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My instructor has a fondness for asking questions regarding the convergence of such integrals: $$ \int_{0}^{1} \frac{\ln(x)}{x^{1/2}}\,\mathrm dx $$

$$ \int_{0}^{1} \frac{\ln(x)}{x^{3/2}}\,\mathrm dx $$

What is the best way to determine the convergence of such integrals? Comparison tests require that $ f(x), g(x) \ge 0 $ so let's consider $-\ln x$.

How can I proceed from here?

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4 Answers 4

up vote 1 down vote accepted

More or less André Nicolas' answer in a slightly different form:

  • We have $f(x) = |\ln x|\,x^{-1/2}= -(\ln x) \,x^{-1/2}$ which we claim to be smaller than $g(x)= \alpha\,x^{3/4}$ for all $x\in[0,1]$ for an appropriate $\alpha>0$. To prove that we introduce $h(x)= g(x)/f(x)$ and seek for its extrema $h'(x^*)=0$ with the only solution $x^*=e^{-4}$. It is easy to show that $h''(x^*) >0$ such that the extremum is in fact a minimum. With $g(x^*)= e \alpha/4$ we see that $g(x) > f(x)$, e.g., for $\alpha=2$. In conclusion, we have $$\int_0^1\frac{|\ln x|}{x^{1/2}} dx \leq \int_0^1 \frac{2}{x^{3/4}}dx =8.$$

  • In the second case, we have $f(x) =|\ln x|\,x^{-3/2}= -(\ln x) \,x^{-3/2}$. We choose $g(x)= (1-x)x^{-3/2}$ and have $g(x)\leq f(x)$ because $(1-x) \leq -\ln x$. Thus, $$\int_0^1\frac{|\ln x|}{x^{3/2}} dx \geq \int_0^1 \frac{1-x}{x^{3/2}}dx \to \infty.$$

With similar arguments, you can show that $$\int_0^1 \frac{|\ln x|}{x^p} dx$$ converges for $p<1$. In fact, these integral can be easily evaluated explicitly via integration by parts. We have $$\int_0^1 \frac{|\ln x|}{x^p} dx = -\int_0^1 \frac{\ln x}{x^p} dx =\underbrace{\lim_{x\downarrow 0} \frac{\ln x}{(1-p) x^{p-1}}}_{0}+ \underbrace{\int_0^1 \frac{1}{(1-p)x^p} dx}_{(1-p)^{-2}} .$$ The results below the underbrace hold for $p<1$ otherwise both terms diverge.

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Hint: For the second one, use this fact that $$\lim_{x\to 0^+}(x-0)^\frac{3}{2}\frac{\ln x}{x^{\frac{3}{2}}}$$ is infinite.

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Another long overdue +1 $\Large\checkmark$ –  amWhy Apr 11 '13 at 1:50
    
@amWhy: Thanks my dearest friend. –  Babak S. Apr 11 '13 at 7:18

Look at the leading order behavior at the singularity. In both cases you gave, the singularity is at $x=0$. The logarithm integrates to something finite near $x=0$, and $x^a$ integrates to something finite near $x=0$ as long as $a>-1$.

Then in your specific cases, $x^{-1/2}\ln x$ will have a finite integral on $(0,1)$ while $x^{-3/2}\ln x$ will diverge.

These comments are more heuristic than anything, but it sounds like your teacher wants you to be able to make an assessment by eye.

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Ah yes, this intuitively makes sense.. thank you. You are correct, he asks this sort of questions as true and false. Would you mind briefly explaining what you meant by "leading order behavior?" –  jp24 Dec 13 '12 at 6:24

The second one is automatic: Already $\displaystyle\int_0^1\dfrac{dx}{x^{3/2}}$ blows up, and the $\log$ makes things worse.

For $\dfrac{\log x}{x^{1/2}}$, I would rewrite it as, say, $\dfrac{x^{1/4}\log x}{x^{3/4}}$.

By L'Hospital's Rule, or otherwise, we can show that $\lim_{x\to 0+}x^{1/4}\log x=0$. But $\displaystyle\int_0^1 \dfrac{dx}{x^{3/4}}$ converges, so our integral does.

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I agree with your first two statements. I'm not sure I understand your third. What information does the numerator going to 0 reveal? –  jp24 Dec 13 '12 at 6:32
    
The numerator is going to $0$, so from some point $a$ down, the absolute value of the top is $\le 1$. So the integral from $0$ to $a$ has absolute value $\le \int_0^a\frac{dx}{x^{3/4}}$. This integral converges ("$p$" less than $1$). –  André Nicolas Dec 13 '12 at 6:38

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