Calculate $\lim_{n\rightarrow \infty} \int_{-\infty}^{\infty}\frac{1}{1+\frac{x^{4}}{n}}dx$ if it exists.

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.

• Please, have a look at an alternative route below. – Olivier Oloa May 16 '17 at 20:23
• In the monotone convergence theorem, it is allowed that the integral of the limit function is $+\infty$. Since it deals with non-negative measurable functions, the integral always exists in $[0,+\infty]$. – Daniel Fischer May 16 '17 at 20:37

One may write, as $n \to \infty$, \begin{align} \int_{-\infty}^{\infty}\frac{1}{1+\frac{x^{4}}{n}}dx&=2\int_0^{\infty}\frac{1}{1+\frac{x^{4}}{n}}dx \\\\&=2n^{1/4}\underbrace{\int_0^{\infty}\frac{1}{1+u^4}du}_{>0}\qquad \left(u=\frac{x}{n^{1/4}}\right) \\\\& \to \infty. \end{align}
$$\int_{-\infty}^{\infty} \frac{1}{1+\frac{x^4}{n}} \; dx >\int_{-\sqrt[4]{n}}^{\sqrt[4]{n}} \frac{1}{1+\frac{x^4}{n}} \; dx > \int_{-\sqrt[4]{n}}^{\sqrt[4]{n}} \frac{1}{2} \; dx = \sqrt[4]{n}.$$
$$\int_{-\infty}^{\infty}\frac{1}{1+\frac{x^4}{n}}\mathrm{d}x=\pi\frac{\sqrt[4]{n}}{\sqrt{2}}$$
$$\lim_{n\to\infty}\int_{-\infty}^{\infty}\frac{1}{1+\frac{x^4}{n}}\mathrm{d}x=\infty$$