$\int_0^\infty \frac{1}{(x^p+2020)^q} \,dx $

Question

Let $p,q>0$, when $$\int_0^\infty \frac{1}{(x^p+2020)^q} \,dx $$ converges?

I know when $q=1$, $p\ge2$ is the condition, and if $p\ge2$, $q\ge1$　is the condition, but in other case, I have no idea. Thank you for your help in solving this.

Answer 1

Hint $$ \int_0^\infty \frac{1}{(x^p＋2020)^q} \,dx$$ converges if and only if $$\int_1^\infty \frac{1}{(x^p＋2020)^q} \,dx$$

Now apply the Limit Comparison Theorem for $$\int_1^\infty \frac{1}{(x^p＋2020)^q} \,dx \, \mbox{ and } \, \int_1^\infty \frac{1}{x^{pq}} \,dx$$

Answer 2

Convergence of the integral on the interval $ [0,1] $ always occurs by continuity (thanks to the $ 2020 $ term). So we can restrict our attention to the interval $ [1, \infty) $. We see that the integrand $ I_1 = (x^p + 2020)^{-q} $ can be directly compared to $ I_2 = x^{-pq} $. Indeed we have $ I_1 \leq I_2 $; but also it's not hard to find a constant $ C $ such that $ I_2 \leq C I_1 $ for $ x \in [1, \infty) $. Thus, the original integral converges on $ [1,\infty) $ precisely if $ pq > 1 $.