Let $f: [1, \infty ) \to \mathbb{C}$ be a continuous function with a bounded antiderivative $F(x)$ on $[1, \infty)$. Show, that the integral
$$ \int_1^\infty \frac{f(x)}{x^s} dx$$
exists for each $s > 0$.
Now, my idea was to somehow increase the potency of the denominator by integration by parts. But so far, that didn't lead anywhere. Thanks in advance!
We have $f(x) = F'(x)$, i.e., $f(x) dx = dF(x)$. Hence, $$I = \int_1^{\infty} \dfrac{f(x)dx}{x^s} = \int_1^{\infty} \dfrac{dF(x)}{x^s} = \left.\dfrac{F(x)}{x^s} \right\vert_1^{\infty} + \int_1^{\infty} \dfrac{sF(x)}{x^{s+1}}dx$$ Since $F(x)$ is bounded, we have $\lim_{x \to \infty} \dfrac{F(x)}{x^s} = 0$. Further, since $F(x)$ and $\dfrac1{x^{s+1}}$ are integrable on $[1,\infty)$, so is $\dfrac{F(x)}{x^{s+1}}$. Further, since $F(x)$ is bounded, i.e., $\vert F(x) \vert \leq M$ for all $x \in [1,\infty)$, we have that $$I \leq sM \int_1^{\infty} \dfrac{dx}{x^{s+1}} = M$$