Substitution in integral, how shall I proceed

Question

Say we have $\int_2^\infty \frac{1}{(\log n)^{\log n}}dn.$ Let $u=\log n.$ We have the boundaries become $u=\log 2$ and $u=\infty.$ How should I proceed with $dn.$ I have $du=\frac{1}{n}dn,$ hence, $dn=n\cdot du.$ But I don to want $n$ there. Any hints will be helpful, thanks

Answer 1

Hint : $$\begin{align}\int_2^\infty\frac1{\big(\log n\big)^{\log n}}dn&~=~\int_2^{\exp\big(e^2\big)}~\frac1{\big(\log n\big)^{\log n}}dn~+~\int_{\exp\big(e^2\big)}^\infty~\frac1{\big(\log n\big)^{\log n}}dn~\le\\&~\le~\int_2^{\exp\big(e^2\big)}~\frac1{2^{\log n}}dn~+~\int_{\exp\big(e^2\big)}^\infty~\frac1{\big(e^2\big)^{\log n}}dn~=~\\&~=~\int_2^{\exp\big(e^2\big)}~\frac1{n^{\log2}}dn~+~\int_{\exp\big(e^2\big)}^\infty~\frac1{n^2}dn,\end{align}$$ since $a^{\log b}=b^{\log a}.~$ Can you take it from here ?