $\alpha$) and uniform convergence follows by the Cauchy criterion.
