Evaluate $\lim_{n\to\infty}\int^2_0\frac{dx}{1+x^{2n}}$

Question

Evaluate $\lim_{n\to\infty}\int^2_0\dfrac{dx}{1+x^{2n}}$

First, I find $f(x):=\lim_{n\to\infty}\dfrac{dx}{1+x^{2n}}$ \begin{align} 0&\leq x\leq 2\\ 0&\leq x^{2n}\leq 2^{2n}\\ \dfrac{1}{1+2^{2n}}&\leq \dfrac{1}{1+x^{2n}}\leq 1\\ \end{align} By the squeeze lemma, $f_n\to 1$ as $n\to 0\implies f(x)=1$

Also, $\lVert f_n\rVert_\infty$ is bounded since: \begin{align} \lVert f_n\rVert_\infty&=\sup_{[0,2]}\left\lvert \dfrac{1}{1+x^{2n}}\right\rvert\\ -1&\leq\dfrac{1}{1+2^{2n}}\leq \dfrac{1}{1+x^{2n}}\leq 1\implies \left\lvert \dfrac{1}{1+x^{2n}}\right\rvert\leq1 \end{align} Additionally, $f_n$ and $f$ are integrable on $[1,2]$ Now, by the dominated convergence theorem, we can swap the limit and the integration and hence, $$\lim_{n\to\infty}\int^2_0\dfrac{dx}{1+x^{2n}}= \int^2_0f(x) dx=\int^2_0 dx=2 $$

Answer 1

Note that for $x \in [0,2]$ the function $\frac{1}{1+x^{2n}}$ is strictly decreasing, so it attains its maximum value at $x=0$ for all $n$, and we can apply the bounded convergence theorem instead (Which is a bit easier.)

Also, you should look at the limit again. When $x<1$ the function tends to $1$, when $x>1$, it tends to $0$, and $x=1$, it is $\frac{1}{2}$, but this is a set of measure zero.

Answer 2

We can avoid more powerful theorems using a direct proof by splitting the integral into three parts, on $[0,1-\epsilon)$, $[1-\epsilon,1)$, and $[1, 2]$, for some fixed $0<\epsilon<1$.

First, consider $x\in[0,1-\epsilon)$, then $$x^{2n} < (1-\epsilon)^{2n} \leq (1-\epsilon)^{2N},\quad \text{for $n\geq N$}$$ furthermore if we pick any $N > \frac{\log(\epsilon)}{2\log(1-\epsilon)}$ then $$(1-\epsilon)^{2N} < \epsilon$$ so for $n\geq N > \tfrac{\log(\epsilon)}{2\log(1-\epsilon)}$, $$\frac{1-\epsilon}{1+\epsilon} = \int_0^{1-\epsilon}\frac{dx}{1+\epsilon} \leq \int_0^{1-\epsilon} \frac{dx}{1+x^{2n}} \leq \int_0^{1-\epsilon} (1)\;dx = 1-\epsilon$$

Next, $$\frac{\epsilon}{2} = \int_{1-\epsilon}^1\frac{dx}{1+(1)} \leq \int_{1-\epsilon}^{1} \frac{dx}{1+x^{2n}} < \int_{1-\epsilon}^1 (1)\;dx = \epsilon$$ Finally, $$0 <\int_1^2 \frac{dx}{1+x^{2n}} < \int_1^2 \frac{dx}{x^{2n}} = \frac{1- 2 \cdot 2^{-2n}}{1+2n}$$

Giving $$ \frac{1-\epsilon}{1+\epsilon} + \frac{\epsilon}{2} < \int_0^2 \frac{dx}{1+x^{2n}} < (1-\epsilon)+\epsilon + \frac{1-2^{-2n}}{1+2n}$$ for all $n\geq N$, so let $n\to \infty$ gives $$\frac{1-\epsilon}{1+\epsilon} + \frac{\epsilon}{2} \leq \lim_{n\to\infty} \int_0^2 \frac{dx}{1+x^{2n}} \leq 1$$ Now let $\epsilon \to 0^+$, the left hand side approaches $1$ and so by the squeeze theorem $$\lim_{n\to\infty}\int_0^2 \frac{dx}{1+x^{2n}} = 1$$