Csdrhrt

For $a>0$, does
$$f_b(s)=int_{a}^{b}{frac{sin x}{x^s}}dx$$
converge uniformly for all compact subsets of ${sinmathbb C|Re(s)>0}$ when $btoinfty$
?

For $Re(s)>1$ the integral converges absolutely, and since
$g_s(z)=frac{sin z}{z^s}$ is holomorphic on ${xinmathbb R|x>0}$, by applying the Weierstrass theorem (which states that for a family of holomorphic functions converging uniformly, the family of derivative of those functions converges uniformly and equals the derivative of the converging function of the given family) it is not difficult to prove.

However, for $0<Re(s)leq 1$ the integral only seems to converge conditionally at best, which makes it more difficult. Any good ways of proving convergence?(or maybe divergence?)

Let $s=sigma+iomega$ where $sigma=text{Re}(s)$ and $omega=text{Im}(s)$. Then, we can write the integral of interest as

$$int_a^infty frac{sin(x)}{x^s},dx=int_a^infty frac{sin(x)cos(omega log(x))}{x^sigma},dx-i int_a^infty frac{sin(x)sin(omega log(x))}{x^sigma},dx$$

Now show that there are numbers $M_1$ and $M_2$ such that
$$left|int_a^L sin(x) sin(omega log(x)) , dxright|le M_1$$ and

$$left|int_a^L sin(x) cos(omega log(x)),dxright|le M_2$$

for all $Lge a$.

Finish by applying Dirichet's Test (aka Abel's Test) for uniform convergence.