# Does $\int_{0}^{\int_{0}^{\int_{0}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~} dx} dx} dx$ converge? [closed]

In another question, I have asked about the "value" of $\int_{0}^{\int_{0}^{\int_{0}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~} dx} dx} dx$.

Concerning the issue of convergence, I think I have a proof that this expression does not converge.

The reasoning is the following: since the expression can take the value of any real number, we can write both: $$\pi=\int_{0}^{\int_{0}^{\int_{0}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~} dx} dx}dx$$ and $$\sqrt{2}=\int_{0}^{\int_{0}^{\int_{0}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~} dx} dx} dx$$ Since this expression cannot be equal to both $\pi$ and $\sqrt{2}$, we have a contradiction, which means that the integral does not converge. Is this line of reasoning correct ?

## closed as unclear what you're asking by Paolo Leonetti, Travis, Watson, Jonas Meyer, user99914 Jun 8 '16 at 0:31

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• You could've begun with "since the expression can take the value of any real number", cut out the part in the middle, finished with "the integral does not converge", and you'd be done. – Arthur Jun 7 '16 at 11:46
• In my opinion, it is not an issue of "convergence": it's just the fact that the problem is ill-posed. In fact, the iterated sequence $$\begin{cases} a_0=\alpha\\ a_{n+1}=\int_0^{a_n}\,dx\end{cases}$$ not only converges, but it is also constant. The "takes all real values" stuff comes from the wrong assumption (suggested by a not-thoroughly thought notation) that $\lim_n a_n$ does not depend on $\alpha$. – user228113 Jun 7 '16 at 11:59
• @Tos : related: math.stackexchange.com/questions/1623403/… – Watson Jun 7 '16 at 12:00
• Perhaps you should describe the expression as ill-defined. The usual approach of defining a value in terms of fixed-points of some iteration fails to uniquely specify a value, precisely because (as you observe) any value is a fixed-point of the obvious iteration. – hardmath Jun 7 '16 at 17:25

It all depends on what you intend the $\cdots$ in your expression to mean.

A typical way to define the meaning would be as the limit of the sequence $$a, \int_0^a dx, \int_0^{\int_0^a dx}dx, \int_0^{\int_0^{\int_0^a dx}dx}dx, \int_0^{\int_0^{\int_0^{\int_0^a dx} dx}dx}dx, \ldots$$

Note that in order even to define which sequence we're talking about, we need to decide what to put instead of the not-yet-unfolded part of the "infinite expression" in order to get one of the terms in the sequence -- here I've arbitrarily called it $a$.

And with the way your infinite-stack-of-integrals work, the outcome is that at each finite cut-off the entire stack just collapses into the placeholder you put at the top of it -- so the sequence you're now looking for the limit of is simply $$a, a, a, a, a, \ldots$$

This obviously converges -- to $a$ -- so it is not true when you claim that it doesn't. The problem with your infinite expression is not that it fails to converge, but that the choice of $a$ influences what the limit is.

Using limits to give meaning to things with "$\cdots$" in them only works well if either there is a natural "neutral" choice for the placeholder value, or the placeholder value matters less and less the more of the "ideal" infinite expression we include.

The former is the case for the typical case of infinite sums (where it is arguable that $0$ is a natural placeholder); the latter is the case for continued fractions or some cases of infinitely nested radicals.

• I think this is the most extreme contrast of answer upvotes and question downvotes that I've seen yet :-) – joriki Jun 7 '16 at 16:55
• – Henning Makholm Jun 7 '16 at 18:05
• Wow, nice, I didn't know about that! One downvote and $13$ upvotes to go... – joriki Jun 7 '16 at 18:08
• @Henning Makholm : are you aware of references that discuss what you write about "giving meaning to things with "$...$" " ? – Tos Jun 17 '16 at 15:48
• @Tos, no -- it's one of those things students generally seem to be expected to learn through osmosis. – Henning Makholm Jun 17 '16 at 16:05