Alternating integral convergence

Question

This test exist for alternating series:

https://en.wikipedia.org/wiki/Alternating_series_test

I want to know if there is an equivalent for integrals of this form

$$ \int\limits_a^\infty \mathrm{d}x\ f(x) \sin{x} $$

Where $f$ is monotonically decreasing, positive, does not diverge and when x goes to $\infty$, $f$ goes to zero.

I do not have much experience with real analysis so I have no idea how to proceed. I thought of using something like Riemann integral to be able to use the test above, but had no success.

Answer 1

Yes: such an integral always converges under the conditions you described.

For clarity of the argument, I'll assume that $a = 0$, but it can be adapted for any $a$. Break up the interval $[0, \infty)$ into a union: $$[0, \infty) = [0, \pi] \cup [\pi, 2 \pi] \cup [2 \pi, 3 \pi] \cup \dots$$ Then, compare the value of the integrals on adjacent regions. Note that \begin{align*} \left|\int_{\pi}^{2 \pi} f(x) \sin(x) \, \textrm{d} x \right| &= \int_{\pi}^{2 \pi} \left| f(x) \sin(x) \right| \, \textrm{d} x\\ &= \int_0^{\pi} \left|f(u + \pi) [\sin(u + \pi)] \right| \, \textrm{d} u \\ &= \int_0^{\pi} \left|f(u + \pi) [-\sin(u)] \right| \, \textrm{d} u \\ &= \int_0^{\pi} f(u+ \pi) \sin(u) \, \textrm{d} u \\ & \leq \int_0^{\pi} f(u) \sin(u) \, \textrm{d} u \\ &= \left| \int_0^{\pi} f(x) \sin(x) \, \textrm{d} x \right| \end{align*} and the same argument can be applied to any adjacent intervals. Thus, the integral terms $a_n := \int_{n \pi}^{(n+1)\pi} f(x) \, \textrm{d} x$ for $n \geq 0$ form a sequence that satisfies the alternating series test hypotheses.

One caveat: these integrals will converge if considered as Riemann integrals -- but as Lebesgue integrals, they may not. In order for the Lebesgue integrals to converge, you also have to guarantee that $a_0 + a_2 + a_4 + \dots$ and $a_1 + a_3 + a_5 + \dots$ each converge to a finite value.