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I am trying to evaluate this integral or at least get bounds for its aboslute 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,

$$\int_{1}^{\infty} \frac{f(t)}{t^\sigma} d(-\cos(\tau \log t)) = \left[ -\frac{f(t)}{t^\sigma}\cos(\tau \log t) \right]_{1}^{\infty} + B$$


$$B = \int_{1}^{\infty} \cos(\tau\log t) g'(t) dt$$

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...

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Nothing special about $f$? – J. M. Nov 6 '10 at 15:27
(1) your integration by parts is a bit odd: shouldn't there be a $t^{\sigma}$ factor somewhere in $B$? (2) the Riemann Lebesgue lemma requires the thing you are integrating against be absolutely integrable. which $f'$ itself is not. Perhaps with the corrected factor $t^{\sigma}$ inserted you can say something about it. – Willie Wong Nov 6 '10 at 15:37
Edited: thanks for pointing that out – Roupam Ghosh Nov 6 '10 at 15:41

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