Prove that $\int_{a}^{\infty}|f(x)\sin(e^x)|$ diverges

Question

Continuing this post: Prove that $\int_a^\infty f(x)\sin(e^x) \, dx$ conditionally converges.

Let $f$ be a bounded and with a continuous derivative on the interval $[a,\infty)$.

$$ \int_a^\infty f(x) \, dx\;\;\text{diverges.}$$

Also, $$ \exists t> a, \forall x>t: f'(x) < f(x) $$

Prove that the integral $\displaystyle \int_a^\infty f(x) \sin(e^x) \, dx$ conditionally converges.

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I understand all that was written in the previous post.

My proof for the divergence of $\int_{a}^{\infty}|f(x)\sin(e^x)|$ that I wrote there was incorrect (you can see what I tried at the link).

Can someone give me a hint please? (This is a homework question so I would prefer hints rather than full written answer.)

Thank you; I am pretty stuck so every word may help.

Answer 1

This problem is surprisingly hard with the hypothesis $f'<f$, which makes me wonder whether $|f'|<|f|$ is what was meant by the problem's author. Nevertheless, I have found a way to prove it for this hypothesis. I've posted a full solution on the linked question, but here's a summary of the part of the proof for divergence of $\int_a^\infty |f(x) \sin(e^x)|dx$.

1. Convergence is determined only by behavior of $f(x)$ at large $x$, so WLOG we can suppose that $f'<f$ for all $x\geq a$, and we can also take $e^a$ to be an integer multiple of $2\pi$.
2. Note that $f'<f$ and $f$ bounded implies that $f$ is positive (otherwise it would diverge to $-\infty$).
3. Note that $\int_0^y e^x |\sin(e^x)| dx = 2n(y) + 1 - (-1)^{n(y)} \cos(e^y)$, where $n(y) = \lfloor\frac{e^y-e^a}{\pi}\rfloor$ counts how many times $\sin(e^x)$ changes sign. Note that $n(y)=e^y/\pi$ plus a bounded term.
4. We have $$\begin{align} \int_a^b |f(x) \sin(e^x)| dx = & \int_a^b e^{-x} f(x) e^x |\sin(e^x)| dx \\ = & \left[ e^{-x} f(x) \left(2n(x) + 1 - (-1)^{n(x)} \cos(e^x)\right) \right]_a^b \\ & - \int_a^b e^{-x} \left(f'(x)-f(x)\right)\left(2n(x) + 1 - (-1)^{n(x)} \cos(e^x)\right) dx \end{align}$$ Almost everything in the preceding expression is bounded as $b\rightarrow \infty$. The first term becomes just $\frac{2}{\pi}f(b)$, and the second term becomes $\int_a^b \frac{2}{\pi}\left(f'(x)-f(x)\right) dx$ plus absolutely convergent terms. $\int_a^b f'(x) dx$ cancels the $\frac{2}{\pi}f(b)$ from the first term, and $\int_a^b f(x) dx$ diverges. So all in all we have a bunch of convergent terms plus one divergent term, and thus the whole thing diverges.
5. The last thing of note: In the preceding point I glossed over showing that $\int_a^b e^{-x} f'(x) dx$ converges absolutely. See my answer to the linked question for how this works. The problem is that $f'(x)$ need not be bounded, but we can obtain an effective bound by using $f'<f$ and boundedness of $f$.