For what functions does $\int\limits_{-\infty}^{\infty}x \sin(f(x))\,dx$ converge? 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.
 A: 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.
A: 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.
