is it true that
if $f(x)$ is periodic, non-constant and bounded $$\text{when } T \to \infty ,\qquad\qquad\sup_{|t| \ <\ T}\ \ \left|\ \int_1^\infty f(x) x^{-\sigma-it} dx\ \right| \ \sim \ C\left|\ \int_1^\infty \sin(x) x^{-\sigma-it} dx\ \right|$$
$$ $$
EDIT : I just added the $\ \displaystyle\sup_{t \ <\ T}\ $ because it is false that
$$\text{when } t \to \infty,\qquad\qquad\left|\int_1^\infty f(x) x^{-\sigma-it} dx\right| \ \sim \ C\left|\int_1^\infty \sin(x) x^{-\sigma-it} dx\right| $$
the Dirichlet $\eta$ and $\beta$ functions are counter-examples :
$$\eta(s) = \sum_{n=1}^\infty n^{-s} (-1)^{n+1} = s \int_1^\infty {}_{\_\_}\!|^{^{\!\_\_\!\_}}\!|\!{}_{\_\_}\!|^{^{\!\_\_\!\_}}(x) \ x^{-s-1} dx$$
$$\beta(s) =\sum_{n=1}^\infty (2n-1)^{-s} (-1)^{n+1} = s \int_1^\infty {}_{\_\_}\!|^{^{\!\_\_\!\_}}\!|\!{}_{\_\_}\!|^{^{\!\_\_\!\_}}{\textstyle\left(\frac{x+1}{2}\right)} \ x^{-s-1} dx$$
on $\text{Re}(s) = 1$, the former has an infinity of zeros while the latter doesn't have any zero has says the prime number theorem for Dirichlet characters, so $\frac{\eta(1+it)}{1+it}$ cannot be equivalent to $\frac{\beta(1+it)}{1+it}$.
I'm aware that for $Re(s) > 0$ : $\Gamma(s) = \int_0^\infty e^{-x} x^{s-1} dx = \int_0^\infty e^{-ax} (ax)^{s-1} d(ax)$ for any $Re(a) > 0$ because the function $h(z) = e^{-z} z^{s-1}$ is analytic on $Re(z) \ge 0$ hence $\int_0^\infty + \int_\infty^{a\infty}+\int_{a\infty}^0 h(z) dz = 0$, and $h(z)$ decreases exponentially fast when $|z| \to \infty$ hence $\int_\infty^{a\infty} h(z) dz = 0$. by continuity, if $Re(s) < 1$ we can go until $a = \pm i = e^{\pm i \pi /2}$ and we get $\Gamma(s) = e^{i \pi s/2}\int_0^\infty e^{-ix} x^{s-1} dx = e^{-i \pi s/2}\int_0^\infty e^{ix} x^{s-1} dx$, hence where it converges, i.e. for $Re(s) \in ]0;2[$ : $\int_0^\infty \sin(x) x^{-s} dx = \sin{\textstyle\left(\frac{\pi(1-s)}{2}\right)} \Gamma(1-s)$ and $$\int_1^\infty \sin(x) x^{-s} = \sin{\textstyle\left(\frac{\pi(1-s)}{2}\right)} \Gamma(1-s) + \mathcal{O}(1)$$
I'm also aware that with the Fourier series expansion $f(x) = c_0 + \sum_{n=1}^\infty a_n e^{ inx} + b_n e^{-inx}$ if everything is convergent, we get $$\int_1^\infty f(x) x^{-s} dx = \frac{c_0}{s-1} + \sum_{n=1}^\infty n^{s-1} \int_n^\infty (a_n e^{ix}+b_n e^{-ix}) x^{-s} dx = \frac{c_0}{s-1} + \int_1^\infty (A_{1-s}(x) e^{ix} + B_{1-s}(x) e^{-ix}) dx$$
where $A_{1-s}(x) = \sum_{n \le x} a_n n^{s-1}$ and $B_{1-s}(x) = \sum_{n \le x} b_n n^{s-1}$ are the partial Dirichlet series.