How to prove improper integral $\int_0^1 {\frac{1}{{{{(-\ln x)}^p}}}dx} $ diverges when $p>=1$?

Question

I can prove $\int_0^1 {\frac{1}{{{{(-\ln x)}^p}}}dx} $ converges when $p<1$, since $ - \log x \ge 1 - x$ when $x\in[0,1]$ and $\int_0^1 {\frac{1}{{{{(-\ln x)}^p}}}dx} < \int_0^1 {\frac{1}{{{{(1 - x)}^p}}}dx} = \int_0^1 {\frac{1}{{{x^p}}}dx} $ and $\int_0^1 {\frac{1}{{{x^p}}}dx} $ converges when $p<1$. My question is how to prove $\int_0^1 {\frac{1}{{{{(-\ln x)}^p}}}dx} $ diverges when $p>=1$? Thank you.

Answer 1

Let $u = \dfrac{1}{x}, I = \displaystyle \int_{1}^\infty \dfrac{1}{u^2(\ln u)^p}du$. Next let $v = \ln u \to u = e^v \to du = e^vdv \to I = \displaystyle \int_{0}^\infty \dfrac{e^{-v}}{v^p}dv \geq \displaystyle \int_{0}^1 \dfrac{e^{-v}}{v^p}dv \geq \dfrac{1}{e}\cdot \displaystyle \int_{0}^1 v^{-p}dv=+\infty$

Answer 2

Compare with $$ \frac{1}{(1-x)^p}. $$