I have to find the limit:

$$\lim\limits_{n \to \infty} n \displaystyle\int_2^e (\ln x)^n dx$$

I tried using the Squeeze Theorem but it didn't work (or at least I didn't use it correctly):

$$2 \le x \le e$$

I took the log of this inequality

$$\ln(2) \le \ln(x) \le 1$$

Raised to the $n$:

$$\ln^n(2) \le \ln^n(x) \le 1$$

Took the integral from $2$ to $e$:

$$\int_2^e \ln^n(2)dx \le \int_2^e \ln^n(x) dx \le \int_2^e 1 dx$$

$$\ln^n(2) \cdot (e-2) \le \int_2^e \ln^n(x) dx \le (e-2)$$

Multiplied by n:

$$n \ln^n(2) \cdot(e - 2) \le n\int_2^e \ln^n(x) dx \le n(e-2)$$

And now, since $\ln(2) < 1$, so $\ln^n(2) = 0$ as $n \to \infty$, if I take the limit with $n \to \infty$ I get:

$$\infty \cdot 0 \le n\int_2^e \ln^n(x) dx \le \infty$$

So I didn't get anywhere trying to use the Squeeze Theorem.

  • $\begingroup$ I wonder, does it matter that the lower bound is $2$? I mean, if the lower bound were to be $3$, would that change tho outcome of the limit? $\endgroup$ – imranfat Mar 10 '20 at 14:14
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    $\begingroup$ @imranfat Yes, I think it would change the answer. If you look at Atticus' answer you will see that he used the Squeeze Theorem. If you look at his last inequality you will see that he has a limit involving $\ln^{n+1}2$ on one side and $\ln^{n}2$ on the other side. Because $\ln2<0$, we have $\lim_{n \to \infty} (\ln 2)^n = 0$ and we can evaluate both the limits and use the Squeeze Theorem, but if we would have $\ln 3$, that wouldn't go to $0$ as $n$ goes to $\infty$. Does that make sense? $\endgroup$ – user592938 Mar 10 '20 at 21:57

As you said $\lim\limits_{n\to \infty}\ln^n 2=0$. Let:

$$I_n = \int_2^e\ln^n x\,dx$$

$I_n$ is decreasing because:

$$I_{n}-I_{n+1}=\int_2^e\ln^n x(1-\ln x)\,dx \geq 0$$

Integrating by parts:

$$ \begin{aligned} \displaystyle\int_2^e \ln^n x\, dx &= \displaystyle\int_2^e \ln^n x \cdot (x)'\, dx\\ &=x\ln^n x\bigg|_2^e -n\int_2^e\ln^{n-1}x\,dx\\ &= e-2\ln^n 2-nI_{n-1} \end{aligned} $$

Therefore $I_n+nI_{n-1}=e-2\ln^n2$. Now, since $I_n$ is decreasing:

$$I_n+nI_{n-1}\geq I_n+nI_n\Rightarrow I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$


$$I_{n+1}+(n+1)I_n\leq I_n+(n+1)I_n\Rightarrow I_n\geq \frac{1}{n+2}(e-2\ln^{n+1}2)$$

Combining the two:

$$\frac{1}{n+2}(e-2\ln^{n+1}2) \leq I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$


$$\frac{n}{n+2}(e-2\ln^{n+1}2) \leq nI_n\leq \frac{n}{n+1}(e-2\ln^n 2)$$

and squeezing $\lim\limits_{n\to \infty}nI_n=e$.

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    $\begingroup$ The line before "combining the two". It should start with $I_{n+1}+(n+1)I_{n}$ $\endgroup$ – bjorn93 Mar 4 '20 at 22:31
  • $\begingroup$ @bjorn93, Great spotting. It took me a while to realize it even after you pinpointed. Edited accordingly. $\endgroup$ – LHF Mar 4 '20 at 22:32

Use $t = \log^{n+1}(x) \implies \frac{1}{n+1}e^{t^{\frac{1}{n+1}}}dt = \log^n(x)dx$:

$$I = \lim_{n\to\infty} \frac{n}{n+1}\int_{\log^{n+1}(2)}^1 e^{t^{\frac{1}{n+1}}}\:dt \longrightarrow \int_0^1 e\:dt = e$$

by dominated convergence.


Let $\epsilon $ be any fixed number such that $0<\epsilon <e-2$ and we split the integral as a sum of two integrals using intervals $[2,e-\epsilon]$ and $[e-\epsilon, e] $. The first integral tends to $0$ as integrand uniformly tends to $0$.

For the second integral we need to bound it suitably. We have via integration by parts $$(n+1)\int_{e-\epsilon} ^{e} \dfrac{(\log x) ^n} {x} \cdot x\, dx=\left. x(\log x) ^{n+1}\right|_{x=e-\epsilon} ^{x=e} - \int_{e-\epsilon} ^{e}(\log x) ^{n+1}\,dx$$ (note that the factor $n$ before integral can be safely replaced by $(n+1)$ as $n/(n+1)\to 1$). The first term on right tends to $e$ and second term is bounded in absolute value by $\epsilon$ (as integrand lies in $[0,1]$). It follows that the desired limit is $e$.


Fixed $\epsilon\in(0,1-\ln2)$. Let $u=\ln x$ and then $$ n\int_2^e (\ln x)^n dx=n \int_{\ln 2}^1 u^ne^u du=n \int_{\ln 2}^{1-\epsilon} u^ne^u du+n \int^1_{1-\epsilon} u^ne^u du.$$ For the 1st part, $$ 0\le n \int_{\ln 2}^{1-\epsilon} u^ne^u du\le 3n(1-\epsilon)^n $$ and one has $$ \lim_{n\to\infty} n \int_{\ln 2}^{1-\epsilon} u^ne^u du=0. $$ For the 1st part, by the Mean Value Theorem for integrals, there is $\xi\in(1-\epsilon,1)$ such that $$ n \int_{1-\epsilon}^1 u^ne^u du=n e^\xi\int_{1-\epsilon}^1 u^n du=\frac{n}{n+1}e^\xi(1-(1-\epsilon)^{n+1}). $$ Noting that $$ e^{1-\epsilon}(1-(1-\epsilon)^{n+1})\le e^\xi(1-(1-\epsilon)^{n+1})\le e $$ one has $$ \frac{n}{n+1}e^{1-\epsilon}(1-(1-\epsilon)^{n+1})\le n \int_{1-\epsilon}^1 u^ne^u du=n e^\xi\int_{1-\epsilon}^1 u^n du\le\frac{n}{n+1}e $$ and hence $$ e^{1-\epsilon}\le \underline{\lim}_{n\to\infty}n \int_{1-\epsilon}^1 u^ne^u du\le \overline{\lim}_{n\to\infty}n \int_{1-\epsilon}^1 u^ne^u du\le e. $$ So $$ e^{1-\epsilon}\le \underline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le \overline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le e. $$ Letting $\epsilon\to0$ one has $$ e\le \underline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le \overline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le e $$ or $$ \lim_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du= e $$


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