Problem: Prove that the improper integral $\displaystyle\int_{-\infty}^\infty \frac{x^{1/3}}{\sqrt{1+x^2}}\,dx$ does not converge, but it has Cauchy principal value $0$.
I proved the second part using the fact that the function is symmetric, but I can't prove the first part. I split the integral and tried to prove the nonconvergence of the plus infinity part by showing that it's bigger than another integral that is known to be infinite but I couldn't come up with a suitable function.
Note that$$\lim_{x\to\infty}\frac{\frac{x^{1/3}}{\sqrt{1+x^2}}}{x^{-2/3}}=\lim_{x\to\infty}\frac x{\sqrt{1+x^2}}=1$$and that therefore your integral converges if and only if the integral $\displaystyle\int_1^\infty x^{-2/3}\,\mathrm dx$ converges. But it does not: $x^{-2/3}>x^{-1}$ and $\displaystyle\int_1^\infty x^{-1}\,\mathrm dx$ diverges.