Say, we are interested in deriving
$$\int_{-\infty}^{\infty}e^{-x^2}=\sqrt{\pi}\tag{1}$$
There are many well known ways to do it, for example:
After coming across this limit $$\lim\limits_{n\to\infty} \frac{\sqrt{\pi n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = e^{-x^2}\tag{2}$$ I wonder how can we derive $(1)$ by using $(2)$ as an approximation via step functions.
Here is a picture of $n=4$:
On
My hint: This is Gauss integration that is famous. Laplace proof equal $\sqrt{\pi}$. Also the technique to calculate like this: $$I=\int_{-\infty}^{+\infty}e^{-x^2}dx$$ Squared two sides and change variable:
$$I^2=\left ( \int_{-\infty}^{+\infty}e^{-x^2}dx \right )^2=\int_{-\infty}^{+\infty}e^{-x^2}dx\int_{-\infty}^{+\infty}e^{-y^2}dy=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}dxdy$$
$dxdy$ is the differential surface area in $Oxy$, integrating domain is the whole plane should we switch to polar coordinates: $$\left\{\begin{matrix} x=r \cos\varphi\\ y=r\sin\varphi \end{matrix}\right.$$ In which $$\left\{\begin{matrix} \varphi: 0 \rightarrow 2\pi\\ r: 0 \rightarrow +\infty \end{matrix}\right.$$
$$I^2=\int_{0}^{+\infty}\int_{0}^{2\pi}e^{-r^2}rdrd\varphi=\int_{0}^{+\infty}e^{-r^2}rdr\int_{0}^{2\pi}d\varphi=\frac{1}{2}.2\pi=\pi$$
Hence $$I=\sqrt{\pi}$$
For every $n\geqslant1$, consider the function $u_n$ defined by $$ u_n(x)=\frac{\sqrt{\pi n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor}. $$ It happens that, for every $n$, $$ \int_\mathbb Ru_n(x)\,\mathrm dx=\sqrt\pi. $$ To see this, note that $u_n(x)=0$ for every $x\lt-\sqrt{n}$ and every $x\geqslant\sqrt{n}$ and that $u_n(x)=u_n(k/\sqrt{n})$ for every $x$ in the interval $[k/\sqrt{n},(k+1)/\sqrt{n})$, for every integer $-n\leqslant k\leqslant n$. The length of each of these $2n+1$ intervals is $1/\sqrt{n}$, hence $$ \int_\mathbb Ru_n(x)\,\mathrm dx=\sum_{k=-n}^nu_n(k/\sqrt{n})/\sqrt{n}=\sqrt{\pi}\,2^{-2n}\sum_{k=-n}^n{2n\choose n+k}=\sqrt{\pi}\,2^{-2n}\sum_{i=0}^{2n}{2n\choose i}. $$ By the binomial theorem, the last sum on the RHS is $2^{2n}$ hence we are done.