Integrate $\int_0^{\int_0^{\vdots}\frac{1}{\sqrt{x}}\text{d}x}\frac{1}{\sqrt{x}}\text{d}x$ and monotonicity of integrals I must evaluate
$$\int_0^{\int_0^{\vdots}\frac{1}{\sqrt{x}}\text{d}x}\frac{1}{\sqrt{x}}\text{d}x$$
My idea is that if we set
$$L:=\int_0^{\vdots}\frac{1}{\sqrt{x}}\text{d}x$$
Then the integral must satisfy the equation
$$\int_0^L \frac{1}{\sqrt{x}}\text{d}x=L$$
And we have that
$$\int_0^L \frac{1}{\sqrt{x}}\text{d}x=\lim_{\varepsilon \to 0^+}\int_\varepsilon^L\frac{1}{\sqrt{x}}\text{d}x=2\sqrt{L}$$
So we have the equation $2\sqrt{L}=L$, that leads to the solutions $L_1=0$ and $L_2=4$; now I suspect that
$$\int_0^{\int_0^{\vdots}\frac{1}{\sqrt{x}}\text{d}x}\frac{1}{\sqrt{x}}\text{d}x > 0$$
So the only choice left if $L_2=4$, but I don't actually know how to prove it rigorously; I'm sure that the last integral is $\geq0$ because $\sqrt{x}\geq0$, but maybe it is $>0$ because the square root can't be $0$ being at the denominator.
Two questions:
1) is my argument right? I'm not sure if it is rigorous, especially when I "substitute" the upper bound with $L$; maybe I can approach it with sequences.
2) In this case the square root was at the denominator so somehow I've excluded the fact that the integrand could be $\geq0$ (if my argument is correct), but in general how can I prove that if $f(x)>0$ then $\int_a^b f(x) \text{d}x >0$ and not $\int_a^b f(x)\text{d}x \geq 0$ (if this is true)?
Thanks.
 A: 1)
I'll try to make it more rigorous.
Let
$$
a_0= \int_0^\alpha \frac{1}{\sqrt{x}} \mathrm{d}x = 2 \sqrt{\alpha}\\[3ex]
a_{n+1}=\int_0^{a_n}\frac{1}{\sqrt{x}} \mathrm{d}x = 2 \sqrt{a_n}
$$
for some $\alpha \ge 0$. We are interested in evalulate
$$L=\lim_{n \to \infty} a_n. $$
From the general equation we have
$$L =2 \sqrt{L} $$
so if we suppose the limit exists, then a priori we have three possibilities:


*

*$L=0$;

*$L=4$;

*$L= +\infty$.


Notice that for the first terms of the sequence we have
$$
a_0 = 2 \alpha^{\frac{1}{2}}\\
a_1 = 2 \sqrt{a_0}= 2 \sqrt{2 \alpha^{\frac{1}{2}}} = 2 \cdot 2^{\frac{1}{2}}  \cdot\alpha^{\frac{1}{2^2}}\\
a_2 =  2 \sqrt{a_1} = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2^2}}\cdot\alpha^{\frac{1}{2^3}}
$$
and, in general,
$$
a_n=2^{\sum_{i=0}^n \big(\frac{1}{2}\big)^i} \cdot \alpha^{\frac{1}{2^{n+1}}}
$$
(note that if $\alpha = 0$ then $a_n = 0$ for every $n > 0$ ).
So, since
$$
\sum_{i=0}^n \bigg(\frac{1}{2}\bigg)^i = \frac{1-\frac{1}{2^{n+1}}}{1 - \frac{1}{2}} = 2 - \frac{1}{2^n}
$$
(see Formula for geometric series ), we get
$$
a_n = 4 \cdot 2^\frac{1}{2^{n+1}} \cdot  \alpha^{\frac{1}{2^{n+1}}}
$$
Now 
$$\lim_{n \to \infty} 2^\frac{1}{2^{n+1}}=2^0=1$$
and hence if $\alpha \neq 0$,
$$
\lim_{n \to \infty}a_n = 4 \cdot \lim_{n \to \infty} 2^\frac{1}{2^{n+1}}\cdot \lim_{n \to \infty}\alpha^{\frac{1}{2^{n+1}}} = 4 \cdot\lim_{n \to \infty}\exp \left( \frac{\ln \alpha}{2^{n+1}} \right) = 4 \cdot\exp \left(\ln \alpha\cdot \lim_{n \to \infty} \frac{1}{2^{n+1}} \right)=4 \cdot e^0 = 4
$$
finally
$$ \lim_{n \to \infty} a_n = \begin{cases} 0 & \text{if  $\alpha =
 0$}; \\ 4 & \text{if  $\alpha > 0$} \\ \end{cases} $$
2)
Riemann integral has the monotonicity proprierties, see here for more details.
A: 
1) is my argument right? I'm not sure if it is rigorous, especially when I "substitute" the upper bound with L; maybe I can approach it with sequences.

The question isn't even properly posed, as
$$\int_0^{\int_0^\vdots\frac1{\sqrt x}~\mathrm dx}\frac1{\sqrt x}~\mathrm dx$$
isn't a meaningful expression. You want to instead consider something such as:
$$a_0=\alpha\ge0\\a_{n+1}=\int_0^{a_n}\frac1{\sqrt x}~\mathrm dx=2\sqrt{a_n}$$
and ask what
$$L=\lim_{n\to\infty}a_n$$
is. It can easily be seen that if $\alpha=0$, then $L=a_n=0$ for all $n$. If $0<a_n<4$, then we have
$$a_n<2\sqrt{a_n}=a_{n+1}<4$$
and if $a_n>4$, then we have
$$4<a_{n+1}=2\sqrt{a_n}<a_n$$
so for any $\alpha>0$, it converges, since $a_n$ is monotone and bounded. Now that you know that it converges, it suffices to solve
$$L=2\sqrt L$$
restricted to the interval that we know the solution is in (from the bounds), which in this case gives $L=4$.

2) In this case the square root was at the denominator so somehow I've excluded the fact that the integrand could be $\ge0$ (if my argument is correct), but in general how can I prove that if $f(x)>0$ then $\int^b_af(x)dx>0$ and not $\int^b_af(x)dx\ge0$ (if this is true)?

This is not true unless $a<b$ and $f$ is Riemann integrable, in which case you can see Is the Riemann integral of a strictly positive function positive?.
