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Suppose $\lim_{x\rightarrow \infty} f(x) = 0, \int_a^\infty f'(x)dx$ is absolutely convergent and $f'(x)$ is continuous for $x \geq a$. Prove that $\int_a^\infty f(x)\sin x dx$ converges.

I know $\int_a^\infty f(x)dx$ converges as $\lim_{x\rightarrow \infty} f(x) = 0$, but I don't know how to incorporate that here.

Any help is appreciated!

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  • $\begingroup$ Use integration by parts. $\endgroup$ – xpaul May 3 '18 at 12:59
  • $\begingroup$ Note that for $f(x) = 1/x$ we have that $\int f(x) \, \mathrm{d}x$ diverges, so your final claim is false. $\endgroup$ – gj255 May 3 '18 at 13:00
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By integration by parts, $$\int_a^\infty f(x)\sin(x) dx=[-\cos(x)f(x)]_a^{+\infty}+\int_a^\infty f'(x)\cos x dx=\cos(a)f(a)+\int_a^\infty f'(x)\cos x dx.$$ Now the second integral is convergent because it is absolutely convergent: $$\int_a^\infty \left|f'(x)\cos (x)\right| dx\leq \int_a^\infty \left|f'(x)\right| dx<+\infty.$$

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