# If $\lim_{x\rightarrow \infty} f(x) = 0, \int_a^\infty f'(x)dx$ is absolutely convergent, prove that $\int_a^\infty f(x)\sin x dx$ converges

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!

• Use integration by parts. – xpaul May 3 '18 at 12:59
• Note that for $f(x) = 1/x$ we have that $\int f(x) \, \mathrm{d}x$ diverges, so your final claim is false. – gj255 May 3 '18 at 13:00

## 1 Answer

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: $$\int_a^\infty \left|f'(x)\cos (x)\right| dx\leq \int_a^\infty \left|f'(x)\right| dx<+\infty.$$