Dangerous Integral

Let $u = \log \log \log x$ . Then we have:

$\begin{aligned} \log \log \log x & = u \\ \log \log x & = e^u \\ \log x & = e^{e^u} \\ \implies x & = e^{e^{e^u}} & \small \color{#3D99F6} \text{when } x=e^{e^e}, \ u=1; \ x = \infty, \ u = \infty. \\ dx & = e^{e^{e^u}} e^{e^u} e^u \ du \end{aligned}$

Therefore,

$\begin{aligned} I & = \int_{e^{e^{e^e}}}^\infty \frac {dx}{x(\log x)(\log \log x)(\log \log \log x)^\frac 43} \\ & = \int_1^\infty \frac {e^{e^{e^u}} e^{e^u} e^u}{e^{e^{e^u}} e^{e^u} e^u u^\frac 43} du \\ & = \int_1^\infty u^{-\frac 43}\ du \\ & = -3 u^{-\frac 13}\ \bigg|_1^\infty \\ & = \boxed{3} \end{aligned}$

We want to calculate the integral

$I = \int _{ { e }^{ { e }^{ e } } }^{ \infty }{ \frac { dx }{ x\left( \log { x } \right) \left( \log { \log { x } } \right) { \left( \log { \log { \log { x } } } \right) }^{ 4/3 } } },$

where $\log(x)$ stands for the natural logarithm of $x$ . For that we may use the substitution $u = \log (x)$ and therefore $du = dx/x$ , so

$I = \int _{ { e }^{ e } }^{ \infty }{ \frac { du }{ u\left( \log { u } \right) \left( \log { \log { u } } \right)^{4/3} } }.$

Under a new substitution $v = \log(u)$ we have

$I = \int _{ { e } }^{ \infty }{ \frac { dv }{ v \left( \log { v } \right)^{4/3} } }.$

Finally, if $y = \log(v)$

$I = \int _{ { 1 } }^{ \infty }{ \frac { dy }{ y^{4/3} } } = \left(- \frac{3}{y^{1/3}} \right)_1^\infty = 3.$