Find the value of $$\int_0^{\int_0^{\int_{0}^{\int_0^{.^{.^{.^{.^.}}}}} 2x\,dx} 2x\,dx} 2x\,dx$$

My friend and I were arguing over the answer. One way to approach this is to let it be equal to S, then substitute it into the upper bound of the first integral, solve normally, then end up with S=1,0. I don’t think this is correct as an integral cannot be equal to two numbers. Can someone please explain if this is undefined, actually equal to two numbers, or if something completely different is going on??

  • 3
    $\begingroup$ see youtu.be/xnFCncV-288 $\endgroup$
    – Alan
    Jan 10 at 23:47
  • $\begingroup$ The Youtube video @Alan links to is very thorough. To address your specific question, this notation is not "well-defined." This is different than "undefined." The $\dots$ notation here is ambiguous this has different solutions depending on the interpretation of the notation. When all interpretations have the same set of solutions it we say it is well-defined. $\endgroup$
    – ShawSa
    Jan 11 at 1:10

1 Answer 1


$\newcommand{\il}{\int_0^{\int_0^{\int_{0}^{\int_0^{.^{.^{.^{.^.}}}}} 2x\,dx} 2x\,dx} 2x\,dx}$We will probably need a more thorough explanation of the situation here : using mathematical notation. I've seen the video linked in the comments , I'll put it here anyway. The critical difference is that the bottom limit of the integral here is $0$ instead of $1$ : but as pointed out, the video misses some points and doesn't elaborate on others that I think we should be doing.

The idea behind interpretation

Now, the naïve option to take, is that which has already been taken by the author of the post. That is, if we assume that $$ \il = S \implies \int_0^S 2xdx = S $$

The problem with the naïve approach is the same as the problem with approaches in "fake-proofs" that involve divergent summations : if one side isn't a bona fide real number, then that can't be treated as an equality between real numbers, whence all subsequent logic breaks down as well.

So, the approach is instead to use a sequence of real numbers to capture the nesting : I might as well consider a sequence of real numbers like so : let $a_0 \in \mathbb R$ and let $$ a_1 = \int_0^{a_0} 2x\ dx \quad ;\quad a_2 = \int_0^{a_1} 2x\ dx = \int_0^{\int_0^{a_0} 2x\ dx } 2x \ dx \\ a_3 = \int_0^{a_2} 2x\ dx = \int_0^{\int_0^{\int_0^{a_0} 2x\ dx } 2x \ dx } 2x\ dx $$

In other words, $$ a_0 \in \mathbb R , a_{i+1} = \int_0^{a_i} 2x dx \ \forall i \geq 0 $$ defines a sequence, where $a_i$ has $i$ integrals involved. Now, what is clear is that each $a_i$ is a real number, regardless of $a_0$. That can be proven by induction : $a_0$ is real, $2x$ is bounded and continuous in the region between $0$ and $a_0$ hence is integrable(Riemann/Lebesgue etc. , doesn't really matter here) in this region, hence $a_1$ is real : and the same can be said with $a_0,a_1$ replaced by $a_i,a_{i+1}$ to complete the induction step.

So we have a sequence of real numbers $a_i$. If $a_i$ converged to a value as $i$ goes to infinity, it is logical to expect that such a value of the integral can be assigned to $S$.

Pros and cons

The pros are obvious : we've captured the nested nature of the integral with the help of a sequence, and we've also ensured that the sequence of numbers is real, so that we can embed it in existing theory.

There are two big cons : the first, is that there is a dependence on $a_0$. A dependence that we can't quite get rid of, but the point is that it's captured in the expression $\il$ as well. Indeed, we have no clue what is a reasonable upper bound here as well, so what we do is allow that upper bound to fluctuate (it will be $a_0$, as soon as you see the definitions) and see what kind of upper bound works.

The second regards the definition of the quantity $\int_a^b f(x)dx$ as a Riemann integral, when $b<a$. Indeed, according to the video, we treat the integral to be "small to big" all the time, so that the notation $\int_a^b$ only means we are integrating on the set $[a,b]$ and not in any order. In that case, $\int_a^b f(x)dx$ is the same as $\int_b^a f(x)dx$.

The other regards the additive rule $\int_a^b f(x)dx + \int_b^c f(x)dx = \int_a^c f(x)dx$. If we wish that this be respected , then we would have $\int_a^b f(x)dx = - \int_b^a f(x)dx$. We will cover both interpretations although it's clear that the latter approach is more comfortable, since the video requires familiarity with the first approach and is worth watching anyway.

Approach as in the video i.e. $\int_a^b f(x)dx = \int_b^a f(x)dx$

The sequence is very easy to analyse, because we can simplify the integral $$ a_{n} = \int_0^{a_{n-1}} 2x dx = \mbox{sgn}(a_{n-1})a_{n-1}^2 $$

where $\mbox{sgn}(y) = \frac{y}{|y|}$ for $y \neq 0$ and $0$ if $y=0$.

Now, if $a_0 = 0$, then the recursion tells us that $a_n = 0$ for all $n$. Therefore, $\lim_{n \to \infty} a_n = 0$, consequently one can say that $0$ is a reasonable candidate for $\il$.

If $a_0 > 0$, then by induction, $a_i >0$ for all $i$, and by recursion, $$ a_n = a_{n-1}^2 = a_{n-2}^4 \ldots = a_0^{2^n} $$ therefore, we ask ourselves when the limit $\lim_{n \to \infty} a_0^{2^n}$ exists. Indeed, when $0<a_0<1$, it equals $0$. When $a_0 =1$, it equals $1$. Otherwise, it doesn't exist.

On the other hand, if $a_0<0$, then by induction, $a_i < 0$ for all $i$, and yet again by recursion, $|a_n| = |a_0|^{2^n}$ for all $n$. This limit, once again, exists precisely when $|a_0| \leq 1$, and equals $-1$ for $a_0=-1$ and $0$ for $-1<a_0<0$.

So we have three candidates : $0,1,-1$. Which of these should be assigned to $\il$? The answer is : none of them. It isn't reasonable to perform calculations with $\il$ unless $a_0$ is specified prior to doing this.

Approach $\int_a^b f(x)dx = -\int_b^a f(x)dx$

If we do this, then $$ a_{n} = \int_0^{a_{n-1}} 2x dx = a_{n-1}^2 $$

always. By induction, $a_n = a_0^{2^{n}}$, so that using the approach of the previous part, we get that $a_n \to 0$ if $|a_0| < 1$, $a_n \to 1$ if $a_0 = \pm 1$ and $a_n \to \infty$ if $|a_0|>1$.

Finishing the naïve approach

While pursuing the naïve approach, if we take the latter approach ,then indeed from $\int_0^S 2x dx =S$, we get $S^2 = S$ and $S = 0,1$.

If we consider the first interpretation, then the possibility of a negative $S$ will have to be included. Indeed, if $S \geq 0$, then it is true that $$ \int_0^S 2xdx = S \implies S^2 = S \implies S = 0,1 $$ However, if $S \leq 0$, then $$ \int_S^0 2xdx = S \implies S^2 = -S \implies S = -1,0 $$

which is why the interpretation of $S=-1$ also comes into play. In short, the theory helps us place these naïve thoughts into a mathematical framework.

More references
  • $\begingroup$ It's typically a bad idea to define the Riemann integral to be invariant under swap of the limits, because we want $\int_a^b + \int_b^c = \int_a^c$... $\endgroup$
    – user21820
    Jan 13 at 19:25
  • $\begingroup$ @user21820 In that case, the situation is simpler. I want to retain this approach because it was used in the linked video and I didn't want to confuse posters. I'll edit the answer with the better interpretation. $\endgroup$ Jan 13 at 19:30

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