Convergence of $\lim_{n \rightarrow \infty} \int\limits_0^n \frac{1}{1+nx \ln (x)} dm$ I am struggling with the following integral:
\begin{equation}
\lim_{n \rightarrow \infty} \int\limits_0^n \dfrac{1}{1+nx \ln (x)} dm\,,
\end{equation}
where $dm$ is Lebesgue measure.
I was thinking of splitting the integral as follows:
\begin{equation}
\int\limits_0^n \dfrac{1}{1+nx\ln(x)}dm = \int\limits_0^1 \dfrac{1}{1+nx\ln(x)}dm + \int\limits_1^n \dfrac{1}{1+nx\ln (x)}dm.
\end{equation}
Then for the second part I think I can use the dominated convergence theorem,  since
\begin{equation}
\dfrac{1}{1+nx\ln (x)} \leq 1
\end{equation}
and 
\begin{equation}
\lim_{n \rightarrow \infty} \dfrac{1}{1 + nx \ln (x)} = 0,
\end{equation}
but I'm not sure because the upper boundary ($n$) also goes to infinity. For the first part, I have no clue. Could someone help me out?
Thanks!
 A: Assuming $n\gg e$, we have
$$ \int_{e}^{n}\frac{dx}{1+nx\log(x)} = \int_{1}^{\log(n)}\frac{dz}{e^{-z}+nz}=\int_{0}^{\log\log n}\frac{du}{n+\exp(-u-e^u)}\leq \frac{\log\log n}{n}$$
and
$$ \int_{1}^{e}\frac{dx}{1+n x\log(x)}=\int_{0}^{1}\frac{dz}{e^{-z}+n z}\leq \int_{0}^{1}\frac{dz}{1+(n-1)z}=\frac{\log n}{n-1}, $$
not causing any trouble. The troubles come from the integral over $(0,1)$, whose meaning in principal value is 
$$ \text{PV}\int_{0}^{+\infty}\frac{dz}{e^{z}-n z}. $$
All the complex roots of $e^z-nz$ are simple, so this is a convergent principal value. The real roots of $e^z-nz$ are located at $\zeta_0=-W\left(-\frac{1}{n}\right)$, which is approximately $\frac{1}{n}$-apart from the origin, and at a point $\zeta_1$ close to $\log(n)+\log\log(n)$. The convergence of the above principal value as $n\to +\infty$ is quite non-trivial, but
$$\operatorname*{Res}_{z=\zeta_i}\frac{1}{e^z-nz}=\frac{1}{e^{\zeta_i}-n}=\frac{1}{n}\cdot\frac{1}{\zeta_i-1} $$
reduce the problem to the study of
$$ \frac{1}{n}\int_{0}^{+\infty}\frac{1}{(\zeta_0-1)(z-\zeta_0)}+\frac{1}{(\zeta_1-1)(z-\zeta_1)}\,dz $$
which is divergent.
A: The integral in question diverges whenever $n\ge e$.  
To see this, we write $f(x)=1+nx\log(x)$ and note that if $n\ge e$, then $f(1/e)=1-n/e\le 0$.  
Moreover, we see that $f(0^+)=f(1)=1$.  Therefore, there are roots of $f(x)$ at some $x_1\in(0^+,1/e)$ and $x_2\in(1/e,1)$.  
Applying the mean value theorem, we can write $f(x)=f'(\xi_1)(x-x_1)$ for some number $\xi_1\in(x_1,x)$.  Inasmuch as $\frac1{f(x)}=O\left(\frac1{x-x_1}\right)$ for $x\in (x_1,1/e)$, the integral in question diverges.
