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

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?

• This is Fubini's theorem en.wikipedia.org/wiki/Fubini%27s_theorem and the fact that $e^{-x^2-y^2} = e^{-x^2}e^{-y^2}$ Jun 10, 2020 at 20:49
• It should be Fubini's Theorem. Jun 10, 2020 at 20:49
• Worth comparing with the distributive law. If $a_1,\dots,a_n$ and $b_1,\dots,b_m$ are two finite sequences, then $$\sum_{I=1}^n\sum_{j=1}^m a_ib_j =\left(\sum_i a_i\right)\left(\sum_j b_j\right)$$ This doesn’t work for all infinite sequences, but Fubini gives cases where it does work. Jun 10, 2020 at 21:09
• @ThomasAndrews can we conclude $\int_a^b \int_c^d f(x)g(y)dxdy= (\int_a^b f(x)dx)(\int_c^d g(y)dy)$? Jun 10, 2020 at 21:45

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 $$\iint_{S_m}e^{-x^2-y^2}\;dxdy=\int_0^m\int_0^m e^{-x^2-y^2}\;dxdy=(\int_0^{\infty} e^{-x^2}\;dx)(\int_0^{\infty} e^{-y^2}\;dy)$$.