Integral whose upper limit is the integral itself: $\int_{0}^{\int_{0}^{\ldots}\frac{1}{\sqrt{x}} \ \mathrm{d}x} \frac{1}{\sqrt{x}} \ \mathrm{d}x$ I recently encountered the following definite integral:
$$\int_0^{\int_0^\ldots \frac{1}{\sqrt{x}} \ \mathrm{d}x} \frac{1}{\sqrt{x}} \ \mathrm{d}x$$
where "$\ldots$" seems to indicate that the upper limit of $\int_{0}^{\ldots}\frac{1}{\sqrt{x}} \ \mathrm{d}x$ is also $\int_{0}^{\ldots}\frac{1}{\sqrt{x}} \ \mathrm{d}x$, so that the upper limit of the integral repeats itself infinitely, somewhat similarly to a continued fraction.
I tried to solve this by renaming the integral $\int_0^\ldots \frac{1}{\sqrt{x}} \ \mathrm{d}x$ in the upper limit $U$, so that we get
$$\int_0^{\int_0^\ldots \frac{1}{\sqrt{x}} \ \mathrm{d}x} \frac{1}{\sqrt{x}} \ \mathrm{d}x =\int_0^U \frac{1}{\sqrt{x}} \ \mathrm{d}x=\left[2\sqrt{x} \vphantom{\frac 1 1} \right]_0^U = 2\sqrt{U}$$
If we now consider that the integral in the upper limit $U$ is in fact equal to the original definite integral which we are trying to evaluate, we find that $U=2\sqrt{U}\Rightarrow U^2=4U\Rightarrow U^2-4U=0$, which is a quadratic equation with solutions $U_1=4$ and $U_2=0$. This seems to imply that
$$\int_{0}^{\int_{0}^{\ldots}\frac{1}{\sqrt{x}} \ \mathrm{d}x} \frac{1}{\sqrt{x}} \ \mathrm{d}x=4\ \ \vee\ \ \int_{0}^{\int_{0}^{\ldots}\frac{1}{\sqrt{x}} \ \mathrm{d}x} \frac{1}{\sqrt{x}} \ \mathrm{d}x=0$$
My question is:
Is one (or both) of these solutions correct, and is there a way to prove (or disprove) this (assuming that what I've written here is not sufficient proof)?
 A: The only missing part is assumed convergence. You have to find the conditions for the convergence this way or another.
$$a_{1}=\int_{0}^{t}\frac{1}{\sqrt{x}}=2\sqrt{x}|_{0}^{t}=2\sqrt{t}$$
$$a_{2}=\int_{0}^{2\sqrt{t}}\frac{1}{\sqrt{x}}=2\sqrt{x}|_{0}^{2\sqrt{t}}=2\sqrt{2\sqrt{t}}=2^{1+\frac{1}{2}}t^{\frac{1}{4}}$$
It is not difficult to notice the pattern
$$a_{n}=2^{2-\frac{1}{2^{n-1}}}t^{\frac{1}{2^n}}$$
Now you have
$$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}2^{2-\frac{1}{2^{n-1}}}t^{\frac{1}{2^n}}=\lim\limits_{n \to \infty}2^{2-\frac{1}{2^{n-1}}}\lim\limits_{n \to \infty}t^{\frac{1}{2^n}}$$
$$t \neq 0\;\lim\limits_{n \to \infty}t^{\frac{1}{2^n}}=1$$
$$t = 0\;\lim\limits_{n \to \infty}t^{\frac{1}{2^n}}=0$$
As you can notice we have no problem with negative values.
$$\lim\limits_{n \to \infty}a_{n}=2^2=4, t \neq 0$$
$$\lim\limits_{n \to \infty}a_{n}=0, t = 0$$
It is this "universal" convergence that allows you to write the infinite expression. There are cases where this convergence is stricter.
The existence of two solutions is telling you that infinity never goes to both ends. It had started somewhere.
Here is the reason why it is important.
Observe that you have an implicit assumption that initial $t$ is a constant. What if it is not a constant? Take $t$ to be a function $t=t(n)$. Now things get more complicated.
$$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}2^{2-\frac{1}{2^{n-1}}}t(n)^{\frac{1}{2^n}}$$
Obviously in this case the convergence depends on $\lim\limits_{n \to \infty}t(n)^{\frac{1}{2^n}}$, and this could go anywhere you like, it can be even undefined.
This is the reason why you cannot have infinities at both ends.
A: As the others said, you can think about it in terms of recursive sequences $$
u_{n+1} = 2\sqrt{u_n}\\
u_0>0
$$
The issue is that $L = 2\sqrt L$ has two real solutions, $0$ and $4$. And both are possible, depending on the initial value. What I think should be done here is figuring out which one is the limit depending on $u_0$.
If $u_0 = 0$, then the limit is $0$
If $u_0 = 4$, then the limit is $4$
But if $0<u_0<4$, then $u_n$ is increasing and bounded above by $4$, so it converges to $4$
Similarly, if $u_0>4$, $u_n$ is decreasing and bounded below by $4$, and so it converges to $4$.
So, in other words, the sequence only converges to $0$ if the first term is $0$. In any other case, it converges to $4$. So in some sense, it can be said that $4$ is the "true" value of the nested expression.
A: Let's treat this as a sequence such that the upper limit of the $n$th nested integral is $u_n$.  Then it should be clear that the $n$th integral $I_n$ is equal to
$$I_n = 2 \sqrt{2 \sqrt{2 \sqrt{2 \cdots \sqrt{2 u_n}}}} $$
As $n \to \infty$ then, we just get twice an infinite nested square-root sequence of two's, which is equal to $4$.
To see that last statement, note that the expression is equal to
$$2^{1+1/2 + 1/2^2 + 1/2^3+\cdots} = 2^2 = 4$$
ADDENDUM
More precisely, if we take the log of both sides in the finite expression, we get
$$\log{I_n} = \log{2} + \frac{\log{2}}{2} +  \frac{\log{2}}{2^2}+ \frac{\log{2}}{2^3}+ \cdots +  \frac{\log{2}}{2^{n-1}}+ \frac{\log{2}}{2^{n}}+ \frac{\log{u_n}}{2^n}$$
So long as $u_n$ is bounded, the limiting result of $4$ is attained.
A: Sorry for my bad English and very low level layout as it's my first post here.
We could easily  find an answer just with convergence tools, as the result of this integral with upper limit $3$ is above $3$, in fact it's $2 \cdot \sqrt{3} = 3.464$ and if upper limit is $3.464$, the result of the integral is above $3.464 \ (3.722)$ and similarly the result for an upper limit above $4$ for example $5$ is lower than $5 \ (4.472)$.
It’s sure we have a convergence point between them which is the answer of the following equation $x = 2\cdot \sqrt{x}  \Rightarrow  x= 4 $ on a graph here, this solution seems very easy. 
enter image description here
But in fact the more important thing for me is where Michael found that integral and which problem is resolved by this equation.
