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If $f(x)\nearrow\infty$ in $[a, \infty)$ and f'(x) is continuous in $[a, \infty)$ then \int_{a}^{\infty}\frac{f'(x)}{f(x)}\sin(f(x)) converges

I'm not exactly sure what to try here. I thought of Dirichlets test but I can't seem to identify the right functions to make it work.

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    @Adrian: I think it means $f(x)$ is increasing in the interval and $\lim_{x\to\infty}f(x)=\infty$.2011-02-09

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Try a change of variables to reduce it to an integral of a sinc function.

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    I don't think that $f$ being increasing is necessary. Continuity of $f'$ is a standard assumption in the proof that integration by substitution works, because the elementary calculus version of FTC is usually done with continuously differentiable functions. The hypotheses required to generalize are more complicated. E.g., I believe it would suffice for $f$ to be absolutely continuous on bounded intervals, as long as you're willing to trade in Riemann integrals for Lebesgue integrals.2011-02-08
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Integration by parts gives

\int_a^b \frac{f'(x)}{f(x)}\sin(f(x)) \, dx = \frac{-\cos(f(x))}{f(x)}\Bigg \vert_a^b - \int_a^b \frac{f'(x)}{f(x)^2}\cos(f(x)) \, dx

But now

\left| \int_a^b \frac{f'(x)}{f(x)^2}\cos(f(x)) \, dx \right| \le \int_a^b \frac{f'(x)}{f(x)^2} \, dx = \frac{1}{f(x)}\Bigg \vert_b^a

Edit: Looking at it again, there might be some trouble with f being zero somewhere...

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    For this to work, note that $f' \ge 0$ is required. (As opposed to the other answer.)2011-02-09