Show that $\int_{0}^{1}x^pdx$ converges for $p>-1$ and diverges otherwise.
I know how to show that this integral converges for values less than one and diverges for all others, but the question that I've posted seems to say the opposite.
So my question is can I show this the same way that I should show that it converges, or could this have been a typo on my professor's part? I'm not sure why it's -1 instead of 1. Am I missing something obvious? Any thoughts would be appreciated!
Consider the equivalent statement: Show that $\int_{0}^{1}x^{-p}dx$ converges for $p<1$ and diverges otherwise.
$$\int_{0}^{1}x^{-p}dx=\lim_{c\to0}\int_{c}^{1}x^{-p}dx$$
If p<1 then: $$\int_{0}^{1}x^{-p}dx=\lim_{c\to0}\frac{1}{1-p}(1-c^{1-p})=\frac{1}{1-p}$$ We see that for p <1 it converges.
If p=1 then: $$\int_{0}^{1}x^{-p}dx=\lim_{c\to0}-\ln(c)=+\infty$$ Thus for p = 1 it diverges.
If p>1 then: $$\int_{0}^{1}x^{-p}dx=\lim_{c\to0}\frac{1}{1-p}(1-c^{1-p})=+\infty$$ So for p>1 it diverges.
Therefore, since the statements are equivalent $\int_{0}^{1}x^{-p}dx$ converges for p> -1 and diverges otherwise.