Judge the convergence of this integral $\int_{0}^{1} \frac{\sqrt[m]{\ln ^{2}(1-x)}}{\sqrt[n]{x}} d x$ Note: The following questions are from the third question of the 2010 Chinese Graduate Entrance Examination Mathematics (first set):
Suppose m, n are positive integers, which of the following conclusions on the convergence of the abnormal integral $\int_{0}^{1} \frac{\sqrt[m]{\ln ^{2}(1-x)}}{\sqrt[n]{x}} d x$ is correct?
$$\begin{array}{c}
&(A)&\text{It is Only related to the value of m} \\
&(B)& \text{It is Only related to the value of n }\\
&(C)& \text{It is related to the values of m and n}\\
&(D)&\text{It has nothing to do with the value of m, n}\\
\end{array}$$
I used Mathematica to verify this problem, but I didn't get results:
Integrate[Power[Log[1 - x]^2, (m)^-1]/Power[x, (n)^-1], {x, 0, 1}, 
 Assumptions -> Element[m | n, PositiveIntegers]]

 A: The answer is $D)$ for the following reason. First, expand the series around $x=0$. This gives you
$$\frac{\sqrt[m]{ \log ^2(1-x)}}{\sqrt[n]{x}}$$
$$=x^{\frac{2}{m}-\frac{1}{n}}+\frac{x^{\frac{2}{m}-\frac{1}{n}+1}}{m}+\frac{5 x^{\frac{2}{m}-\frac{1}{n}+2}}{12 m}+\frac{x^{\frac{2}{m}-\frac{1}{n}+3}}{4 m}+\frac{x^{\frac{2}{m}-\frac{1}{n}+3}}{6 m^3}+\frac{x^{\frac{2}{m}-\frac{1}{n}+2}}{2 m^2}+\frac{5 x^{\frac{2}{m}-\frac{1}{n}+3}}{12
   m^2}+\cdots$$
Now, it is not too hard to prove that the terms in the exponent will always be of the form
$$k+\frac{2}{m}-\frac{1}{n}\text{ for }k\in\{0,1,2,...\}$$
(I leave the details of this as an exercise). In general, the integral
$$\int_0^1 x^\beta dx$$
converges unless $\beta\leq -1$. Thus, our question becomes, is it ever possible for
$$k+\frac{2}{m}-\frac{1}{n}\leq -1$$
The answer is no. This is obvious as
$$-1=0-1<\frac{2}{m}-1\leq \frac{2}{m}-\frac{1}{n}\leq k+\frac{2}{m}-\frac{1}{n}$$
We conclude that no matter which $n,m\in\mathbb{N}$ we choose, the given integral will always converge.

EDIT: I was going to formally prove the radius of convergence for the power series above. However, I got bogged down in the details so instead here is a more rigorous proof of an entirely different flavor:
First, note that
$$\log^2(1-x)=(-\log(1-x))^2$$
(and that $\log(1-x)<0$ for $x\in(0,1)$). We also know that around $x=\frac{1}{2}$ we have
$$-\log(1-x)=\log(2)+\sum_{i=1}^\infty \frac{2^i}{i}\left(x-\frac{1}{2}\right)^i$$
$$\frac{x}{1-x}=1+\sum_{i=1}^\infty 2^{i+1}\left(x-\frac{1}{2}\right)^i$$
It is a standard exercise to compute both these series and show that the radius of convergence of both is $r=\frac{1}{2}$. Since $\log(2)\approx0.69<1$, we can compare term by term and see that for $x\in (0,1)$
$$-\log(1-x)<\frac{x}{1-x}$$
This implies
$$\int_0^1 \frac{\sqrt[m]{ \log ^2(1-x)}}{\sqrt[n]{x}}dx\leq \int_0^1 \frac{\left(\frac{x}{1-x}\right)^{2/m}}{\sqrt[n]{x}}dx$$
For $m\geq 3$, this integral evaluates to
$$=\frac{\Gamma \left(\frac{m-2}{m}\right) \Gamma \left(-\frac{1}{n}+1+\frac{2}{m}\right)}{\Gamma \left(2-\frac{1}{n}\right)}$$
For $m=2$ and $n\geq 2$, the integral evaluates to
$$\frac{n H\left(\frac{n-1}{n}\right)}{n-1}$$
where $H(x)$ is the function given by
$$H(x)=\int_0^1\frac{1-t^x}{1-t}dt$$
This converges for all $x>-1$. For $m=2$ and $n=1$ we get $\frac{\pi^2}{6}$. For $m=1$ and $n=1$, the integral becomes $2\zeta(3)$ where
$$\zeta(x)=\sum_{i=1}^\infty \frac{1}{i^x}$$
This converges for $x>1$. For $m=1$ and $n=2$, the integral becomes
$$2 \left(8+\log ^2(4)-8 \log (2)\right)-\frac{2 \pi ^2}{3}$$
Finally, for $m=1$ and $n\geq 3$ we will use Holder's Inequality. For positive functions, the inequality states
$$\int f(x)g(x)dx\leq \left(\int f(x)^pdx\right)^{1/p}\left(\int g(x)^qdx\right)^{1/q}$$
where $q,p\in [1,\infty)$ and $\frac{1}{q}+\frac{1}{p}=1$. In our case, we will use
$$f(x)=(-\log(1-x))^2$$
$$g(x)=\frac{1}{x^{1/n}}$$
$$q=p=2$$
Then the inequality states
$$\int_0^1 \frac{ \log ^2(1-x)}{\sqrt[n]{x}}=\int_0^1 \frac{(-\log(1-x))^2}{x^{1/n}}\leq \left(\int_0^1(-\log(1-x))^4dx\right)^{1/2}\left(\int_0^1\frac{1}{x^{2/n}}dx\right)^{1/2}$$
$$=\left(24\right)^{1/2}\left(\frac{n}{n-2}\right)^{1/2}$$
Having exhausted all cases, we conclude the integral converges for all $m,n\in\{0,1,2,...\}$.
