$\int_a^\infty f(x)\sin(e^x)\text{d}x$ conditionally convergent 
Let $f$ be a bounded $C^1$ function on $[a,\infty)$ such that
  $\int_a^\infty f$ diverges. Assume there is $t>a$ s.t $f'(x)<f(x)$ on
  $[t,\infty)$. prove $\int_a^\infty f(x)\sin(e^x)\text{d}x$ is 
  conditionally convergent.

I couldn't really get anything done. It seems like a Dirichlet test result but $f$ doesn't have to approach $0$ at infinity, nor it is monotonic. The given property $f'<f$ seems related to the $e^x$ in the integral but I can't connect the two. The fact that  $\int_a^\infty f$ diverges seems to hint some sort of comparision test, but that too does not yield anything as $\underbrace{f(x)\sin(e^x)}_\text{????}\leq \underbrace{f(x)}_\text{integral will diverge}$ 
 A: Let $\text{Si}(z)=\int_{0}^{z}\frac{\sin u}{u}\,du$. By integration by parts we have:
$$\begin{eqnarray*}\int_{a}^{M}\sin(e^x)\,f(x)\,dx &=& \int_{e^a}^{e^M}\frac{\sin u}{u}\,f(\log u)\,du\\&\stackrel{\text{IBP}}{=}&\left[\text{Si}(u)\,f(\log u)\right]_{e^a}^{e^M}-\int_{e^a}^{e^M}\frac{\text{Si}(u)}{u}\,f'(\log u)\,du\end{eqnarray*} $$
where $\text{Si}(u)\to\frac{\pi}{2}$ as $u\to +\infty$. In particular there is some $b\in\mathbb{R}^+$ such that $\text{Si}(u)\in\left[\frac{3}{2},2\right]$ for any $u\geq b$. Let $c=\max(e^a,b)$. Disregarding what happens on the bounded interval $[e^a,c]$, for any $M$ such that $e^M\geq c$ it is simple to bound both $[\text{Si}(u)\,f(\log u)]_{c}^{e^M}$ and
$$ \int_{c}^{e^M}\frac{\text{Si}(u)}{u}\,f'(u)\,du$$
(by the mean value theorem for integrals) with an absolute constant, only depending on $\sup_{x\in\mathbb{R}^+}\left|f(x)\right|$ and $b$. This proves that $\int_{a}^{M}\sin(e^x)\,f(x)\,dx$ is uniformly bounded with respect to $M$. In order to finish the proof, it is enough to show that
$$ \lim_{M\to +\infty}\int_{a}^{M}e^{-x} f(x)\cdot \underbrace{e^x \sin(e^x)}_{\text{bounded primitive}}\,dx $$
exists. Invoking Dirichlet's test, it is enough to show that $f(x)e^{-x}$ is decreasing towards zero from some point on. Can you finish the proof by yourself, noting that $f'(x)<f(x)$ for any $x\geq t$ is among the assumptions? Futher hint: consider the derivative of $f(x)e^{-x}$ and the fact that $f$ is bounded.
Side note: it looks to me that the assumption about the unbounded-ness of $\int f$ never comes into play.
