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?

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    $\begingroup$ Why the downvote? It seems the OP made honest efforts. $\endgroup$ Commented Nov 29, 2023 at 13:23
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    $\begingroup$ Looking at online graphs suggests to me that we could look for a bound for $0<x<1$ and another bound for $x>1.$ $\endgroup$ Commented Nov 29, 2023 at 13:41
  • $\begingroup$ Find $\lim_{n \to \infty}{\frac{1}{x^{x^{n}}}}$ for $0 \leq x \leq 1$ then for $x > 1$. $\endgroup$ Commented Nov 29, 2023 at 13:55
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    $\begingroup$ @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 $x>1.$ $\endgroup$ Commented Nov 29, 2023 at 14:01
  • $\begingroup$ Also, something like Holder inequality or Minkowski inequality could be useful... $\endgroup$ Commented Nov 29, 2023 at 14:10

3 Answers 3


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.

  • $\begingroup$ 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). $\endgroup$ Commented Nov 29, 2023 at 18:53
  • $\begingroup$ 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. $\endgroup$
    – Mittens
    Commented Nov 30, 2023 at 6:17
  • $\begingroup$ 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. $\endgroup$ Commented Nov 30, 2023 at 19:02
  • $\begingroup$ @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. $\endgroup$ Commented Nov 30, 2023 at 19:03
  • $\begingroup$ 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? $\endgroup$ 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)$")) 

enter image description here

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, $\lim I_n^n = \infty.$

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    $\begingroup$ I've changed my answer accordingly. Those parts with Dominated Convergence Theorem were remains of previous (wrong) approach. $\endgroup$ Commented Nov 30, 2023 at 21:10
  • $\begingroup$ Thanks. I hope to understand it in the future. My level is too low to understand any of the solutions provided $\endgroup$ Commented Dec 1, 2023 at 3:04
  • $\begingroup$ @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 every $x,$ $1-x \le e^{-x}$, 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. $\endgroup$ Commented Dec 1, 2023 at 5:09
  • $\begingroup$ @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 :) $\endgroup$ Commented Dec 1, 2023 at 5:12
  • $\begingroup$ @DominikKutek Thanks for pointing that out, I've edited :) $\endgroup$ Commented Dec 1, 2023 at 5:12

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