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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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    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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    Yes, I meant $|\sin f(x)|\le f(x)$.2011-02-03