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In here, i want to show this entropy exist or not exist, namely i should calculate the integral of $\int_0^c\frac{1}{x\log^2\frac{e}{x}}\frac{1}{2} \log\frac{e}{x}\,dx$. If the result is $ <\infty$, we can say the entropy exists, otherwise it does not exist. \begin{equation*} \int_0^1f(x)\log f(x)\,dx \geq \int_{0}^c \frac{1}{x\log^2\frac{e}{x}}\frac{1}{2} \log\frac{e}{x} \, dx \end{equation*} where $x \in (0,c)$

Michael Hardy
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mhmt
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    $\int_{b}^c\frac{1}{x\log\frac{e}{x}}\frac{1}{2}dx = -\log(\log(1/x)+1)\big\vert_b^c$ this won't converge when $b=0$ for a proof you might try to change variables $t=e/x$ – Sextus Empiricus Sep 15 '19 at 14:44

2 Answers2

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By u-substitution,

\begin{aligned} & \int_0^c \frac{1}{x\log^2 \frac{e}{x}} \frac{1}{2} \log\frac{e}{x} \, dx = \int_0^c \frac{1}{2} \left(\log\frac{e}{x} \right)^{-1} \frac{1}{x} dx \\[8pt] & u=\log\frac{e}{x}=1-\log(x), \qquad du=-\frac{dx}{x}\\[8pt] = {} & -\frac{1}{2}\int u^{-1} \, du \quad \text{(ignore limits for now)} \\[8pt] = {} & -\frac{1}{2} \log u \\[8pt] = {} & \left. -\frac{1}{2} \log(1-\log x) \right|_a^c, ~~ \text{in } \lim a \rightarrow 0\\[8pt] = {} & -\frac{1}{2} \lim_{a\rightarrow 0} \log(1-\log c) -\log(1-\log a) \end{aligned} And we see the limit doesn't exist. So, no, the entropy doesn't exist.

And I see that Martijn Weterings beat me to the punch 12 minutes ago! OK, he had it first. :)

Michael Hardy
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Peter Leopold
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    I find the change of variables u = log(e/x) more elegant. But this is not really integration by parts. – Sextus Empiricus Sep 15 '19 at 15:04
  • @MartijnWeterings. Quite right. It is u-sub not by-parts. Of course! – Peter Leopold Sep 15 '19 at 15:11
  • Thanks Martijn Weterings and Peter Leopold, for your clearly explaining – mhmt Sep 15 '19 at 15:36
  • See my edits for proper MathJax usage. – Michael Hardy Sep 15 '19 at 17:37
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    @MichaelHardy, thank you for the edits. I'm a bit astonished that it is considered "proper MathJax usage" to left justify a set of sequentially-derived equations on the '=' sign. Aligning equations on the central '=' sign is considered proper $AMS ~\LaTeX$ style, I believe. (See 117 of https://www.ams.org/publications/authors/AMS-StyleGuide-online.pdf.) But I guess your point is that MathJax is not $\LaTeX$. OK! – Peter Leopold Sep 15 '19 at 17:55
  • @PeterLeopold : I think that alignment issue depends on the context. But it is certainly not proper usage to write $log(x)$ instead of either $\log(x)$ or $\log x,$ nor $lim$ instead of $\lim,$ nor $lim~a\rightarrow0$ instead of $\lim\limits_{a\to0},$ nor, within the MathJax code, "(ignore~limits~for~now)" instead of "\text{(ignore limits for now)}", nor $f(x)dx$ instead of $f(x)\,dx$ (that last is a very frequent mistake), and I wonder if you would not agree that $\left.-\dfrac 1 2 \log(\cdots\cdots)\right|_a^c$ looks better than $-\dfrac 1 2 \log(\cdots\cdots)|_a^c. \qquad$ – Michael Hardy Sep 17 '19 at 01:00
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    @MichaelHardy I agree. It looks much better! And for the rest: correctness! I won't be as sloppy in the future. At least I'll try not to be. – Peter Leopold Sep 17 '19 at 03:48
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$$\int_{0}^c\frac{1}{x\log\frac{e}{x}}\frac{1}{2}dx $$

substitute $t=e/x$ (and use $dt/dx=-e/x^2$)

$$\int_{e/c}^\infty\frac{e^2}{t\log t}\frac{1}{2}dt = \log(\log(t)) \big\vert_{e/c}^\infty$$

which diverges because $\log(\log(t))$ becomes infinite as $t \to \infty$

Sextus Empiricus
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