Does the integral $$ \int_0^\infty \left| \frac{\cos(2x)+x \sin(2x)-1}{x^3} \right| dx $$ converge?
The obvious upper bound $$ \left| \frac{\cos(2x)+x \sin(2x)-1}{x^3} \right| \leq \frac{|\cos(2x)|+|x \sin(2x)| + 1}{|x^3|} \leq \frac{2+|x|}{|x^3|} $$ does not work, since $\int_0^\infty \frac{2+x}{x^3} dx$ is not convergent.
Mathematica says this integral is convergent. How can I find a convergent upper bound? Thanks.
Split the integral up into $$ \int^\infty_0 \left \lvert \frac{\cos(2x)+x\sin(2x)-1}{x^3} \right \rvert = \underbrace{\int^1_0 \left \lvert \frac{\cos(2x)+x\sin(2x)-1}{x^3} \right \rvert~\mathrm{d}x}_{=:A} + \underbrace{\int^\infty_1 \left \lvert \frac{\cos(2x)+x\sin(2x)-1}{x^3} \right \rvert~\mathrm{d}x}_{=:B} $$ Now prove $$ \lim_{x \downarrow 0} \left \lvert \frac{\cos(2x)+x\sin(2x)-1}{x^3} \right \rvert = 0 $$ using e.g. l'Hopital. Therefore $A< \infty$, because the integrand stays bounded on a compact set.
For $B$ you can now use that $\displaystyle\int^\infty_1 \frac{2 + x}{x^3}~\mathrm{d}x$ does indeed converge. So $B< \infty$.