Convergence or divergence of integral $\int_1^\infty \frac{x \sin x}{\sqrt{1+x^5}}dx$

Question

I'm struggling with how to show that

$$ \int_1^\infty \frac{x \sin x}{\sqrt{1+x^5}}dx $$ either diverges or converges.

If we call the integrand $f(x)$ then $$ f(x)\leq g(x)=\frac{x}{\sqrt{1+x^5}}\forall x\in[1, \infty) $$ so I tried to show that $g(x)$ converges (and hence that $f(x)$ must converge), but with no luck. Another idea I had was to show that

$$ h(x)=\frac{\sin x}{\sqrt{1+x^5}} $$ diverges, which would mean that $f(x)$ diverges too since

$$ h(x)\leq f(x)\forall x\in[1, \infty) $$ but I got nowhere.

I also tried to use some partial integration and get two expressions rather than one, but that didn't help.

Can I get some hints? Please note that it's homework so no solutions, thanks!

Answer 1

Note that your integrand is not positive, so you have to use absolute values. For the estimate, you have to get rid of the $1$.

Answer 2

It is (absolutely) convergent by the comparison test since $\int_1^\infty|\frac{x\sin x}{\sqrt{1+x^5}}|dx\leq \int_1^\infty\frac{1}{x^{3/2}}dx$ (convergent by $p$ test since $p=3/2>1$.