Find a real value $y$ for that the identity holds Question:  Find a real value $y$ such that the following identity holds
$$\log(\log(\log(x)))=\int^x_y\frac{dx}{x\log(x)\log(\log(x))}\quad (x>y)$$
Then, find an algorithm to calculate
$$f_n(x)\quad (x>e^{n-1})$$
where $f_0(x)=x$ and $f_n(x)=\log(f_{n-1}(x))$.
Thanks in advanced!
Help appreciated!!
 A: Make the substitution $u=\log(\log x)$, We find that one antiderivative is $\log(|\log(\log x)|)$. 
Substitute the two endpoints. So we want $\log(\log(\log y))=0$. That means $\log(\log y)=1$, so $\log y=e$, and therefore $y=e^e$. 
Note that if we want the iterated logarithm to be defined, in general $x\gt e^{n-1}$ will not be enough. We need an iterated exponential, as in the calculation of $y$.  
A: If you find the suitable lower bounds $y_n$ for $x$, then :
$$f_n(x)=\log(\log(\log\dots(\log x)\dots))=\int_{y_n}^x\dfrac{dt}{\prod\limits_{k=1}^{n-1}f_k(t)}$$
for all $x>y_n$, and for all $n\geq1$
However, as the above said, if you want to define the iterated logarithm, then taking $x>e^{n-1}$ is not enough, but you have to consider :$$x>e^{e^{\dots ^e}}$$ 
For example take $\log(\log(\log(\log(\log x))))$. In order define it, you need :
$$\log(\log(\log(\log x)))>0\Rightarrow \log(\log(\log x)) >1$$
$$\Rightarrow \log(\log x)>e \Rightarrow \log x >e^e$$
$$\Rightarrow x>e^{e^e}\thickapprox 3.8\times10^6$$
while $e^{n-1}=e^{5-1}=e^4\thickapprox 54$
