Double Integral Proof of the Gaussian Integral without Fubini’s theorem

There are several ways of proving I=∫∞−∞e−x2dx=√πI=\int_{-\infty}^{\infty} \mathrm{e}^{-x^2}\, dx = \sqrt \pi, of which one of the ways is to consider
I2=(∫+∞−∞e−x2dx)(∫+∞−∞e−y2dy)=∫+∞−∞∫+∞−∞e−(x2+y2)dxdyI^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy and then convert to polar coordinates.

The step changing the product of two single variable integrals into a double integral can be justified using Fubini’s theorem, the proof of which is beyond the standard undergraduate curriculum of many universities. Is there a more elementary way to justify this step?

Clarification: This is not a duplicate of the question requiring a proof of the integral. I am asking for justification of a step used in the accepted answer without using Fubini’s theorem which is not something addressed there.

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Won’t the undergraduate calculus courses show this variant of Fubini’s theorem: ∫(x,y)∈[a,b]×[c,d]f(x,y)dA=∫bx=a(∫dy=cf(x,y)dy)dx?\int_{(x,y)\in[a,b]\times [c,d]} f(x,y)\,dA=\int_{x=a}^b\left(\int_{y=c}^df(x,y)\,dy\right)\,dx? I do explain this, and justify it by the usual slicing argument (the proof is often a bit lacking in rigor). How do you calculate 2D- (or 3D-)integrals if not as iterated integrals, i.e. by applying Fubini?
– Jyrki Lahtonen♦
2 days ago

  

 

Anyway, that simplest case of Fubini is all you need to show that the improper integral ∫∞−∞e−x2dx=√π.\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt\pi.
– Jyrki Lahtonen♦
2 days ago

  

 

@JyrkiLahtonen: My question is asking why product of two integrals (not iterated integral) equals the double integral.
– Shahab
2 days ago

  

 

Can someone (preferably the people who closed this question) explain why this is a duplicate?
– Shahab
2 days ago

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Assuming the existence of the one-dimensional integrals, simply the linearity of the integral converts the product of integrals to an iterated integral, (∫baf(x)dx)(∫dcg(y)dy)=∫dc(∫baf(x)dx)g(y)dy=∫dc(∫baf(x)g(y)dx)dy.\biggl(\int_a^b f(x)\,dx\biggr) \biggl(\int_c^d g(y)\,dy\biggr) = \int_c^d \biggl(\int_a^b f(x)\,dx\biggr) g(y)\,dy = \int_c^d \biggl( \int_a^b f(x) g(y)\,dx\biggr)dy.
– Daniel Fischer♦
2 days ago

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