Substitute log(t) = x
$\therefore$ $\frac{1}{t}dt$ = dx
$\therefore$ dt = ex dx
t=1 $\Rightarrow$ x=0
t=n $\Rightarrow$ x=logn
So, integral
= $\int_{0}^{logn}$ $\frac{e^{x}}{x}dx$
Now, if we expand the Mclaurin series for ex, then denominator will cancel out. I think this will lead to infinity after integration.
See here