I am currently on this problem $$\int_{2}^{N}\frac{dt}{t\,(\log t)^{\frac{t+1}{t}}}$$
I tried to make substitution but it wouldn't work. That is, $x=\log t$ and my result was
$$\int_{2}^{N}\frac{dt}{t\,(\log t)^{\frac{t+1}{t}}}=\int_{\log 2}^{\log N}x^{-1-e^{-x}}dx.$$
I think I got it wrong somewhere. Please, could anyone help me out?
It is useful to notice that $$ \int_{2}^{N}\frac{dt}{t\log t}=\left[\log\log t\right]_{2}^{N} = \log\log N-\log\log 2\tag{1}$$ $$\forall \alpha>0,\qquad \int_{2}^{N}\frac{dt}{t\left(\log t\right)^{1+\alpha}}=\left[-\frac{1}{\alpha\left(\log t\right)^{\alpha}}\right]_{2}^{N}=-\frac{1}{\alpha\left(\log N\right)^{\alpha}}+\frac{1}{\alpha\left(\log 2\right)^{\alpha}}\tag{2} $$ hence by evaluating the RHS of $(2)$ at $\alpha=\frac{1}{2}$ and $\alpha=\frac{1}{N}$ we have that the given integral is convergent.
On the other hand, $\frac{1}{t\left(\log t\right)^{1+\frac{1}{t}}}$ does not have an elementary primitive.
What is the actual statement of the original problem?