2
$\begingroup$

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...

  • 0
    Edited: thanks for pointing that out2010-11-06

0 Answers 0