Evaluating $\lim_{n\to \infty}\int^n_{-n}e^{-x^2}dx$

Question

How do I evaluate the following limit/integral:

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

Answer 1

$$\int_{-n}^ne^{-x^2}dx=\int_{\mathbb R}\chi_{[-n,n]}e^{-x^2}dx$$ Notice that $\chi_{[-n,n]}e^{-x^2}\leq e^{-x^2},\forall x\in\mathbb R$, i.e. the function $f_n(x)=\chi_{[-n,n]}e^{-x^2}$ is dominated by the function $f(x)=e^{-x^2}$, which is integrable, since $\int_{\mathbb R}e^{-x^2}dx=\sqrt{\pi}<\infty$ (in other words $f\in L^1)$.

From the Dominated Convergence Theorem: $$\lim\limits_{n\rightarrow\infty}\int_{-n}^ne^{-x^2}dx=\lim\limits_{n\rightarrow\infty }\int_{\mathbb R}\chi_{[-n,n]}e^{-x^2}dx$$$$=\lim\limits_{n\rightarrow\infty }\int_{\mathbb R}f_n(x)dx=\int_{\mathbb R}\lim\limits_{n\rightarrow\infty }f_n(x)dx$$ $$=\int_{\mathbb R}f(x)dx=\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt {\pi}$$