I am reading "Analysis on Manifolds" by James R. Munkres.
- (a) Show that $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=[\int_{\mathbb{R}} e^{-x^2}]^2,$$ provided the first of these integrals exists.
I don't understand what answer the author expects.
I showed $$\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\left[\int_{\mathbb{R}} e^{-x^2}\right]^2$$ holds as follows.
Theorem 13.6. Let $S$ be a bounded set in $\mathbb{R}^n$; let $f:S\to\mathbb{R}$ be a bounded continuous function; let $A=\operatorname{Int}S$. If $f$ is integrable over $S$, then $f$ is integrable over $A$, and $\int_S f=\int_A f$.
Theorem 15.6. Let $A$ be open in $\mathbb{R}^n$; let $f:A\to\mathbb{R}$ be continuous. Let $U_1\subset U_2\subset\cdots$ be a sequence of open sets whose union is $A$. Then $\int_A f$ exists if and only if the sequence $\int_{U_n} |f|$ exists and is bounded; in this case $$\int_A f=\lim_{N\to\infty}\int_{U_N} f.$$
My solution:
Let $U_n:=(-n,n)\times (-n,n)$.
Let $Q_n:=[-n,n]\times [-n,n]$.
Then, $U_n=\operatorname{Int}Q_n$ and $\int_{Q_n}e^{-(x^2+y^2)}$ exists.
By Theorem 13.6 above, $\int_{U_n}e^{-(x^2+y^2)}$ exists and $\int_{Q_n}e^{-(x^2+y^2)}=\int_{U_n}e^{-(x^2+y^2)}$ holds.
By Fubini's theorem, $$\int_{Q_n}e^{-(x^2+y^2)}=\int_{x\in [-n,n]}e^{-x^2}\cdot \int_{y\in [-n,n]}e^{-y^2}=\left[\int_{x\in [-n,n]}e^{-x^2}\right]^2.$$
By Theorem 13.6 above, $\int_{[-n,n]}e^{-x^2}=\int_{(-n,n)}e^{-x^2}$.
For all $x\in\mathbb{R}$, $1+x\leq e^x$ holds. (I don't prove this. But it is easy to prove this.)
So, if $x>-1$ then, $e^{-x}\leq\frac{1}{1+x}$ holds.
So, $e^{-x^2}\leq\frac{1}{1+x^2}$ holds for all $x\in\mathbb{R}$.
So, $\int_{(-n,n)}e^{-x^2}=\int_{[-n,n]}e^{-x^2}\leq\int_{[-n,n]}\frac{1}{1+x^2}=2\arctan n\leq\lim_{n\to\infty}2\arctan n=\pi$ for any $n\in\{1,2,\dots\}$.
So, by Theorem 15.6 above, $\int_{\mathbb{R}}e^{-x^2}$ exists and $\int_{\mathbb{R}}e^{-x^2}=\lim_{n\to\infty}\int_{(-n,n)}e^{-x^2}$.
And $$\int_{U_n}e^{-(x^2+y^2)}=\int_{Q_n}e^{-(x^2+y^2)}=\left[\int_{x\in [-n,n]}e^{-x^2}\right]^2\leq\pi^2$$ for any $n\in\{1,2,\dots\}$.
So, by Theorem 15.6 above, $\int_{\mathbb{R}^2}e^{-(x^2+y^2)}$ exists and $\int_{\mathbb{R}^2}e^{-(x^2+y^2)}=\lim_{n\to\infty}\int_{U_n}e^{-(x^2+y^2)}$.
From the equation $$\int_{U_n}e^{-(x^2+y^2)}=\left[\int_{(-n,n)}e^{-x^2}\right]^2,$$ $$\int_{\mathbb{R}^2}e^{-(x^2+y^2)}=\left[\int_{\mathbb{R}}e^{-x^2}\right]^2$$ follows.
I proved both of $\int_{\mathbb{R}^2}e^{-(x^2+y^2)}$ and $\int_{\mathbb{R}}e^{-x^2}$ exist and $\int_{\mathbb{R}^2}e^{-(x^2+y^2)}=\left[\int_{\mathbb{R}}e^{-x^2}\right]^2$ holds.
I don't understand what answer the author expects.
The author says "you do not need to prove that $\int_{\mathbb{R}^2}e^{-(x^2+y^2)}$ exists but you need to prove that $\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=[\int_{\mathbb{R}} e^{-x^2}]^2$ holds if $\int_{\mathbb{R}^2}e^{-(x^2+y^2)}$ exists.".
We don't need to evaluate the value of $\int_{(-n,n)}e^{-x^2}$ in the following answer.
So, the following answer is easier.
If $\int_{\mathbb{R}^2} e^{-(x^2+y^2)}$ exists, then by Theorem 15.6, the sequence $\int_{U_n} |e^{-(x^2+y^2)}|=\int_{U_n} e^{-(x^2+y^2)}$ exists (this is obvious directly) and is bounded and $\int_{\mathbb{R}^2} e^{-(x^2+y^2)}=\lim_{n\to\infty}\int_{U_n} e^{-(x^2+y^2)}$.
By Fubini's theorem and Theorem 13.6, $$\int_{U_n}e^{-(x^2+y^2)}=\int_{Q_n}e^{-(x^2+y^2)}=\int_{x\in [-n,n]}e^{-x^2}\cdot \int_{y\in [-n,n]}e^{-y^2}=\left[\int_{x\in [-n,n]}e^{-x^2}\right]^2=\left[\int_{x\in (-n,n)}e^{-x^2}\right]^2.$$
So $$\lim_{n\to\infty}\int_{x\in (-n,n)}|e^{-x^2}|=\lim_{n\to\infty}\int_{x\in (-n,n)}e^{-x^2}=\lim_{n\to\infty}\sqrt{\int_{U_n} e^{-(x^2+y^2)}}=\sqrt{\int_{\mathbb{R}^2} e^{-(x^2+y^2)}}.$$
So $\int_{x\in (-n,n)}|e^{-x^2}|$ exists (this is obvious) and is bounded and $\int_{\mathbb{R}}e^{-x^2}=\lim_{n\to\infty}\int_{x\in (-n,n)}e^{-x^2}=\sqrt{\int_{\mathbb{R}^2} e^{-(x^2+y^2)}}$.
So $$\left[\int_{\mathbb{R}}e^{-x^2}\right]^2=\int_{\mathbb{R}^2} e^{-(x^2+y^2)}.$$