Considering the behaviour of the integrant at both integration limits, study the convergence of the integral:
$$\int_{0}^\infty x \sin\left(\frac{1}{x^\frac{3}{2}}\right). $$
I was trying to integrate it but I don't know if that is even possible and also to use some comparison test but it wasn't really helpful.
I know that I can rewrite it as:
$$\int_{0}^c x \sin\left(\frac{1}{x^\frac{3}{2}}\right) + \int_{c}^\infty x \sin\left(\frac{1}{x^\frac{3}{2}}\right)$$ where $c$ is some positive number, but I don't know what to do next.
How should I test the convergence of that integral?
As $\lim_{x\to0}{\sin x\over x}=1$,$\;x\to\infty\implies x\sin(x^{-1.5})\to x\cdot x^{-1.5}=x^{-0.5}$. If $c>0,$ $\int_c^\infty x^{-p}\,dx<\infty\iff p>1.$ So the $2$nd integral doesn't converge. As $|x\sin(x^{-1.5})|\le |x|$ and $\int_0^c|x|dx=0.5c^2,$ the $1$st integral converges. So the integral $\int_0^\infty x\sin(x^{-1.5})$ doesn't converge.