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How to show that:

$$I:=\lim_{\epsilon \to 0^+} \int_{\epsilon}^{2\epsilon} \frac{1}{\ln{(1+x)}}dx=\ln{2}$$

Using the mean value theorem for integrals I can show that $\frac{1}{2 }\leq I \leq 1$, but I'm not able to show that $I=\ln{2}$. Any hints?

6 Answers 6

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Consider the function $$f(x) = \frac{1}{\log(1+x)}-\frac{1}{x}, f(0)=\frac{1}{2}$$ then $f$ is continuous at $0$. Let $$F(x) =\int_{0}^{x}f(t)\,dt,x\geq 0$$ Then we can see by continuity of $F$ that $F(2\epsilon)-F(\epsilon)\to 0$ as $\epsilon\to 0^{+}$. Your job is now complete and desired limit is $\log 2$ because $$\int_{\epsilon} ^{2\epsilon}\frac{dt}{t}=\log 2$$

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    How do you justify value $1/2$ in $0$ ? – 2017-02-24
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    @JeanMarie: the value at $0$ is chosen to ensure that $f$ becomes continuous. Also it is not necessary to know the value $1/2$. It is only necessary to know that $\lim_{x\to 0}f(x)$ exists. The limit value $1/2$ can be obtained easily by usual techniques. – 2017-02-24
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For $0, we can use a classical "bracketting":

$$0

$$\tag{1}\dfrac{1}{x}< \dfrac{1}{\ln(1+x)}< \dfrac{1}{x-\dfrac{x^2}{2}}$$

Remark: the last fraction can be written : $$\dfrac{1}{x-\dfrac{x^2}{2}}=\dfrac{2}{2x-x^2}=\dfrac{2}{x(2-x)}=\dfrac{1}{x}+\dfrac{1}{2-x}.$$

By integration of (1) between $\varepsilon$ and $2 \varepsilon$, using the ascending property of the integral:

$$\underbrace{\ln(x)|_{\varepsilon}^{2 \varepsilon}}_{= \ \ln{2 \varepsilon}-\ln{\varepsilon} \ = \ \ln(2)}

where $I$ is the looked for integral, establishing the result.

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    (+1) I think it's pretty intuitive for a beginner like me. – 2017-02-24
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Hint: If you are allowed to use that for small $x$ it holds that

$$ \log(1+x) =x + O(x^2), $$

try proceeding that way.

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As the function $f(x)=\ln(1+x)$ is concave and has tangent $y=x$ at origin, if $x>0$, we have $$(1-\varepsilon)x \le \ln(1+x)\le x\quad\text{for any }\varepsilon \; \text{such that}\;0<\varepsilon<1,$$ We deduce that $$\frac1x\le\frac1{\ln(1+x)}\le \frac 1{1-\varepsilon}\frac1x,$$ whence $$\int_{\varepsilon}^{2\varepsilon}\frac{\mathrm dx}{x}=\ln 2\le\int_{\varepsilon}^{2\varepsilon}\frac{\mathrm dx}{\ln(1+x)}\le \int_{\varepsilon}^{2\varepsilon}\frac{\mathrm dx}{(1-\varepsilon)x}=\frac{\ln 2}{1-\varepsilon} $$ Apply the squeezing principle when $\varepsilon\to 0\;$ to get the limit $\ln 2$.

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    I think people who down vote should justify their vote. Brave, but not reckless… – 2017-02-24
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    I not down voted you but your first inequality is flawed, you have that $$x\le(1-\epsilon)x$$ for $\epsilon\in(0,1)$ what cannot be possible for $x>0$. – 2017-02-24
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    @Masacroso:the inequalities should be reversed I think. And I guess this is a typo. – 2017-02-24
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    @Masacroso: I've got it! Messed up in copy-pasting fragments of formulae,, and didn't check before posting. Thank you for pointing the typos! – 2017-02-24
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Make the substitution $x=\epsilon t$, to give $$ I=\lim_{\epsilon\to0^+}\int_1^2 \frac\epsilon{\ln(1+\epsilon t)}\ dt = \int_1^2 \lim_{\epsilon\to0^+}\frac\epsilon{\ln(1+\epsilon t)}\ dt $$ Note that moving the limit inside the integral can fail in some situations (even with constant limits)... fortunately, we can do it if the function converges uniformly on the interval, and this function does.

Evaluating the limit, we get $$ I = \int_1^2 \frac1t\ dt = \big[\ln t\big]_1^2= \ln 2-\ln 1 = \ln 2 $$

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Lucky day, you are allowed to integrate a Taylor expansion.

$\frac{1}{ln(1+x)}=\frac1x+O(1)$

$\displaystyle{\int_{\varepsilon}^{2\varepsilon}\frac{1}{ln(1+x)}=\left[C+ln(x)+O(x)\right]_{\varepsilon}^{2\varepsilon}=(C-C)+\ln(2\varepsilon)-\ln(\varepsilon)+O(\varepsilon)=\ln(\frac{2\varepsilon}{\varepsilon})+O(\varepsilon)\to\ln(2)}$