Cfrgtkky

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

$$J=lim_{arightarrow +infty} frac{a^3}{(1+a^2)(2a-1)}cos{(a^2-a)} -\
int_{0}^{infty}sin{(x^2-x)}
left( frac{3x^2}{(1+x^2)(2x-1)}-\
frac{2x^4}{(1+x^2)^2(2x-1)}-frac{2x^3}{(1+x^2)(2x-1)^2}right)dx$$

If we integrate by parts again, we find that the latter integral reduces to
a sum of absolutely convergent integrals plus
the boundary term

$$ -lim_{arightarrow +infty},
cos{(a^2-a)}left(frac{3a^2}{(1+a^2)(2a-1)^2}
-\
frac{2a^4}{(1+a^2)^2(2a-1)^2}-frac{2a^3}{(1+a^2)(2a-1)^3}
right)=0.$$

So the question simplifies to the existence of the limit
$lim_{arightarrow +infty} frac{a^3}{(1+a^2)(2a-1)}cos{(a^2-a)}$ which I believe does not exist.

Correct ?

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 $a_{n}=:frac{1+sqrt{1+8pi n}}{2}$
that diverges to infinity and satisfies $a_{n}^2-a_{n}=2npi$.
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
$f(b_{n})rightarrow -frac{1}{2}$.

Also you can do that: an improper integral $lim_{rtoinfty}int_0^r f(x), dx$ converges if and only if for each $epsilon>0$ there is some $M>0$ such that $left|int_a^b f(x), dxright|<epsilon$ for all pairs $a,bge M$.

Now note that $cos(x)gesqrt 2/2$ when $xin(-pi/4+2kpi,pi/4+2kpi)$ for any $kinBbb Z$. Now observe that

$$x^2-x=alpha,land, x,alpha>0iff x=frac{1+sqrt{1+4alpha}}2$$

Now for $alpha=2pi npmpi/4$ we set $a_{npm}:=frac12(1+sqrt{1+pi(4npm 1)})$ and in your case you have that

$$begin{align}I_n&:=left|int_{a_{n-}}^{a_{n+}}f(x), dxright|gefrac{sqrt2}2int_{a_{n-}}^{a_{n+}}frac{x^3}{1+x^2}, dx\
&gefrac{sqrt 2}2cdot(a_{n+}-a_{n-})min_{xin[a_{n-},a_{n+}]}frac{x}2,&text{because }frac{x^3}{1+x^2}ge frac{x}2text{ when }xge 1\
&=frac{sqrt 2}2(a_{n+}-a_{n-})cdotfrac{a_{n+}+a_{n-}}{a_{n+}+a_{n-}}cdotfrac{a_{n-}}2\
&gefrac{sqrt 2}2cdotfrac{a^2_{n+}-a^2_{n-}}{2a_{n+}}cdotfrac{a_{n-}}2,&text{ because } 2a_{n+}ge a_{n-}+a_{n+}\
&gefrac{pisqrt 2}{16}cdotsqrt{frac{1+pi(4n-1)}{1+pi(4n+1)}}end{align}$$

Hence $lim_{ntoinfty}I_ngepisqrt 2/16$, so we can conclude that the integral doesn't converge.