I'm dealing with a problem for improper integrals reads the following:
Let $f(x)$ be a function defined on $[a,+\infty)$, monotonic down to $0$ as $x \to +\infty$. Prove that the following integrals $\int_{a}^{+\infty} f(x)dx$ and $\int_{a}^{+\infty} \cos^2 x.f(x)dx$ are convergent or disvergent simultaneously.
I'm quite confused with some words monotonic down to $0$. Once I thought that $f(x)$ is simply non-increasing (means if $a \le b$ then $f(a) \ge f(b)$) so that $f(x) \ge \lim_{x \to +\infty}f(x)=0, \quad \forall x \in [a,+\infty)$. Therefore, $$0\le \cos^2 x.f(x) \le f(x), \quad \forall x \in [a,+\infty).$$ This means that, the integrals $\int_{a}^{+\infty} f(x)dx$ and $\int_{a}^{+\infty} \cos^2 x.f(x)dx$ are convergent or disvergent simultaneously.
Is my solution correct? Otherwise, if is just equipped the assumption that $\lim_{x \to +\infty}f(x)=0$ without the non-increasing property. How can I come up with another proof?
Outline: For the harder direction, without loss of generality we may assume that we are integrating from $0$. Let $I_n=\int_{n\pi}^{(n+1)\pi} f(x)\cos^2 x \,dx$.
The sum $\sum I_n$ converges. Let $J_n=\int_{n\pi}^{n\pi+\pi/4} f(x)\cos^2 x\,dx$. Then $\sum J_n$ converges.
Let $K(n)=\int_{n\pi}^{(n+1)\pi} f(x)\,dx$. Conclude that $\sum K(n)$ converges.