How to prove $\int_0^\infty\int_0^\infty e^{-x^2 -y^2}dxdy= (\int_0^\infty e^{-x^2}dx)(\int_0^\infty e^{-y^2}dy)$?

Question

Why this statement is true? $$\int_0^\infty\int_0^\infty e^{-x^2 -y^2}dxdy= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy\right)$$ does it have any name or something? how can I use this method in other integrals?

Answer 1

Theorem: Let R be unbounded, if there are compact Jordan-measruable sets $R_n$ such that: $R_1\subset R_2\subset ... \subset R$ and any compact set $S\subset R_N$ for some N; additionally, let $f(x,y)$ be continuous in each $R_n$ and assume$\iint_{R_n} |f(x,y)|dxdy\lt \mu\lt \infty$ for every n, then $\lim_{n\to \infty}\iint_{R_n}f(x,y)dxdy$ exists and is independent of the choice of $\{R_n\}$.

Proof:

Define a sequence by $A_n=\iint_{R_n}|f(x,y)|dxdy$. Since this sequence is strict increasing and, by assumption, bounded above, it has a limit. In other words, $\{A_n\}$ is a Cauchy sequence. Then for m, n sufficiently large with $m\lt n$, $$|\iint_{R_n}f(x,y)dxdy-\iint_{R_m}f(x,y)dxdy|=|\iint_{R_n-R_m}f(x,y)dxdy|\le\iint_{R_n-R_m}|f(x,y)|dxdy=\iint_{R_n}|f(x,y)|dxdy-\iint_{R_m}|f(x,y)|dxdy=A_n-A_m.$$ This final expression may be made as small as we please, and thus so does $|\iint_{R_n}f(x,y)dxdy-\iint_{R_m}f(x,y)dxdy|$, which tells us $\lim_{n\to \infty}\iint_{R_n}f(x,y)dxdy$ exists.

For a fixed m, let $S_m$ be any closed measurable subset of $R_m$ in which $f$ is continuous for some m. $S_m\subset R_{N_m}$ for some $N_m$, so for $n\ge N_m, S_m\cap R_n=S_m$. Therefore, $\lim_{n\to\infty}\iint_{S_m\cap R_n} fdxdy=\iint_{S_m} fdxdy\;(*)$, and by replacing $f$ by $|f|$ we get $\iint_{S_m} |f|dxdy\le\lim_{n\to\infty}\iint_{R_n} |f|dxdy\le \mu$.

We claim that the rate of convergence of $(*)$ is independent of S. Pick $N\gt N_m\gt n$ with n sufficiently large, then $|\iint_{S_m} fdxdy-\iint_{S_m\cap R_n} fdxdy|\le |\iint_{S_m\cap R_N} fdxdy-\iint_{S_m\cap R_n} fdxdy|$$=|\iint_{S_m\cap (R_N-R_n)} fdxdy|\le \iint_{R_N}|f|dxdy-\iint_{R_n}|f|dxdy$, which may be made as small as we please independently of $S_m$, and thus the claim is proven.

Now let $\{S_m\}$ be a sequence of subset each satisfying the same properties of $S_m$ from before (with some trivial modification). Suppose $A=\lim_{n\to\infty}\iint_{R_n}f(x,y)dxdy$ and $B=\lim_{m\to\infty}\iint_{S_m}f(x,y)dxdy$. We have $|A-B|\le |A-\iint_{R_n\cap S_m}f(x,y)dxdy|+|\iint_{R_n\cap S_m}f(x,y)dxdy-B|$ and $$|B-\iint_{R_n\cap S_m}f(x,y)dxdy|\le |B-\iint_{S_m}f(x,y)dxdy|+|\iint_{S_m}f(x,y)dxdy-\iint_{R_n\cap S_m}f(x,y)dxdy|.$$

By interchanging the roles of $R_n$ and $S_m$, a similar inequality may be acquired. Hence, $\forall \varepsilon \gt 0, \exists N_{\varepsilon}$ such that $|B-\iint_{S_m}f(x,y)dxdy|\lt \varepsilon$ and $|\iint_{S_m}f(x,y)dxdy-\iint_{R_n\cap S_m}f(x,y)dxdy|$ for all $m\gt N_{\varepsilon}$, and such that $|A-\iint_{R_n}f(x,y)dxdy|\lt \varepsilon$ and $|\iint_{R_n}f(x,y)dxdy-\iint_{R_n\cap S_m}f(x,y)dxdy|$ for all $n\gt N_{\varepsilon}$. Thus, by the above inequalities, we have $$|A-B|\le |A-\iint_{R_n\cap S_m}f(x,y)dxdy|+|\iint_{R_n\cap S_m}f(x,y)dxdy-B|\lt 4\varepsilon$$

Since $\varepsilon$ is arbitrary, $A=B$ and therefore the limit in the theorem does not depend on the particular choice of the sequence of subsets.

Let $R_n=\{x^2+y^2\le n^2, x\ge 0,y\ge 0\}$, then by changing to polar coordinates we have $I_n=\iint_{R_n} e^{-x^2-y^2}\;dxdy=\int_0^{\pi/2}(\int_0^n re^{-r^2} dr)d{\theta}=\dfrac{\pi}2(\dfrac12-\dfrac{e^{-n^2}}2)$, which is less than $ \dfrac{\pi}4$ for every n. If we let R be the entire first quadrant of the x-y plane, we see that $R_1\subset R_2 \subset \;... \subset R$. Here we may apply the theorem as all its conditions are satisfied. Notice that another possible choice for the sequence is $\{S_m\}$, defined by $S_m=[0,m]\times [0,m]$. Then by Fubini's theorem and the theorem above, $$\lim_{n\to\infty}I_n=\lim_{m\to\infty} \iint_{S_m}e^{-x^2-y^2}\;dxdy=\lim_{m\to\infty}(\int_0^m e^{-x^2}\;dx)(\int_0^{m} e^{-y^2}dy)=(\int_0^{\infty} e^{-x^2}\;dx)(\int_0^{\infty} e^{-y^2}\;dy).$$

Since $I_n$ converges, so does $\int_0^{\infty} e^{-x^2}\;dx$, and thus $$\iint_{S_m}e^{-x^2-y^2}\;dxdy=\int_0^m\int_0^m e^{-x^2-y^2}\;dxdy=(\int_0^{\infty} e^{-x^2}\;dx)(\int_0^{\infty} e^{-y^2}\;dy)$$.