I have a question from one of my analysis courses which I am not too sure how I am supposed to answer
$"Let f(R)=\int_{-R}^{R}e^{-x^2}dx$. Find $\lim_{R\to\infty} f(x)"$
The suggestion is to represent $f^2(R)$ as a two-dimensional integral over the square $$S=\{(x,y) \in \mathbb{R^2}:|x| \leq R, |y| \leq R\},$$ but I don't understand how I am supposed to do this. It then says as part of the suggestion to show that the same function computed over the disk of radius R converges to the same limit using polar co-ordinates, which I think I know how to do but I am unsure why this is necessary in addition to the first part.
Help appreciated thanks.
We have \begin{equation} f(R)^2=\left(\int_{-R}^R e^{-x^2} dx\right)\left(\int_{-R}^R e^{-y^2} dy\right)=\int_{[-R,R]^2}e^{-(x^2+y^2)}dxdy \end{equation} Let $g(R)$ be the integral of $e^{-r^2}$ over a disk $D_R$ of radius $R$ at the origin. Let $E_R=[-R,R]\setminus D_R$. We can bound the difference: \begin{align} \left|g(r)-f(r)^2\right|&=\left|\int_{E_R} e^{-r^2} d\vec{r}\right|\\ &\leq \text{Area}(E_R) \sup_{r\in E_R}\left|e^{-r^2}\right|\\ &= (4-\pi)R^2 e^{-R^2} \to 0 \quad \text{as}\quad R\to \infty. \end{align} Computing $g(R)$ in polar coordinates isn't to difficult: \begin{equation} g(R)=2\pi\int_0^R e^{-r^2}rdr=\pi\left(1-e^{-R^2}\right)\to\pi\quad \text{as} \quad R\to\infty. \end{equation}