TITLE: Calculate $\lim_{n\rightarrow \infty} \int_{-\infty}^{\infty}\frac{1}{1+\frac{x^{4}}{n}}dx$ if it exists.
QUESTION [6 upvotes]: Calculate $$\lim_{n\rightarrow \infty} \int_{-\infty}^\infty \frac{1}{1+\frac{x^4}{n}} \, dx$$
if it exists. If this limit does not exist, show why it does not exist.
My attemp: Consider $f_n(x):=\frac{1}{1+\frac{x^{4}}{n}}$,  since $f_n$ is continuous in $\mathbb{R}$, then $f_n$ is $f_n$ are Lebesgue measurable in $\mathbb{R}$, futhermore, note that $f_n\leq f_{n+1}$ for all $n\in\mathbb{N}$. Moreover, we have
$$\lim_{n\rightarrow \infty }f_n(x)=1.$$
Therefore,. by Monotone convergence Theorem we have
$$\lim_{n\rightarrow \infty} \int_{-\infty}^\infty\frac{1}{1+\frac{x^4}{n}} \, dx=\int_{-\infty}^\infty\lim_{n\rightarrow \infty}\frac{1}{1+\frac{x^4}{n}} \, dx= \int_{-\infty}^\infty 1\,dx=\infty.$$
Quiestion: This last conclusion brought me doubts. Does the monotone convergence theorem guarantee that the integral exists? I have read over and over the theorem and I do not find guarantee the existence, it only allows to enter the limit in the integral. I would like to know if I am correct. Additionally, know if there is any error in my attempt and know another way to do it.

REPLY [5 votes]: Note that 
$$\int_{-\infty}^{\infty} \frac{1}{1+\frac{x^4}{n}} \; dx >\int_{-\sqrt[4]{n}}^{\sqrt[4]{n}} \frac{1}{1+\frac{x^4}{n}} \; dx  > \int_{-\sqrt[4]{n}}^{\sqrt[4]{n}} \frac{1}{2} \; dx = \sqrt[4]{n}.$$