Find $\lim_{n\to \infty} \int_{\mathbb R}\frac{n \log^{4}(x)}{n+nx+x^2}\chi_{[0,\infty[}d\lambda(x)$

Question

$\lim_{n\to \infty} \int_{\mathbb R}\frac{n \log^{4}(x)}{n+nx+x^2}\chi_{[0,\infty[}d\lambda(x)$

This seems like an apt situation to utilize dominating convergence.

$f_{n}(x):=\frac{n \log^{4}(x)}{n+nx+x^2}$ is continuous on $]0,\infty[$ and therefore measurable $\forall n \in \mathbb N$. But I am struggling to show $\int_{[0,\infty[}\frac{n \log^{4}(x)}{n+nx+x^2}d\lambda(x)=\int_{]0,\infty[}\frac{n \log^{4}(x)}{n+nx+x^2}d\lambda(x)<\infty$ to ensure $(f_{n})_{n}\subseteq\mathcal{L}^{1}(\mu)$.

$|\frac{n \log^{4}(x)}{n+nx+x^2}|\leq...?$

Any hints?

Answer 1

To apply LDCT you need an $L^1$ majorant $g$ satisfying $|f_n(x)| \le g(x)$ almost everywhere for all $n$. If $f_n \to f$ almost everywhere then you have $|f(x)| \le g(x)$ too.

You can write $$f_n(x) = \frac{n \log^4 x}{n + nx + x^2} = \frac{\log^4 x}{1 + x + x^2/n}$$ so that $$f_n(x) \to \frac{\log^4 x}{1+x}.$$ This function is not integrable, so you won't find an integrable majorant $g$ with $|f_n| \le g$ for all $n$. It looks like LDCT may not be the way to go.

Answer 2

Note that for $x > 0$ we have $$f_n(x):= \frac{\log(x)^4}{1+x+x^2/n}.$$ Thus for all $x >0$ $$f_n(x) \le f_{n+1}(x) \le f(x):=\frac{\log(x)^4}{1+x}$$ and pointwise $f_n \rightarrow f$. Therefore, we can apply the monotone convergence theorem. However, the limes $f$ is not integrable, because $$\int_{e}^\infty f(x) \, dx \ge \int_{e}^\infty \frac{1}{1+x} \, dx =\infty,$$ and we cannot apply the dominated convergence theorem.