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Find $\lim_{x\to 1}f(x)$, where $$f(x) = \int_{x}^{x^2}\frac{1}{\ln {t}}\mathrm dt$$

I tried splitting it into two integrals, one from 1 to $x^2$ and the other one from $x$ to $1$. Doesn't matter how I split it, I got zero. Wolfram Alpha has some kind of a different, more sophisticated answer with functions we didn't cover at the lecture.

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  • $\begingroup$ What's with all these limit integral questions? $\endgroup$ – Simply Beautiful Art Aug 13 '17 at 16:59
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    $\begingroup$ Perhaps you can use the mean value theorem for definite integrals. en.wikipedia.org/wiki/… $\endgroup$ – md2perpe Aug 13 '17 at 17:00
  • $\begingroup$ Use inequality for $\ln$. $\endgroup$ – Nosrati Aug 13 '17 at 17:01
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    $\begingroup$ Could you explain how you are getting $0$? As is, noone knows what you've tried, which could be of help. $\endgroup$ – Simply Beautiful Art Aug 13 '17 at 17:04
  • $\begingroup$ @md2perpe: Have you tried? With the mean value theorem, you get $\ln \xi$ between $\ln x$ and $\ln(x^2)=2\ln x$, so you don't get an exact answer, you only bound the limit within a factor of 2 (provided it exists). $\endgroup$ – Hans Lundmark Aug 13 '17 at 17:12
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For $t>0$: $$\frac{t-1}t\leq\ln t\leq t-1$$ so $$\int_x^{x^2}\frac{1}{t-1}dt\leq\int_x^{x^2} \frac{1}{\ln t}\,dt\leq \int_x^{x^2}\dfrac{t}{t-1}dt$$ with limit $x\to1$ you have answer $\ln2$.

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  • $\begingroup$ (+1) This is one of the ways I would have approached it. $\endgroup$ – Mark Viola Aug 13 '17 at 17:11
  • $\begingroup$ @MarkViola Thanks. $\endgroup$ – Nosrati Aug 13 '17 at 17:13
  • $\begingroup$ Thank you for your edits. $\endgroup$ – Nosrati Aug 13 '17 at 17:13
  • $\begingroup$ No problem :-) No idea why \le fails today, unless someone did some funky sneaky mathjax >.> $\endgroup$ – Simply Beautiful Art Aug 13 '17 at 17:14
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    $\begingroup$ Hm, what a nice inequality you have there. Just noticed the lower bound is equivalent to the upper bound after $t\mapsto1/t$. Very simple and nice. $\endgroup$ – Simply Beautiful Art Aug 13 '17 at 17:22
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Substituting $t=e^u$, $\mathrm dt=e^u\mathrm du$, we have $$ f(x)=\int_{\ln x}^{2\ln x}\frac{e^u\mathrm du}{u}.$$ For $x>1$ we then have $1\le e^u\le x^2$, hence $$ \int_{\ln x}^{2\ln x}\frac{\mathrm du}{u}\le f(x)\le x^2\int_{\ln x}^{2\ln x}\frac{\mathrm du}{u}$$ (and similar for $x<1$) with a now well-known integral: $\int \frac{\mathrm du}u=\ln |u|+C$. We conclude that $$ \ln 2\le f(x)\le x^2\ln 2$$ and so $$\lim_{x\to 1}f(x)=\ln 2.$$

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HINT:

Write $\frac{1}{\log(t)}=\left(\frac{1}{\log(t)}-\frac{1}{t-1}\right)+\frac{1}{t-1}$. Then note that $\frac{1}{\log(t)}-\frac{1}{t-1}$ has a removable discontinuity at $t=1$, which once removed, is analytic in a neighborhood of $t=1$.

Then write,

$$\begin{align}\int_x^{x^2}\frac{1}{\log(t)}\,dt&=\int_x^{x^2}\left(\frac{1}{\log(t)}-\frac{1}{t-1}\right)\,dt+\int_x^{x^2}\frac{1}{t-1}\,dt\tag1 \end{align}$$

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  • $\begingroup$ You mean $1/\log(t)$ on the left-hand side? $\endgroup$ – Hans Lundmark Aug 13 '17 at 17:13
  • $\begingroup$ @HansLundmark Thanks Hans! Yes, I sure did. I've edited accordingly. $\endgroup$ – Mark Viola Aug 13 '17 at 17:15
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Use the fact that $\ln$ is a concave function, and thus, for $1\leq x\leq t\leq x^2$,

$$2(t-1)\frac{\ln(x)}{(x-1)(x+1)}\leq\ln(t)\leq t-1$$

Invert this:

$$\int_x^{x^2}\frac1{t-1}~\mathrm dt\leq\int_x^{x^2}\frac1{\ln(t)}~\mathrm dt\leq\frac{(x-1)(x+1)}{2\ln(x)}\int_x^{x^2}\frac1{t-1}~\mathrm dt$$

These are easy to integrate:

$$\int_x^{x^2}\frac1{t-1}~\mathrm dt=\ln(x+1)$$

And so by the squeeze theorem, one can show that

$$\lim_{x\to1}\int_x^{x^2}\frac1{\ln(t)}~\mathrm dt=\ln(2)$$

Noting the right integral mess can be fixed up with the first inequality applied one more time and using the squeeze theorem.

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$\frac{1}{\ln(t)}=\frac{t}{t\ln(t)}=\frac{(t-1)+1}{t\ln(t)}=\frac{t-1}{t\ln(t)}+\frac{1}{t\ln(t)}$

Let $f : t \mapsto \frac{t-1}{t\ln(t)}$ and $g : t \mapsto \frac{1}{t\ln(t)}$.

  • $f$ can be continuously extended in 1, thus $\int f(t) ~\mathrm{d}t$ converges. Then, $\int_x^{x^2} f(t) ~\mathrm{d}t \underset{x=1}{\rightarrow}0$.
  • $g(t) = \frac{d}{dt}(\ln(\ln(t))$, and $\ln(\ln(x^2))-\ln(\ln(x))=\ln(2\ln(x))−\ln(\ln(x))=\ln(2)$

Conclusion: the limit is $\ln(2)$.

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  • $\begingroup$ $\ln\ln(x^2) - \ln \ln x = \ln(2 \ln x) - \ln \ln x = \ln 2 + \ln \ln x - \ln \ln x = \color{red}{\ln 2}$ $\endgroup$ – Cauchy Aug 13 '17 at 17:11
  • $\begingroup$ You're right corrected $\endgroup$ – Antoine C. Aug 13 '17 at 17:13
  • $\begingroup$ Note that you can write common functions using \ln or \sin. This makes it much easier to read. $\endgroup$ – Simply Beautiful Art Aug 13 '17 at 17:14
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This seems like a good problem for generalization, since the proof for the general result might actually be easier. Suppose $f$ is continuous near $1,$ $f(1)=0,$ and $f'(1)\ne 0.$ Then

$$\tag 1 \lim_{x\to 1} \int_x^{x^2} \frac{1}{f(t)}\,dt = \frac{\ln 2}{f'(1)}.$$

With $f(t)= \ln t,$ we get the answer of $\ln 2$ for the given problem.

Rough sketch of proof of $(1)$:

$$\frac{1}{f(t)} = \frac{1}{t-1}\cdot \frac{t-1}{f(t)}.$$

Now the second fraction has limit $1/f'(1)$ as $t\to 1.$ So given $\epsilon>0,$ this fraction lies between the constants $(1-\epsilon)/f'(1)$ and $(1+\epsilon)/f'(1)$ for $t$ close enough to $1.$ Move those constants outside the integral, evaluate the easy integral remaining, and the result follows.

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  • $\begingroup$ There is an error in the anlaysis. $$\int_x^{x^2} \frac{1}{t-1}\,dt=\log(x+1)$$ The limit as $x\to 1$ is $\log(2)\ne 2$. The result of the generalization is $\frac{\log(2)}{f'(1)}$. $\endgroup$ – Mark Viola Aug 13 '17 at 18:49
  • $\begingroup$ Right, thank you, I actually knew that, but $\ln 2$ magically and tragically turned into $2.$ $\endgroup$ – zhw. Aug 13 '17 at 19:17
  • $\begingroup$ And $f'(1)$, not $f'(0)$, right? ;-)) ... Two magical transforms don't make a right, but three rights do make a left. $\endgroup$ – Mark Viola Aug 13 '17 at 19:20
  • $\begingroup$ Yes, this is hilarious. Thanks again. $\endgroup$ – zhw. Aug 13 '17 at 19:21
  • $\begingroup$ You're welcome. Shall we keep these comments for comical effect? Or shall I delete like a good old MSE user. $\endgroup$ – Mark Viola Aug 13 '17 at 19:22
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Setting $t = e^s$ we get $$ f(x) = \int_{\ln x}^{2 \ln x} \frac{e^s}{s}\ ds = \int_{\ln x}^{2 \ln x} \frac{1+s+O(s^2)}{s}\ ds = \int_{\ln x}^{2 \ln x} \left(\frac{1}{s}+1+O(s)\right) \ ds \\ = (\ln(2 \ln x) - \ln(\ln x)) + (2\ln x - \ln x) + O((\ln x)^2) \\ = \ln 2 + \ln x + O((\ln x)^2) \to \ln 2 $$ as $x \to 1$.

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