Show that the following improper integral with parameter converges uniformly on $N = [c,d], 1 < c < d < 2$: $$\int_{0}^{\infty} \frac{\sin{x}}{x^y} dx.$$
I thought I could use the Weierstrass M - test, where I find a function $M(x)$ such that $\int_{0}^{\infty}M(x) dx$ converges and $\left | f(x,y) \right | = \left | \frac{\sin{x}}{x^y} \right | \leq M(x).$
I think that $\left| \frac{\sin{x}}{x^y} \right |\leq \frac{1}{x}$ since $|\sin{x}| \leq 1$ and $x^y \geq x$ because $y \in [c,d]$ (I am not sure about this). But then this integral does not converge for $0 \to \infty$. Is there another way?
Hint. Split it into two cases: take $M(x)=\frac{1}{x^c}$ over $[1,+\infty)$and take $M(x)=\frac{x}{x^d}=\frac{1}{x^{d-1}}$ over $(0,1]$.