# Find $\lim_{n\to\infty}\left(\int _{0}^{n}\frac{\mathrm dx}{x^{x^n}}\right)^n$

I had to solve the following limit :$$\lim_{n\to\infty}\left(\int _{0}^{n} \frac{\mathrm dx}{x^{x^n}}\right)^n$$

So far I've made no progress and the integral looks unsolvable. With online calculators, it seems like the inner integral approaches to $$1$$, yielding the indeterminate form of $$1^{\infty}$$.

How can I approach to solve it?

• Why the downvote? It seems the OP made honest efforts. Commented Nov 29, 2023 at 13:23
• Looking at online graphs suggests to me that we could look for a bound for $0<x<1$ and another bound for $x>1.$ Commented Nov 29, 2023 at 13:41
• Find $\lim_{n \to \infty}{\frac{1}{x^{x^{n}}}}$ for $0 \leq x \leq 1$ then for $x > 1$. Commented Nov 29, 2023 at 13:55
• @FishDrowned For $x>1,\ \lim_{n\to\infty}\frac{1}{x^{x^n}}=0,\$ but that doesn't tell you about the integral for $x>1.$ So I think you need bounds (bounded function of) of the integrand for $x>1.$ Commented Nov 29, 2023 at 14:01
• Also, something like Holder inequality or Minkowski inequality could be useful... Commented Nov 29, 2023 at 14:10

Clearly, $$\lim_{n \to \infty} \frac{1}{x^{x^n}} = 1_{(0,1]}(x)$$ for any $$x>0$$. Moreover, $$\frac{1}{x^{x^n}} \le 1_{[0,1]}(x) + \frac{1}{x^x} 1_{(1,\infty)}(x)$$ and the latter is integrable, hence $$I(n) =\int_0^n \frac{1}{x^{x^n}}dx \to 1$$ as $$n \to \infty$$. This would allow us to pass from exponential type limit to product type limit. Indeed, write $$\begin{split} I(n)^n & = \exp \left( n \cdot \ln I(n) \right) \\ & = \exp \left( \frac{\ln ( 1 + [I(n) - 1])}{\frac{1}{n}}\right) \\ &= \exp \left( \frac{\ln(1 + [I(n)-1])}{I(n)-1} \cdot n \big(I(n)-1\big) \right). \end{split}$$ Since we already know that $$I(n)-1 \to 0$$, in particular that $$\frac{\ln(1 + I(n)-1)}{I(n)-1} \to 1$$, it's enough to find the limit of $$n(I(n)-1)$$. We decompose $$n(I(n)-1) = n\int_0^1 (e^{-x^n\ln(x)} - 1) dx + \int_1^e ne^{-x^n \ln(x)}dx + \int_e^\infty ne^{-x^n\ln(x)}dx$$ which in particular implies $$n(I(n)-1) \ge \int_1^e ne^{-x^n \ln(x)}dx$$

We consider substitution $$x^n = y$$. Then $$x \in (1,e)$$ implies $$y \in (1,e^n)$$ and $$n x^{n-1} dx = dy$$, hence $$\begin{split} \int_1^e n e^{-x^n \ln(x)} dx & = \int_1^{e^n} e^{-\frac{y}{n} \ln (y)} y^{\frac{1}{n}-1} dy \\ & = \int_1^{e^n} \frac{1}{y} e^{ - \frac{y-1}{n} \ln(y)} dy \\ & = \int_1^\infty \frac{1}{y} \cdot e^{-\frac{(y-1)\ln(y)}{n}} 1_{(1,e^n)}(y)dy. \end{split}$$ We consider $$g_n(y) = \frac{1}{y} e^{-\frac{(y-1)\ln(y)}{n}} 1_{(1,e^n)}(y)$$. Since $$-(y-1)\ln(y) < 0$$, the sequence $$(g_n(y))_{n \ge 1}$$ is monotonically increasing for any given $$y \in (1,\infty)$$ and converges to $$\frac{1}{y}$$ pointwise. Hence, by Monotone Convergence Theorem, $$\lim_n \int_1^{\infty} g_n(y) dy = \int_1^\infty \frac{1}{y}dy = +\infty.$$

It follows that $$n(I(n)-1) \to +\infty$$ and so $$I(n)^n \to +\infty$$ as well.

• The limit of integration is from $0$ to $n\to\infty$ (and not from $0$ to $1$ ). Your answer is quite high level for me to understand. But thanks for your efforts which others may understand ( or maybe me in future). Commented Nov 29, 2023 at 18:53
• your bound over the interval $[1,e]$ is in fact false, for example near $x=1$ one as the $f_n(x):=ne^{-x^n\log x}\approx n$ while for the propose upper bound $g(x):=-\frac{\log(\log x)}{\log x}e^{-\tfrac{1}{\log x}}\approx 0$. In fact, $g$ is bounded, while the sequence $f_n$ is unbounded. Commented Nov 30, 2023 at 6:17
• Thank you very much, the bound was indeed wrong. The correct one is of order $-\frac{\ln(\ln(x))}{\ln(x)} e^{-\frac{1}{\ln \ln(x)}}$ which is not integrable, so I had to change that part a little bit. Commented Nov 30, 2023 at 19:02
• @DominikKutek Thanks sir for your efforts. I hoped that it would be understandable and a easy answer to me but it isn't. But others users can understand it. Commented Nov 30, 2023 at 19:03
• What is your background? Limits involving integral (and the limiting random variable under integral sign) often boils down to some sort of monotone/dominated convergence theorem from measure theory. Presumably, you have not covered it yet in your class, right? What is the source of this exercise? Commented Nov 30, 2023 at 19:05

Let $$f_n(x)=x^n\log x$$. It is not difficult to show that $$I_n=\int^n_0e^{-f_n}\xrightarrow{n\rightarrow\infty}1$$ This may be achieved by analyzing the integral over $$(0,1]$$ and $$(1,\infty)$$ separately (as we do below).

As in my previous comment, if in addition one can show that $$\lim_{n\rightarrow\infty}n(I_n-1)=a$$ exists in $$\mathbb{R}\cup\{\pm\infty\}$$, then the problem reduces to $$(I_n)^n=\big(1+\frac{(I_n-1)n}{n}\big)^n\xrightarrow{n\rightarrow\infty}e^a$$

1. Over the interval $$(0,1]$$, $$f_n\leq f_{n+1}\leq 0$$ and, with a little bit of elementary Calculus, we see that $$f_n$$ attains it minimum value at $$x=e^{-1/n}$$; hence,
$$-\frac1{ne}\leq f_n(x)\leq 0, \quad 0\leq x\leq 1$$ and so, $$1\leq e^{-f_n}\leq e^{\tfrac1{ne}}.$$ It then follows that $$1\leq \lim_n \int^1_0x^{-x^n}\,dx\leq e^{\tfrac1{ne}}\xrightarrow{n\rightarrow\infty}1.$$ By the mean value theorem $$0\leq n\big(e^{-f_n(x)}-1\big)=-e^{-\xi_n(x)}nf_n(x)\leq e^{\tfrac1{ne}}\frac1{e}$$ where $$f_n(x)<\xi_n(x)<0$$. Since $$nf_n(x)\xrightarrow{n\rightarrow\infty}0$$ pointwise in $$[0,1]$$, to follows from dominated convergence that $$\int^1_0 n\big(e^{-f_n(x)} -1\big)\,dx\xrightarrow{n\rightarrow\infty}0.$$

2. Over the interval $$[1,\infty)$$, each $$f_n$$ is monotone decreasing, $$0\leq f_n\leq f_{n+1}$$, and $$0\leq e^{-f_{n+1}}\leq e^{-f_n}\xrightarrow{n\rightarrow\infty}0$$ point wise. In particular, $$e^{-f_n}\mathbf{1}_{[1,n]}\leq e^{-f_1}$$ Since $$e^x\leq x^x$$ for $$x\geq e$$ and the map $$x\mapsto e^{-x}$$ is integrable over $$[0,\infty)$$, another application of dominated convergence yields $$\int^n_1e^{-f_n(x)}\,dx\xrightarrow{n\rightarrow\infty}0.$$

3. We now pay out attention to the limit $$\lim_nn\int^n_1 e^{-f_n}$$. Rough numerical experiments suggest that for any $$a>1$$, \begin{align} n\int^{\sqrt[n]{a}}_1e^{-x^n\log x}\,dx\xrightarrow{n\rightarrow\infty}\log a\tag{1}\label{one} \end{align}

# F_n(x)=n*x^{-x^n}
myfun <- function(x,n){
exp(log(n)-x^n* log(x))}

# n*I_n(a)=\int^{a^{1/n}}_1 x^{-x^n} dx
N <- 300
a <- 5
I_n <- sapply(1:N, function(k){integrate(myfun, 1, a^(1/k), n=k)$$value})/log(a) plot(1:N, I_n, xlab = 'n', ylab = '', type = 'l', col = 'blue', main = TeX("$$I_n(a)/log(a)"))  As $$\int^a_1 e^{-f_n} >\int^{\sqrt[n]{a}}_1 e^{-f_n}$$, it would then follow that $$n\int^n_1 x^{-x^n}\,dx \xrightarrow{n\rightarrow\infty}\infty$$ To verify \eqref{one}, let's fix $$a>1$$. Observe that if $$1\leq x\leq a^{1/n}$$, then $$\log x \leq f_n(x)\leq a\log x$$. Consequently \begin{align} n\int^{a^{1/n}}_1 e^{-f_n}&\geq n\int^{a^{1/n}}_1\frac{1}{x^a}\,dx=\frac{n}{a-1}\big(1-a^{\tfrac{1-a}{n}}\big)\\ &=\frac{1}{a-1}\frac{1-e^{-\frac1n(a-1)\log a}}{\frac1n}\xrightarrow{n\rightarrow\infty}\log a, \end{align} and \begin{align} n\int^{a^{1/n}}_1 e^{-f_n}&\leq n\int^{a^{1/n}}_1\frac{dx}{x}=\log(a) \end{align} All this shows that $$\lim_n\Big(\int^n_0 x^{-x^n}\,dx\Big)^n=\infty$$ Commet: If integration is restricted to the unit interval $$[0,1]$$, then we do have that $$\lim_n\Big(\int^1_0 x^{-x^n}\,d\Big)^n=e^0=1.$$ Similarly, for any $$a>1$$, $$\lim_n\Big(1+\int^n_a x^{-x^n}\,dx\Big)^n=e^0=1$$ The last assertion follows from noticing that for $$x>a>1$$, $$a^n\log a^n\leq f_n(x)$$ and so $$n\int^n_a e^{-f_n}\leq \frac{n(n-a)}{a^{a^n}}\xrightarrow{n\rightarrow\infty}0$$ The analysis below is morally similar to Dominik's answer (+1 of course), but simplified (or at least simplified compared to the version of the answer I had seen when writing this, I understand it has since been edited). In particular, I think the problem can be approached with elementary tools, and very precise estimation of the integral is not needed. Let $$I_n$$ be the integral in the question. Observe that $$\int_0^1 \exp(- x^n \log x) \mathrm{d}x \ge \int_0^1 (1-x^n \log x)\mathrm{d}x = 1+\frac{1}{(n+1)^2}.$$ Now, consider $$J_n := \int_1^n \exp(-x^n \log n)\mathrm{d}x,$$ and substitute $$u = x^n$$ in the above to get $$J_n = \int_1^{n^n} \frac{1}{n u} \exp( - (u-1) \log u/n)\mathrm{d}u.$$ Observe that for $$u \in [1,\sqrt{n}],$$ the exponential term in the integrand is lower bounded by $$\exp( - n^{-1/2} \log n),$$ which in turn is $$\ge 1/2$$ for $$n \ge 16$$. Since the integrand of $$J_n$$ is nonnegative, we conclude that for $$n \ge 16,$$ $$J_n \ge \frac{\log n}{4n},$$ and in turn that for $$n \ge 16,$$ $$I_n \ge 1 + \frac{\log n }{4n}.$$ So, we see that $$I_n^n \ge K_n := (1 + \log n/4n)^n.$$ But $$K_n$$ diverges. Indeed, we have $$\log K_n = n \log\left(1 + \frac{\log n}{4n}\right),$$ and since $$\log n /4n < 1,$$ we can use that $$\log(1+x) > x/2$$ over $$[0,1]$$ to lower bound $$\log K_n \ge n \cdot \frac{\log n}{8n} = \log(n)/8,$$ which diverges. Therefore, $$\lim I_n^n = \infty.$$ • I've changed my answer accordingly. Those parts with Dominated Convergence Theorem were remains of previous (wrong) approach. Commented Nov 30, 2023 at 21:10 • Thanks. I hope to understand it in the future. My level is too low to understand any of the solutions provided Commented Dec 1, 2023 at 3:04 • @An_Elephant I think the solution is elementary (i.e., doesn't use very many special tools), so try it again after taking a break maybe. There's maybe three things that need some showing: a. For everyx,1-x \le e^{-x}$, b. For$x \in [0,1], \log(1+x) \ge x/2,$and c. for large$n$,$\exp(-\log(n)/\sqrt{n}) > 1/2$. These are calculus exercises, really, and you should have a go at proving these if you come across them before. Commented Dec 1, 2023 at 5:09 • @An_Elephant After that the strategy is breaking the integral as$I_n = \int_0^1 f + \int_1^n f,$and arguing that the first bit is near$1$, and that the second bit is significantly larger than$1/n$(really all we need is that$n J_n \to \infty$). The reason for this strategy is that this lets us use the fact that if$n a_n \to \infty,$then$(1 + a_n)^n \to \infty\$. Of course, if all this is too much, then you might need more background, but I think that is kind of unavoidable for this problem, so hopefully this'll all make sense when you've had more classes :) Commented Dec 1, 2023 at 5:12
• @DominikKutek Thanks for pointing that out, I've edited :) Commented Dec 1, 2023 at 5:12