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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.

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    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.2011-02-03
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    @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 $0$.2011-02-03
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    @Arturo: Why does $f(x)\to 0^+$?2011-02-03
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    @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.2011-02-03
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    @Joe Johnson: In general if $\int_0^\infty f(x)dx$ converges we need not have $f(x) \to 0$. ($f$ could have infinitely many bumps of large height but very small width.) But for decreasing functions this does hold...2011-02-03
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    @Joe: not sure I follow. You're saying $\int_0^\infty f(x) dx \implies \lim_{x\to \infty}f(x)=0$? What about $g(x)=x$ if $x \in N,\ g(x)=0$ otherwise?2011-02-03
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    @daniel.jackson: The statement holds if $f(x)$ is continuous, though, which is your third assumption (and the reason it's necessary!)2011-02-03
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    @Steven: could you give a hint as to why $f(x)\to 0$? I understand why intuitively but having some difficulty showing it... I'm familiar with a similar result that had $\lim_{x\to \infty}f(x)$ existed, then $\lim_{x\to \infty}f(x)=0$ must be true.2011-02-03
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    @daniel.jackson You can show that $f(x)$ stays positive by noting that if it ever becomes negative, its decreasing nature implies that it's bounded away from zero from that point on (and prove that this means its improper integral from $0$ to $\infty$ doesn't converge); since it's bounded below, decreasing, and continuous, then it has a limit. Once you know that it has a limit then as you say it's straightforward to show that the limit is $0$.2011-02-03
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    Just to clarify: Without the assumption that $f$ is decreasing, you can not conclude from $\int_0^\infty f(x)dx < \infty$ that $f(x)\to 0$ and even continuity does not help. Imagine a function which consists of bumps at every natural number which are getting narrower fast enough such that the integral stays finite. However, what you can conclude is that $f$ can not be bouded away from zero from some point on.2011-02-03

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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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    $|\sin(y)|\le|y|$ 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.2011-02-03
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    Well it's $|\sin(f(x)| \leq f(x)$ so I think you still have to use absolute convergence.2011-02-03
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    Yes, I meant $|\sin f(x)|\le f(x)$.2011-02-03