How do I estimate $int_{2}^{infty} frac{|log log t|}{t, (log t)^2} , dt$

Integration by substitution $t=e^u;;dt=e^u du;;log t = u$ $$int frac{log (log t)}{t log ^2 t} , dt=intfrac{ log u}{u^2},du=-frac{log u+1}{u}+C$$

$$int frac{log (log t)}{t log ^2 t} , dt=-frac{log (log t)+1}{log t}+C$$ $$int_{2}^{infty} frac{|log log t|}{t, (log t)^2} , dt=int_{2}^{e} left(-frac{log log t}{t, (log t)^2}right) , dt+int_{2}^{infty} frac{log log t}{t, (log t)^2} , dt=$$ $$=left[frac{log (log t)+1}{log t}right]_2^e+left[-frac{log (log t)+1}{log t}right]_e^{infty}=$$ $$=1-frac{1+log (log 2)}{log 2}-underset{mto infty }{text{lim}}frac{log (log m)+1}{log m}+1=$$ $$=2-frac{1+log (log 2)}{log 2}approx 1.086$$

Observe that begin{align} int^infty_2 frac{|loglog t|}{t(log t)^2} dt = int^infty_{loglog 2} |u|e^{-u} du. end{align}

You can even evaluate this by hand.