Integral from MIT Integration Bee Qualifiers 2019

Question

$$\lim_{n\to\infty} \int_{-\infty}^{\infty} e^{-x^{2n}} dx$$

This question is the 12th question from the MIT Integration Bee Qualifiers 2019.
Can someone provide a full solution to this problem?

The answer is 2.

Answer 1

Remark. Based on the comments to your question, I will mark a great part of my answer as a spoiler so you can figure it out yourself.

Hint. Use the Dominated Convergence Theorem by Lebesgue

Full Solution.

Let $-1<x<1$. Then $$\lim_{n\to\infty} x^{2n}=0$$ so that $$\lim_{n\to\infty} \exp(-x^{2n})=1.$$ Also, when $x<-1$ or $1<x$, $x^{2n}$ goes to $\infty$ so that $$\lim_{n\to\infty} \exp(-x^{2n})=0.$$ It follows that $$\lim_{n\to\infty} \exp(-x^{2n})=\begin{cases}1,&-1<x<1\\0,&x<-1\text{ or }1<x\end{cases}.$$ Since all the integrands are uniformly bounded by $$\begin{cases}1,&-1\le x\le 1\\\exp(-x^2), &x<-1\text{ or }1<x\end{cases},$$ which is even a Schwartz function, we have by Lebesgue that $$\lim_{n\to\infty} \int_{-\infty}^{\infty} e^{-x^{2n}} \,\mathrm dx=\int_{-1}^11\,\mathrm dx=2.$$