Is there a double integral lurking in this proof? There are many proofs that $\int_\mathbb{R}e^{-x^2}dx=\sqrt{\pi}$. What they all seem to have in common is a dependence on computing a double integral, at least in the sense of a product of two single integrals. In some cases, such as the double integral needed to prove $\Gamma(x+y)\text{B}(x,\,y)=\Gamma(x)\Gamma(y)$ before you compute $\Gamma^2(\frac{1}{2})$, the reliance on the double integral can be easily missed if you forget how you know certain theorems. (The sixth proof in the link above spells this out.)
I was going to ask if there's any reason a double integral is necessary, but the ninth proof (ibid.) makes me wonder whether it is. I'll summarise it here. Define $f(z):=\frac{\exp -z^2/2}{1+\exp -\tau z}$ with $\tau:=\sqrt{\pi}(1+i)$, so $\tau^2=2\pi i$ and $f(z+\tau)=-f(z)\exp -\tau z$. The poles of $f$ are all first-order, at $\sqrt{\pi}(1+i)(n+\frac{1}{2})$ with $n\in\mathbb{Z}$. Then $$\int_\mathbb{R}\exp-\frac{z^2}{2}dz=\int_\mathbb{R}(f(z)-f(z-\tau))dz=\int_\mathbb{R}(f(z)-f(z-i\sqrt{\pi}))dz$$can be evaluated by the residue theorem. Has this proof found a way of avoiding a double integral, or is one needed in proving certain relied-on results in complex analysis?
If I didn't know many other proofs, I'd have no reason to ask this specific question; but because I do, part of me still suspects there's some innate reason a double integral is generally needed. However, the other proofs mostly avoided complex numbers; maybe that makes all the difference. However, there is a subtle use of a double integral in the final proof (ibid.), since without knowing the Gaussian integral one would usually normalise the Fourier transform by first using a double integral to compute $\int_\mathbb{R}\operatorname{sinc}xdx$. So perhaps complex numbers don't circumvent the requirement after all.
 A: Yes, it can be done with a single integral and without complex analysis. First, we can deduce from the reduction formulas$$(\forall n\in\mathbb{N}\setminus\{1\}):\int(1-x^2)^n\,\mathrm dx=\frac{x(1-x^2)^n}{2n+1}+\frac{2n}{2n+1}\int(1-x^2)^{n-1}\,\mathrm dx$$and$$(\forall n\in\mathbb{N}\setminus\{1\}):\int\frac{\mathrm dx}{(x^2+1)^n}=\frac x{(2n-2)(x^2+1)^{n-1}}+\frac{2n-3}{2n-2}\int\frac{\mathrm dx}{(x^2+1)^{n-1}}$$that, for each $n\in\mathbb N$,$$\int_0^1(1-x^2)^n\,\mathrm dx=\frac23\times\frac45\times\cdots\times\frac{2n}{2n+1}$$and$$\int_0^\infty\frac{\mathrm dx}{(1+x^2)^n}=\frac\pi2\times\frac12\times\frac34\times\cdots\times\frac{2n-3}{2n-2}.$$Then, one can prove that$$(\forall x\in[0,1]):1-x^2\leqslant e^{-x^2}\tag1$$and that$$(\forall x\in[0,+\infty)):e^{-x^2}\leqslant\frac1{1+x^2}.\tag2$$Integrating inequalities $(1)$ and $(2)$ from $0$ to $1$ and from $0$ to $\infty$, respectively, and then using the substitution $y=\sqrt nx$, one gets\begin{align}\sqrt n\frac23\times\frac45\times\cdots\times\frac{2n}{2n+1}&\leqslant\int_0^{\sqrt n}e^{-y^2}\,\mathrm dy\\&\leqslant\int_0^{\infty}e^{-y^2}\,\mathrm dy\\&\leqslant\frac\pi2\sqrt n\frac12\times\frac34\times\cdots\times\frac{2n-3}{2n-2}.\end{align}And from this, one can deduce that$$\int_0^\infty e^{-y^2}\mathrm dy=\frac{\sqrt\pi}2.$$I didn't invent this. What I described is exercise 41 from the chapter 19 of Michael Spivak's Calculus.
