Problem with a solution to the integral $\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{dx}$ I am an undergrad in my first year of college. Today, our mathematics professor solved the integral $\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{dx}$ which he called "One of the most important integrals in all of mathematics". Since our topic of study for the day was 'Applications of change of variables in double integrals', this is how he solved the problem:
$$Define\;f:[0,\infty]\to\mathbb{R}\;by\;f(R)=\int_{-R}^{+R}e^{-x^2}\mathrm{dx}\\
\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{dx}=\lim_{R\to\infty}f(R)=I\;(say)\\
Assume\;this\;limit\;exists\;for\;now,\;then\;I^2=\lim_{R\to\infty}f(R).\lim_{R\to\infty}f(R)=\lim_{R\to\infty}f(R)^2\\
f(R)^2=\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{dx}\int_{-\infty}^{+\infty}e^{-y^2}\mathrm{dy}\\
=\iint_{[-R,R]\mathrm{x}[-R,R]}e^{-(x^2+y^2)}\mathrm{dx}\mathrm{dy}\\
So\;basically\;we\;have\;to\;calculate\;this\;integral\;over\;a\;rectangle\;given\;[-R,R]\mathrm{x}[-R,R]$$
Now this next step is the one that not only flummoxed me but also seemed a trifle ad-hoc. Our professor says that as $R\to\infty$, it doesn't matter whether we are integrating over a rectangle or a disc as both areas will tend to $\infty$and therefore he changed the region of integration from a square of side 'R' to a disc of radius 'R'.
$$\iint_{[-R,R]\mathrm{x}[-R,R]}e^{-(x^2+y^2)}\mathrm{dx}\mathrm{dy}=\iint_{D}e^{-R^2}.R\;\mathrm{dR}\mathrm{d\theta}\\(where\;D\;is\;the\;region\;defined\;by\;a\;disc\;with\;radius\;R)\\
=\int_{0}^{2\pi}\left[\int_{0}^{\infty}R.e^{-R^2}\;\mathrm{dR}\right]\mathrm{d\theta}\\\implies\;I^2=\pi\\\implies I=\sqrt{\pi}\\\therefore\\\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{dx}=\sqrt{\pi}$$
Is there any other method to calculate the above integral. Also could you tell me the 'importance' of this integral.
P.S: The class actually broke out in laughter upon seeing the change of region step, so it's clear I was not the only one who was staring in disbelief today, just FYI.
 A: Let $I(R)$ denote the value of the integral on the disk $D_R$ centered at $(0,0)$ with radius $R$ and $J(R)$ denote the value of the integral on the square $S_R=[-R,R]\times[-R,R]$. The function one integrates is positive everywhere and, for every $R$, $$D_R\subset S_R\subset D_{\sqrt2R},$$ hence $$I(R)\leqslant J(R)\leqslant I(\sqrt2R).$$
In particular, $I(R)$ has a limit when $R\to\infty$ if and only if $J(R)$ has a limit when $R\to\infty$, and the limits, if they exist, coincide.
A: How about the following?
As integrals over $\mathbb{R}^2$ the equality is just the change of variable formula so
\begin{equation}
\begin{split}
I^2&=\int_{-\infty}^\infty \int_{-\infty}^\infty\exp^{-(x^2+y^2) }dx dy=
\int_0^{2\pi}\int_0^\infty \exp^{-r^2 }r dr d\theta
=2\pi\lim_{R\rightarrow\infty}\int_0^R\exp^{-r^2 } r dr \\ &=-\pi\lim_{R\rightarrow\infty} \left (\exp(-R^2) -1\right)=\pi
\end{split}
\end{equation}
A: The way your professor derived this result is the standard method, but some others are shown here. As for the importance, Gaussian integrals show up in many areas of math and physics (probability, quantum mechanics, statistical mechanics, and quantum field theory come to mind).
A: Your professor (or your notes) are somehow confused.
Even the laziest mathematician does not write down an obviously wrong statement like this
$$\iint_{[-R,R]\mathrm{x}[-R,R]}e^{-(x^2+y^2)}\mathrm{dx}\mathrm{dy}=\iint_{D}e^{-R^2}.R\;\mathrm{dR}\mathrm{d\theta}\\
$$
make it an $\approx$, or put limits on both sides or whatever.
