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Prove that

$$\lim_{x\to 1^{+}} \int_{x}^{x^{3}}\frac{1}{\ln t}\, dt=\ln3$$

I have never seen something like this before.

I noticed that $\int_{1}^{3} \frac{1}{x}dx=\ln x|_{1}^{3}$ and with the change of variable in the initial integral I obtain, for $x=\ln(t)$, $\int_{\ln t}^{3\ln t} \frac{1}{\ln(\ln(t))}dt$ and for $t=e^s$, it's $\int_{s}^{3s} \frac{1}{\ln(s)}ds$ and from here I am unable to obtain the required limit.

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    $\begingroup$ It makes no sense to have both x in the boundaries and the integrand. Which one is it? $\endgroup$ – user370967 Mar 5 at 7:37
  • $\begingroup$ This is why I said I never saw something like this before $\endgroup$ – Septimiu Cristian Mar 5 at 7:39
  • $\begingroup$ I suppose just in the boundaries. $\endgroup$ – Septimiu Cristian Mar 5 at 7:39
  • $\begingroup$ Yes, I think so too. Having the x in the integrand makes no sense. $\endgroup$ – user370967 Mar 5 at 7:41
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    $\begingroup$ A Romanian Magazine, called ,,Gazeta Matematica ". It's the supliment of the January 2016's issue. $\endgroup$ – Septimiu Cristian Mar 5 at 7:43
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Substitute $t=x^v$, then we obtain $$\begin{align*} I(x)&=\int_{x}^{x^3}\frac1{\ln t}\ dt\\&=\int_1^3 \frac{x^{v}}{v} \ dv. \end{align*}$$ Now, we can see $\lim_{x\to 1^+}I(x)=\ln 3$ by Lebesgue's dominated convergence theorem; for every sequence $x_n>1$ converging to $1$, we have that $$\begin{align*} I(x_n)&=\int_1^3 \frac{x_n^{v}}{v} \ dv\\&\xrightarrow{n\to \infty}\int_1^3 \frac 1 v\ dv=\ln 3 \end{align*}$$ by LDCT. (For $1<x\le 2$, $0\le \frac{x^v}{v}\le \frac{2^v}{v}$.)

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  • $\begingroup$ Shouldn't we use the sequential criterion of limits to use lebesgue dominated convergence? It's a statement about sequences, after all. $\endgroup$ – user370967 Mar 5 at 7:36
  • $\begingroup$ You are absolutely right. I should have been more careful applying it. I'll fix it soon but it's hard to do on mobile... $\endgroup$ – Song Mar 5 at 7:38
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    $\begingroup$ I appreciate your helpful comment and generosity! $\endgroup$ – Song Mar 5 at 7:47
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    $\begingroup$ Since the integrand $f(x, v) =x^v/v$ is continuous on $[1,1]\times [2,3]$ one can switch between limit and integral operation. One does not really need Lebesgue DCT. $\endgroup$ – Paramanand Singh Mar 5 at 16:56
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    $\begingroup$ There was a typo in my previous comment. $f(x,v)$ is continuous on $$[1,2]\times[1,3]=\{(x,v)\mid x\in[1,2],v\in[1,3]\}$$ $\endgroup$ – Paramanand Singh Mar 6 at 2:03
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Just added for your curiosity.

@Song provided a nice and simple solution.

Sooner or later, you will learn that $$\int\frac{dt}{\ln t}=\text{li}(t)$$ where appears a special function , namely the logarithmic integral function.

What is interesting is that, assuming $t>1$, the series expansion is given by $$\text{li}(t)=\gamma +\log (t-1)+\frac{t-1}{2}-\frac{(t-1)^2}{24} +O\left((t-1)^3\right)$$ which means that, for $x$ close to $1$, using the binomial expansion $$\int_x^{x^n}\frac{dt}{\ln t}=\log (n)+(n-1) (x-1)+\frac{(n-1)^2}{4} (x-1)^2+O\left((x-1)^3\right)$$

For example, using $n=5$ and $x=\frac{11}{10}$, the above approximation would give $\frac{11}{25}+\log (5)\approx 2.04944$ while, using numerical integration, you would get $\approx 2.05173$.

Going a bit further, suppose that you want to compute $$I=\int_{g(x)}^{f(x)}\frac{dt}{\ln t}$$ where you can expand the bounds as series around $x=1$ that is to say $$f(x)=1+\sum_{i=1}^\infty a_i (x-1)^i \qquad \text{and} \qquad g(x)=1+\sum_{i=1}^\infty b_i (x-1)^i$$ you would get, as an approximation, $$I=\log \left(\frac{a_1}{b_1}\right)+ \left(\frac{a_1}{2}+\frac{a_2}{a_1}-\frac{b_1}{2}-\frac{b_2}{b_1}\right)(x-1)+O\left((x-1)^2\right)$$

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The key is to observe that $$f(x) =\frac{1}{\log x} - \frac{1}{x-1},x>0,x\neq 1,f(1)=\frac{1}{2}$$ is continuous on $(0,\infty) $. The desired limit is thus equal to the limit of $$\int_{x} ^{x^3}f(t)\,dt+\int_{x}^{x^3}\frac{dt}{t-1}\tag{1}$$ as $x\to 1^{+}$. The second term in $(1)$ clearly equals $$\log\frac {x^3-1}{x-1}$$ and therefore tends to $\log 3$. Our job is done if we can show that the first term in $(1)$ tends to $0$. This is easy as the integrand is bounded, say by $M$, in a neighborhood of $1$ and hence is no greater than $M(x^3-x)$ and thus tends to $0$ as $x\to 1^{+}$.

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Just to give another approach,

$$\int_x^{x^3}{1\over\ln t}\,dt=\int_{\ln x}^{3\ln x}{e^u\over u}\,du=\int_{\ln x}^{3\ln x}\left(e^u-1\over u\right)\,du+\int_{\ln x}^{3\ln x}{1\over u}\,du$$

Now $(e^u-1)/u\to1$ as $u\to0$, so the first integral tends to $0$ as $x\to1$, while

$$\int_{\ln x}^{3\ln x}{1\over u}\,du=\ln u\Big|_{\ln x}^{3\ln x}=\ln(3\ln x)-\ln(\ln x)=\ln 3+\ln(\ln x)-\ln(\ln x)=\ln3$$

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