Convergence/Divergence of Integral, can P-test be used here?

Question

I have an integral like this:

How do I check its convergence? As far as I know, P-test can be used for integrals from 0 to 1, or A to infinity, what would I do in this case?

Answer 1

The "$p$-test" is a special case of comparison tests for integrals; it's easy to remember and very useful, but it doesn't apply to every integral. One should be willing to go back directly to the comparison tests themselves.

In this case, for convergence we need a larger function whose integral still converges on $[0,2]$. Since $x+2>2$ for $x\in[0,2]$, we have $$ 0< \frac2{\sqrt[3]x(x+2)} < \frac1{\sqrt[3]x} \quad (0<x< 2); $$ and since $$ \int_0^2 \frac1{\sqrt[3]x} \,dx = \lim_{\varepsilon\to0+} \int_\varepsilon^2 \frac1{\sqrt[3]x} \,dx = \lim_{\varepsilon\to0+} \bigg( \frac{x^{2/3}}{2/3} \bigg|_\varepsilon^2 \bigg) = \lim_{\varepsilon\to0+} \bigg( \frac{2^{2/3}}{2/3} - \frac{\varepsilon^{2/3}}{2/3} \bigg) = \frac3{2^{1/3}} $$ is a convergent integral, the comparison test shows that the original integral also converges.