Show $\int_{0}^{\infty}\sin(f(x))dx$ converges if $\int_{0}^{\infty}f(x)dx$ converges and $f(x)$ is a decreasing, continuous function in $[0, \infty)$ 
Let $f(x)$ be a decreasing, continuous function in $[0, \infty)$.
  If $\int_{0}^{\infty}f(x)dx$ converges then $\int_{0}^{\infty}\sin(f(x))dx$ converges.



*

*The only improper point is $\infty$ since $f(x)$ is continuous in $[0, \infty)$.

*$f(x)$ has a bounded anti-derivate: if $\int_{0}^{\infty}f(x)dx$ converges, then $F(x)=\int_{0}^{x}f(t)dt$ is bounded.


If I set $g(x)=\frac{\sin(f(x))}{f(x)}$ and look at $\int_{0}^{\infty}g(x)f(x)dx$ then Dirichlets test seems like a possible candidate but I'm not sure if $g(x)\searrow 0$.
I feel close but seems like I'm missing something obvious here. Hints are appreciated.

As the commenters above pointed out, the only way for $\int_0^{\infty} f(x)$ to converge is for $\lim_{x \rightarrow \infty} f(x)$ to be zero: Since $f(x)$ is decreasing it converges to a lower bound $L = \inf_x f(x)$ which may be $-\infty$. If $L$ were negative or $-\infty$, then there'd be some $N$ and some $\epsilon > 0$ such that for $x > N$, we would have $f(x) < -{\epsilon}$. In this case $\int_{N}^{\infty} f(x)$ would be $-\infty$, which can't happen. Similarly, if $L$ were positive, there would be some $N$ and some $\epsilon > 0$ such that for $x > N$, we would have $f(x) > {\epsilon}$. In this case we'd similarly have  $\int_{N}^{\infty} f(x) = \infty$, impossible. We conclude that $L = 0$.
So $f(x)$ decreases to zero. Thus $f(x) = |f(x)|$ and we have that $\int_0^{\infty} |f(x)|$ converges. Since $|\sin(f(x))| = |{\sin(f(x)) \over f(x)}| |f(x)|$ and since $|{\sin(y) \over y}| < M$ for all $y$ for some $M$, $\int_0^{\infty}|\sin f(x)|$ must also be finite. Hence $\int_0^{\infty} \sin f(x)$ converges.
