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.

  • $\begingroup$ Note that for $u$ very close to $0$, $\sin(u)\approx u$; since $\int_0^{\infty}f(x)\,dx$ converges and $f(x)$ is decreasing and continuous, there will be an $M$ such that for all $x\geq M$, $f(x)$ is "very close" to $0$, so that $\sin(f(x))\approx f(x)$. So you would expect $\int_M^{\infty}\sin(f(x))\,dx \approx \int_M^{\infty}f(x)\,dx\lt\infty$. This is just a heuristic, but maybe you can turn it into a proof. $\endgroup$ – Arturo Magidin Feb 3 '11 at 16:44
  • $\begingroup$ @daniel.jackson: As $x\to\infty$, you have $f(x)\to 0^+$, so$$\lim_{x\to\infty}\frac{\sin(f(x))}{f(x)} = \lim_{u\to 0^+}\frac{\sin(u)}{u} = 1,$$so your $g$ does not decrease to $0$. $\endgroup$ – Arturo Magidin Feb 3 '11 at 16:48
  • $\begingroup$ @Arturo: Why does $f(x)\to 0^+$? $\endgroup$ – daniel.jackson Feb 3 '11 at 16:59
  • $\begingroup$ @daniel.jackson: Since $\int_0^\infty f(x)\, dx$ converges you must have $f(x)\rightarrow 0$. Since $f$ is decreasing, it must go to 0 from the right. $\endgroup$ – Joe Johnson 126 Feb 3 '11 at 17:04
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    $\begingroup$ @daniel.jackson: The statement holds if $f(x)$ is continuous, though, which is your third assumption (and the reason it's necessary!) $\endgroup$ – Steven Stadnicki Feb 3 '11 at 17:12

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.

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    $\begingroup$ $|\sin(y)|\le|y|$ for every $y$ hence, once you know that $f(x)\ge0$ for every $x$, you can simply use the inequality $\sin(f(x))\le f(x)$ to conclude. $\endgroup$ – Did Feb 3 '11 at 19:22
  • $\begingroup$ Well it's $|\sin(f(x)| \leq f(x)$ so I think you still have to use absolute convergence. $\endgroup$ – Zarrax Feb 3 '11 at 19:41
  • $\begingroup$ Yes, I meant $|\sin f(x)|\le f(x)$. $\endgroup$ – Did Feb 3 '11 at 19:48

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