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.



Note that for $x\to 1$

$$x^{p}=(1+(x-1))^p\sim1+p(x-1)\implies \sqrt[3]{x^{p}-1}\sim \sqrt[3] p \sqrt[3]{x-1}$$


$$\frac{1}{\sqrt[3]{x^{p}-1}} \sim \frac1{\sqrt[3] p \sqrt[3]{x-1}}$$

and for $x\to \infty$

$$\frac{1}{\sqrt[3]{x^{p}-1}} \sim \frac1{x^{p/3}}$$

  • $\begingroup$ what about x = 1 ? $\endgroup$
    – CSDude101
    May 25 '18 at 11:25
  • $\begingroup$ Ops yes of course you are right, I update $\endgroup$
    – user
    May 25 '18 at 11:27
  • $\begingroup$ Thanks for the help but could you please explain the situation when x goes to 1 ? it is not very clear to me $\endgroup$
    – CSDude101
    May 25 '18 at 11:40
  • $\begingroup$ The expression $\frac{1}{(x-1)^{1/3}}$ is integrable. $\endgroup$
    – Paichu
    May 25 '18 at 11:53
  • $\begingroup$ I get that the expression that you wrote is integrable but why does the power of x go from p to 1 when x approaches 1 ? $\endgroup$
    – CSDude101
    May 25 '18 at 11:57

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