Get stuck in judging convergence of a improper integral

Question

I get stuck in judging convergence of the following formula:$$\int_0^1{\frac{\sqrt[m]{\ln^2(1-x)}}{\sqrt[n]{x}}}dx$$ $m$ and $n$ are both integers

When $x\to0^+$, I can use equivalent infinitesimal to figure it out. it's convergent in $x=0$.

But when $x\to1^-$, I don't know use what kind of theorem to judge.(for it seems really complicated to do manual calculation)

Answer 1

As $x \to 1^-$, you have $$ \frac{\sqrt[m]{\ln^2(1-x)}}{\sqrt[n]{x}}\sim \left(-\ln(1-x)\right)^{2/m} $$ then, with the change of variable $u=-\ln (1-x)$ and $0<\epsilon<1$, $$ \int_{1-\epsilon}^1\left(-\ln(1-x)\right)^{2/m}dx=\int_{\epsilon}^{+\infty}u^{2/m}e^{-u}du<\int_0^{+\infty}u^{2/m}e^{-u}du<+\infty $$ and your initial integral is convergent.

Answer 2

if $x \to 1$ you only care about the logarithm to judge the convergence of the integral. Also note that the argument of the logarithm is going to $0$. So you reduce to the question

When $$\int_0^1 \ln^\alpha (x) dx$$ converges?

With $t = \frac 1x$, the integral reduces to $$=\int_1^\infty \frac 1{t^2}\ln^\alpha \left(\frac 1t\right) dt \le \int_1^\infty \frac 1{t^{2 + \alpha}} dt < \infty$$

Since the last integral converges for $\alpha > -1$, we conclude that $$\int_0^1 \ln^\alpha(x)dx$$ converges for every $\alpha > -1$ so your integral converges in $1$ for every $m,n$