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)?