Assume that p > 0, find the all the values of p so that the following 2 improper integrals converge :-
$$\int_0^2 \frac{1}{\sqrt[3]{x^{p}-1}} dx $$ $$\int_2^\infty \frac{1}{\sqrt[3]{x^{p}-1}} dx $$
I tried solving them and I got stuck. The first integral must be split into 2 improper integrals between 2 intervals, the first one is between 0 and 1 not and the second one is between 1 and 2 because the integrand is not continuous in 1. I tried substituting variables but nothing worked. Thanks for any help.
HINT
Note that for $x\to 1$
$$x^{p}=(1+(x-1))^p\sim1+p(x-1)\implies \sqrt[3]{x^{p}-1}\sim \sqrt[3] p \sqrt[3]{x-1}$$
then
$$\frac{1}{\sqrt[3]{x^{p}-1}} \sim \frac1{\sqrt[3] p \sqrt[3]{x-1}}$$
and for $x\to \infty$
$$\frac{1}{\sqrt[3]{x^{p}-1}} \sim \frac1{x^{p/3}}$$