$\lim_{n\to\infty}\int_0^\infty\frac{n^2\log^2{(x+1)}}{1+(n^3\log^3{(x+1)})}e^{-nx}dx$, use dominated convergence theorem

Question

I have to evaluate the following limit $$\lim_{n\to\infty}\int_0^\infty\frac{n^2\log^2{(x+1)}}{1+(n^3\log^3{(x+1)})}e^{-nx}dx$$ I have thought to apply the dominatated convergence theorem so I have to verify the hypothesis.
1)$f_n(x)=\frac{n^2\log^2{(x+1)}}{1+(n^3\log^3{(x+1)})}e^{-nx}$ are mesurable so I could verify $\int_0^{\infty} |f_n(x)|<\int_0^\infty g(x)$ with $g(x)$ integrable.
2)$\lim_n\to\infty f_n(x)=0$.
So from 2) the integral is null.
I don't know how find g(x)...can you help me?

I have thought $g(x)$ could be something as $xe^{-nx}$ which is integrable...but how can I obtain this?

Answer 1

For $x\le1$, we have

$$\begin{align}\frac{n^2\log^2{(x+1)}}{1+(n^3\log^3{(x+1)})}e^{-nx}\le n^2\log^2{(x+1)}e^{-nx}\le n^2x^2 e^{-nx}\end{align}$$

therefore,

$$\begin{align}\int_0^1 \frac{n^2\log^2{(x+1)}}{1+(n^3\log^3{(x+1)})}e^{-nx}dx&\le \int_0^1 n^2x^2 e^{-nx}dx,~~~~~~~t=nx\\ \\ &=\frac1n\int_0^n t^2 e^{-t}dt\le\frac1n\int_0^\infty t^2 e^{-t}dt\\ \\ &=\frac2n\longrightarrow0\end{align}$$

For $x>1$, we have $$\frac{n^2\log^2{(x+1)}}{1+(n^3\log^3{(x+1)})}e^{-nx}\le \frac{n^2\log^2{(x+1)}}{n^3\log^3{(x+1)}}e^{-x}\le \frac{1}{\log{(x+1)}}e^{-x}$$

where

$$\int_1^\infty \frac{1}{\log{(x+1)}}e^{-x}dx<\infty$$

is integrable.