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

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.

• It makes no sense to have both x in the boundaries and the integrand. Which one is it?
– user370967
Mar 5, 2019 at 7:37
• This is why I said I never saw something like this before
– user318394
Mar 5, 2019 at 7:39
• I suppose just in the boundaries.
– user318394
Mar 5, 2019 at 7:39
• Yes, I think so too. Having the x in the integrand makes no sense.
– user370967
Mar 5, 2019 at 7:41
• A Romanian Magazine, called ,,Gazeta Matematica ". It's the supliment of the January 2016's issue.
– user318394
Mar 5, 2019 at 7:43

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 $$1, $$0\le \frac{x^v}{v}\le \frac{2^v}{v}$$.)

• Shouldn't we use the sequential criterion of limits to use lebesgue dominated convergence? It's a statement about sequences, after all.
– user370967
Mar 5, 2019 at 7:36
• 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... Mar 5, 2019 at 7:38
• I appreciate your helpful comment and generosity! Mar 5, 2019 at 7:47
• 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. Mar 5, 2019 at 16:56
• There was a typo in my previous comment. $f(x,v)$ is continuous on $$[1,2]\times[1,3]=\{(x,v)\mid x\in[1,2],v\in[1,3]\}$$ Mar 6, 2019 at 2:03

@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, $$I=\log \left(\frac{a_1}{b_1}\right)+ \left(\frac{a_1}{2}+\frac{a_2}{a_1}-\frac{b_1}{2}-\frac{b_2}{b_1}\right)(x-1)+O\left((x-1)^2\right)$$

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 $$x\to 1^{+}$$.

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

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