Uniform convergence and integration: $\lim_{n \to \infty} \int_{0}^{1} \frac{1}{1+x^2+\frac{x^4}{n}} dx=?$

Question

For f$_n$(x)=$\frac{1}{1+x^2+\frac{x^4}{n}}$, we need to calculate $\lim_{n \to \infty} \int_{0}^{1} f_n(x) dx$ . I want to prove f$_n$ is Riemann integrable and f$_n$ uniformly converges to f, then I can have $\lim_{n \to \infty} \int_{0}^{1} f_n(x) dx$ = $\int_{0}^{1} f(x)dx$. But how to prove these two steps?

Answer 1

That $\int_0^1 f_n(x)\mathsf dx$ exists follows from $f_n$ being continuous and bounded on $[0,1]$. It's clear that the limit function is $f(x)=\frac1{1+x^2}$. To show that $f_n$ converges uniformly, it suffices to show that $$\sup_{x\in[0,1]}|f(x)-f_n(x)|\stackrel{n\to\infty}\longrightarrow 0.$$ We compute $$|f(x)-f_n(x)| = \left|\frac1{1+x^2}-\frac1{1+x^2+\frac{x^4}n} \right| = \frac{x^4}{n(1+x^2)\left(1+x^2 + \frac{x^4}n\right)}$$ (the absolute value isn't necessary since $f(x)>f_n(x)$ for all $n,x$). Since $0\leqslant x^4\leqslant 1$ and $(1+x^2)\left(1+x^2 + \frac{x^4}n\right)>1$ for all $n,x$, we have $$\sup_{x\in[0,1]}|f(x) - f_n(x)|\leqslant\frac1n\stackrel{n\to\infty}\longrightarrow 0.$$ Therefore \begin{align} \lim_{n\to\infty}\int_0^1 f_n(x)\mathsf dx &= \int_0^1 f(x)\mathsf dx\\ &= \int_0^1 \frac1{1+x^2}\mathsf dx\\ &= \arctan 1 - \arctan 0\\ &= \frac\pi4. \end{align}

As a side note, you don't actually need uniform convergence here. It's enough to observe that each $f_n$ has finite integral, $f_n(x)\stackrel{n\to\infty}\longrightarrow f(x)$ for all $x\in[0,1]$, and $f_n(x)<f_{n+1}(x)$ for all $x\in[0,1]$. Then we may interchange the limit and integral by the monotone convergence theorem.

Answer 2

Hint:

Riemann integrabillity:

$$f_n(x) \text{ is continuous and bounded for all } n$$

Uniform convergence:

$$|f_n(x) - f(x)| = \frac{1}{n}\frac{x^4}{(1+x^2)(1+x^2+x^4/n)} \leq \frac{1}{n},~~~~~{\rm for}~~~~x\in[0,1]$$

Answer 3

You may just observe that, for each $x \in [0,1]$, we have $$ f_n(x)=\frac{1}{1+x^2+\frac{x^4}{n}} \to f(x)=\frac{1}{1+x^2}. $$ But $$ \begin{align} \left|f(x)-f_n(x)\right|&=\left|\frac{1}{1+x^2}-\frac{1}{1+x^2+\frac{x^4}{n}}\right|\\\\ &=\left|\frac{\frac{x^4}{n}}{(1+x^2)(1+x^2+\frac{x^4}{n})}\right|\\\\ &\leq\left|\frac{\frac{1}{n}}{(1+0^2)(1+0^2+\frac{0^4}{n})}\right|\\\\ &\leq\frac{1}{n}\\\\ \end{align} $$ and the convergence is uniform on $[0,1]$. You can deduce that

$$ \lim_{n \to \infty} \int_{0}^{1} f_n(x) dx=\int_{0}^{1} \frac{1}{1+x^2} dx=\color{red}{\frac{\pi}4}. $$