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$$\ln(x)\ln(1-x)\mid^1_0=0$$ I see this result once or twice a day in various forms without proof. Could someone just write the proof so I don't embarrass myself if someone on the street or in a grocery store asks me? One comment for the proof was to expand ln(x) about 1:

I don't see how $\ln(1-x) ((x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\dots)$ helps. It's still $-\infty*0$ at 1, and a non-convergent taylor at 0. Another proof was to use l'hopital's, so I'm still doing something wrong:

$\lim_{x\to{0/1}}\frac{\ln(x)}{1/\ln(1-x)}=\frac{1}{x(1-x)}$ is still indeterminant but no longer l'hoptilable.

I vaguely remember a proof in the margin of a complex variable book using the limit after a substitution involving e^x but the text has 700 pages and I don't know what page it was.

user5389726598465
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3 Answers3

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You've made a mistake in your differentiation:

$$\lim_{x\rightarrow 0^+} \frac{\ln(x)}{\frac{1}{\ln(1-x)}} = \lim_{x\rightarrow 0^+} \frac{\frac{1}{x}}{\frac{1}{(\ln(1-x))^2 (1 - x)}} = \lim_{x\rightarrow 0^+} \frac{(\ln(1-x))^2 (1 - x)}{x}$$

which is finite, as it is the derivative of $(\ln(1-x))^2 (1 - x)$ at $0$. You don't need to know the value as the limit as $x \rightarrow 1^-$ is the same, as a simple substitution of $u = 1 - x$ will tell you. Thus,

$$\ln(x)\ln(1 - x)|_0^1 = \lim_{x\rightarrow 1^-} \ln(x)\ln(1 - x) - \lim_{x\rightarrow 0^+} \ln(x)\ln(1 - x) = 0$$

Theo Bendit
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For $0 < x < 1$,

$$-\ln x = \int_x^1 \frac{dt}{t} < \frac{1-x}{x} \implies -\ln(1-x) < \frac{x}{1-x}$$

As $x \to 0$,

$$0 \leqslant \ln x \ln(1-x) = (-2\ln\sqrt{x})(-\ln(1-x)) < 2 \frac{1- \sqrt{x}}{\sqrt{x}}\frac{x}{1-x} \\= \frac{2(\sqrt{x} - x)}{1-x} \to 0 $$

Thus, by squeezing, $\lim_{x \to 0+} \ln x \ln(1-x) = 0$.

By symmetry, $\lim_{x \to 1-} \ln x \ln(1-x) = \lim_{y \to 0+}\ln(1-y) \ln y = 0. $

RRL
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  • Squeezing between the quasi-secant of the y values is another good method, but you posted after my question fell off the front page. Sorry you don't get any upvotes. – user5389726598465 Jul 17 '17 at 17:31
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$$L=\lim_{x→0}\ln(x)\ln(1-x)=\left(\lim_{x→0}\frac{\ln x}{\frac{1}{x}}\right)\left(\lim_{x→0}\frac{\ln(1-x)}{x}\right)$$ Apply L'Hopital's Rule to both of the limits $$L=\left(\lim_{x→0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}\right)\left(\lim_{x→0}\frac{\frac{-1}{1-x}}{1}\right)=\left(\lim_{x→0}x\right)\left(\lim_{x→0}\frac{1}{1-x}\right)=(0)(1)=0$$ Notice the limit as $x→1$ is equivalent due to the substitution $x→1-x$ $$\lim_{x→1}\ln(x)\ln(1-x)=\lim_{x→0}\ln(1-x)\ln(1-(1-x))=\lim_{x→0}\ln(x)\ln(1-x)=L$$

phi-rate
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