Convergence of $\int_0^\infty f(x)\sin(x)\mathrm dx$

Question

My task is : $ f : [0, \infty[ \to [0, \infty[$, $\lim_{x \to \infty} f(x) =0$

How can I show that this improper integral converges: $ \int_0^\infty f(x) \sin(x) \, dx$?

Is it right that this integral converges to $0$?

Thanks for your help.

Answer 1

Let $a_n=\int_{n\pi}^{(n+1)\pi}f(x)\sin(x)dx$, so we want to prove $\sum_0^\infty a_n$ converges. Note that since $f$ is non-negative and monotonically decreasing, while $\sin$ alternates sign, $a_n$ has the same sign as $(-1)^n$ and $|a_n|$ is monotonically decreasing. Therefore by alternating series test $$\int_0^\infty f(x)\sin(x)dx=\sum_0^\infty a_n=\sum_0^\infty (-1)^n|a_n|$$ converges.