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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]]
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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 $m,n\in\{0,1,2,...\}$.

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  • $\begingroup$ Your answer is great, thank you very much. $\endgroup$ – Please correct GrammarMistakes Aug 12 '20 at 1:48
  • $\begingroup$ I have another question. You have not proved that the convergence region of Taylor series contains [0,1], especially at point 1. How to solve this problem? $\endgroup$ – Please correct GrammarMistakes Aug 12 '20 at 2:44
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    $\begingroup$ It is easy enough to show that the radius of convergence includes $(0,1)$ (just start with the natural log term and simplify from there). For the endpoints, note that the integral is not even defined with these endpoints, and in fact is a limit as $\epsilon\to 0$ (where $\epsilon$ is the lower limit) and $\alpha\to 1$ (where $\alpha$ is the upper limit) of the integral respectively $\endgroup$ – QC_QAOA Aug 12 '20 at 3:04
  • $\begingroup$ Thank you very much. I hope you can attach the proof of convergence region containing (0,1) to the above answer in detail, so that it is easier for others to understand. $\endgroup$ – Please correct GrammarMistakes Aug 12 '20 at 3:23

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