I'm thinking that the class of functions which cause the integral to converge are those whose big O is above $n^3$.

Its not hard to show that if $f(x)$ is $x^3$, $x^4$, etc then the integral converges. I am having trouble with the function $f(x)=x^3+x$ for example and would probably be happy with an explanation of whether it converges or not.

Thinking intuitively, the reason the integral converges for those values of f is that it oscillates quicker than the amplitude grows by just enough. I guess the vague question is then what constitutes fast enough.

  • $\begingroup$ If the function is differentiable, "grows fast enough" seems to imply some condition on $\lim_{x\to \infty} f'(x)$. $\endgroup$ – Bruno Stonek Jul 26 '11 at 23:06
  • $\begingroup$ Yes. I was thinking of well behaved differentiable functions. By "grows fast enough" I assumed that a large enough Big O is what was necessary to guarantee that the integral converges. To make the issue more concrete, what polynomials f cause the integral to converge? $\endgroup$ – Searke Jul 26 '11 at 23:11
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    $\begingroup$ How does your integral converge? $\endgroup$ – Jonas Teuwen Jul 26 '11 at 23:16
  • $\begingroup$ I would think a classification of such functions is rather difficult. It definitely does not consist of only $O(x^3)$ functions. For example $f=0$ is contained, and many ugly functions which oscillate themselves. $\endgroup$ – Sam Jul 26 '11 at 23:21
  • $\begingroup$ It also converges for a lot of other functions besides the rapidly growing ones you have in mind. For example, it converges for $f(x)=7\pi+e^{-|x|}$ and for $f(x)=\pi\lfloor \sqrt{x} \rfloor$. $\endgroup$ – Ben Crowell Jul 27 '11 at 2:25

I think what you mean is something like $f(x) = \Omega(x^3)$, but this is not sufficient. For example, you could have $f(x) = 2 \lfloor x \rfloor^3 \pi + \pi/2$, in which case it's easy to see that the integral diverges.

For $f(x) = x^3 + x$ it does converge, and in fact Maple evaluates it as $\frac{2 \pi}{9} \left( \sqrt{3} I_{1/3}(\frac{2\sqrt{3}}{9}) - I_{2/3} (\frac{2 \sqrt{3}}{9}) + I_{4/3}(\frac{2 \sqrt{3}}{9})\right)$ (where the $I$'s are modified Bessel functions of the first kind). I think the first steps in a proof of convergence might be to look at the integral on $[1,\infty)$, do a change of variables $x = t^{1/3}$, expand the sines, then integrate by parts.

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  • $\begingroup$ Now that you say it Sin(x^3+x) does become Cos(x^3) Sin(x) + Cos(x) Sin(x^3) which helps a ton. Thanks. I guess I was mostly hoping for some nice sufficient condition for a broad class of functions. $\endgroup$ – Searke Jul 27 '11 at 0:05

For polynomials, the integrals converge iff the degree of $f(x)$ is at least 3. To show this, you remove an interval $(-a,a)$ containing the zeroes of $f'(x)$ (over which the integral trivially converges), and rewrite the remaining integral as $$\int_{|x| > a} {x \over f'(x)} (f'(x)\sin f(x))\,dx$$ By definition, the integral converges iff the $x > a$ and $x < -a$ parts both converge. We focus on the $x > a$ part as the $x < -a$ part is done the same way. Next, integrate by parts, obtaining $$-\lim_{a \rightarrow \infty}{a \over f'(a)}\cos(f(a)) + {a \over f'(a)}\cos(f(a)) + \int_a^{\infty}({d \over dx}{x \over f'(x)}) \cos(f(x))\,dx $$ If $f(x)$ is a polynomial of degree $2$ or more, $({d \over dx}{x \over f'(x)})$ is a rational function whose denominator exceeds that of the numerator by at least $2$, so that you have an estimate $|({d \over dx}{x \over f'(x)})| < C{1 \over x^2}$ for some constant $C$. Thus the integral converges by comparison with $C{1 \over x^2}$.

If the degree of $f(x)$ is at least three, then the degree of $f'(x)$ is at least $2$, and -$$\lim_{a \rightarrow \infty}{a \over f'(a)}\cos(f(a)) = \lim_{a \rightarrow \infty}{a \over f'(a)} = 0$$ As a result, if $f(x)$ has degree $3$ or more, the overall integral converges.

If the degree of $f(x)$ is two, then $f'(x)$ has degree $1$ and $\lim_{a \rightarrow \infty}{a \over f'(a)}$ is some nonzero value. Thus the function ${a \over f'(a)}\cos(f(a))$ oscillates and there is no limit; the original integral diverges.

All that remains is the case where $f(x)$ has degree one. For this it's probably easiest just to show that the integral over a given period diverges as $x$ goes to infinity; it diverges faster than the integral of $\cos(f(x))$ which already diverges.

The above technique can be used for nonpolynomials too, but I don't know any sufficient and necessary conditions for it to work offhand.

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    $\begingroup$ Nice technique. +1. $\endgroup$ – Jonas Teuwen Jul 27 '11 at 13:46

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