<s>[INST] Does the integral $\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}$ diverge Does the integral 
$$J:=\int_{0}^{\infty} \frac{x^3\,\cos{(x^2-x)}}{1+x^2}dx $$
diverge ?
If we integrate by parts we find
$$J=\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)} -\\
\int_{0}^{\infty}\sin{(x^2-x)}
\left( \frac{3x^2}{(1+x^2)(2x-1)}-\\
\frac{2x^4}{(1+x^2)^2(2x-1)}-\frac{2x^3}{(1+x^2)(2x-1)^2}\right)dx$$
If we integrate by parts again, we find that the latter integral reduces to 
a sum of absolutely convergent integrals plus
the boundary term
$$ -\lim_{a\rightarrow +\infty}\,
\cos{(a^2-a)}\left(\frac{3a^2}{(1+a^2)(2a-1)^2}
-\\
\frac{2a^4}{(1+a^2)^2(2a-1)^2}-\frac{2a^3}{(1+a^2)(2a-1)^3}
\right)=0.$$
So the question simplifies to the existence of the limit
$\lim_{a\rightarrow +\infty} \frac{a^3}{(1+a^2)(2a-1)}\cos{(a^2-a)}$ which I believe does not exist. 
Correct ?
 [/INST] The integral is divergent because the limit does not exist.
Let $f(a)=\frac{a^3\,\cos{(a^2-a)}}{(1+a^2)(2a-1)}$.
Take the sequence $a_{n}=:\frac{1+\sqrt{1+8\pi n}}{2}$
that diverges to infinity and satisfies $a_{n}^2-a_{n}=2n\pi$.
Then $f(a_{n})\rightarrow \frac{1}{2}$. 
Now, take the sequence $b_{n}:=\frac{1+\sqrt{1+4(2n+1)\pi }}{2}$ that also diverges to infinity and satisfies $b_{n}^2-b_{n}=(2n+1)\pi$. Then 
$f(b_{n})\rightarrow -\frac{1}{2}$.
</s>