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Why does $$\left(\int_{-\infty}^{\infty}e^{-t^2}dt\right)^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2 + y^2)}dx\,dy ?$$

This came up while studying Fourier analysis. What's the underlying theorem?

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    $\begingroup$ Fubini's theorem or if you prefer: Tonelli's theorem. In more words: $$\left(\int_{-\infty}^{\infty} e^{-t^2}\,dt\right)^2 = \left(\int_{-\infty}^{\infty} e^{-t^2}\,dt\right)\left(\int_{-\infty}^{\infty} e^{-\tau^2}\,d\tau\right) \stackrel{\text{Tonelli}}{=} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-t^2}e^{-\tau^2}\,dt\,d\tau.$$ $\endgroup$ – Cameron Williams Sep 8 '15 at 16:42
  • $\begingroup$ What would Fubini's theorem (or Tonelli's theorem, if you prefer) help here if not for the amazing property of the exponential function: $\exp(a+b)=\exp(a)\cdot\exp(b)$ $\endgroup$ – uniquesolution Sep 8 '15 at 16:43
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Hint. One may observe that $$ A\int_{-\infty}^{+\infty}e^{-x^2}dx=\int_{-\infty}^{+\infty}A\:e^{-x^2}dx $$ for any constant $A$. Then putting $A:=\int_{-\infty}^{+\infty}e^{-y^2}dy$ and using Fubini's theorem, which is allowed here, yields the desired result.

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$$\left(\int_a^b f(x)\mathrm dx\right)^2=\left(\int_a^b f(x)\mathrm dx\right)\left(\int_a^b f(x)\mathrm dx\right)\underset{x\ is\ a \ mute\ variable}{=}\left(\int_a^b f(x)\mathrm dx\right)\left(\int_a^b f(y)\mathrm dy\right)$$$$\underset{Fubini}{=}\int_a^b\int_a^bf(x)f(y)\mathrm dx\mathrm dy$$

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$$\left(\int_{-\infty}^{\infty}e^{-t^2}dt\right)^2 = \left(\int_{-\infty}^{\infty}e^{-t^2}dt\right) * \left(\int_{-\infty}^{\infty}e^{-u^2}du\right) $$since $u$ and $t$ varies independently we have$$\left(\int_{-\infty}^{\infty}e^{-t^2}dt\right) * \left(\int_{-\infty}^{\infty}e^{-u^2}du\right)= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(u^2 + t^2)}du\,dt$$

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