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 $x \in (0,c)$
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1$\begingroup$ $\int_{b}^c\frac{1}{x\log\frac{e}{x}}\frac{1}{2}dx = -\log(\log(1/x)+1)\big\vert_b^c$ this won't converge when $b=0$ for a proof you might try to change variables $t=e/x$ $\endgroup$– Sextus EmpiricusSep 15, 2019 at 14:44
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2 Answers
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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. :)
answered Sep 15, 2019 at 14:58
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1$\begingroup$ I find the change of variables u = log(e/x) more elegant. But this is not really integration by parts. $\endgroup$ Sep 15, 2019 at 15:04
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$\begingroup$ @MartijnWeterings. Quite right. It is u-sub not by-parts. Of course! $\endgroup$ Sep 15, 2019 at 15:11
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$\begingroup$ Thanks Martijn Weterings and Peter Leopold, for your clearly explaining $\endgroup$– mhmtSep 15, 2019 at 15:36
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1$\begingroup$ @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 ams.org/publications/authors/AMS-StyleGuide-online.pdf.) But I guess your point is that MathJax is not $\LaTeX$. OK! $\endgroup$ Sep 15, 2019 at 17:55
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1$\begingroup$ @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. $\endgroup$ Sep 17, 2019 at 3:48
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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 $t \to \infty$
answered Sep 15, 2019 at 14:59