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There are several ways of proving $I=\int_{-\infty}^{\infty} \mathrm{e}^{-x^2}\, dx = \sqrt \pi$, of which one of the ways is to consider $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$ 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: $$\int_{(x,y)\in[a,b]\times [c,d]} f(x,y),dA=\int_{x=a}^b\left(\int_{y=c}^df(x,y),dy\right),dx?$$ 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 Oct 21 '16 at 10:26
  • Anyway, that simplest case of Fubini is all you need to show that the improper integral $$\int_{-\infty}^\infty e^{-x^2},dx=\sqrt\pi.$$ – Jyrki Lahtonen Oct 21 '16 at 10:27
  • @JyrkiLahtonen: My question is asking why product of two integrals (not iterated integral) equals the double integral. –  Oct 21 '16 at 10:31
  • Can someone (preferably the people who closed this question) explain why this is a duplicate? –  Oct 21 '16 at 10:33
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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, $$\biggl(\int_a^b f(x),dx\biggr) \biggl(\int_c^d g(y),dy\biggr) = \int_c^d \biggl(\int_a^b f(x),dx\biggr) g(y),dy = \int_c^d \biggl( \int_a^b f(x) g(y),dx\biggr)dy.$$ – Daniel Fischer Oct 21 '16 at 10:38
  • @DanielFischer: That's it! This together with Jyrki Lahtonen's answer is sufficient. –  Oct 21 '16 at 10:44

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