I am trying to evaluate this integral or at least get bounds for its absolute value.
I have where $\tau \to \infty$
$\int\nolimits_{1}^{\infty} f(t) \frac{\tau \sin(\tau\log t)}{t^{\sigma+1}} dt$
$f(t) = (t + 1/2) - \lfloor t + 1/2 \rfloor$ (where $\lfloor.\rfloor$ is the floor function).
Integrating by parts gives me,
$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty} + B$
Where,
$B = \int_{1}^{\infty} \cos(\tau\log t) g'(t) dt$
where $g(t) = f(t)/t^\sigma$ and is piecewise continuous.
Using Riemann-Lebesgue lemma we have $B = 0$ as $\tau \to \infty$, giving us,
$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty}$
Also, since $0 \leq f(t) \leq 1$, I think it makes the absolute value of the integral $\leq 1$.
So, are these steps correct?
Any further insights will be highly appreciated...