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I was looking here: How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$? to find out how to evaluate $\displaystyle\int\limits_{-\infty}^{\infty} e^{-x^2} dx$, and didn't understand why $$\left(\displaystyle\int\limits_{-\infty}^{\infty} e^{-x^2} dx\right)\left(\int\limits_{-\infty}^{\infty} e^{-y^2} dy\right)=\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx\thinspace dy$$ I know you can use Fubini's Theorem from this: Why does $\left(\int_{-\infty}^{\infty}e^{-t^2} dt \right)^2= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2 + y^2)}dx\,dy$?, but I'm still confused about how exactly you can just multiply two integrals together like that.

A detailed answer would be very nice.

Thanks!

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  • $\begingroup$ See Fubini or Tonelli: en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli.27s_theorem $\endgroup$ – gobucksmath Jan 10 '17 at 0:50
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    $\begingroup$ Everyone cites Fubini's theorem, but it's not really needed. If you develop multivariable integration by partitioning the domain into squares, then yes it is needed. But it can also be developed as iterated single variable integration, which is how Rudin does it; this is basically how they initially teach you to integrate multiple variables in calculus. That is, to integrate $\int\int f(x,y)dydx$ hold $x$ constant and integrate along $y$. In this case $f(x,y)=g(x)h(y)$, and since $g(x)$ is being held constant it can be moved outside the integral. $\endgroup$ – juan arroyo Jan 10 '17 at 3:04
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$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2}e^{-y^2} dx dy$$

Because we treat $y$ constant in the first, inner, integral we can pull it out.

$$=\int_{-\infty}^{\infty} e^{-y^2} \int_{-\infty}^{\infty} e^{-x^2} dx dy$$

Now because $\int_{-\infty}^{\infty} e^{-x^2} dx$ is some constant we can pull it out,

$$=\int_{-\infty}^{\infty} e^{-x^2} dx\int_{-\infty}^{\infty} e^{-y^2} dy$$

The result we got is generalizable, given $g(x,y)=f(x)h(y)$ we have,

$$\int_{a}^{b} \int_{c}^{d} g(x,y) dx dy=\int_{a}^{b} h(y) dy \int_{c}^{d} f(x) dx$$

Provided everything we write down exists.

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  • $\begingroup$ This is frobenius, right? $\endgroup$ – Tac-Tics Jan 10 '17 at 1:16
  • $\begingroup$ Sorry, never heard that term before. @Tac-Tics $\endgroup$ – Ahmed S. Attaalla Jan 10 '17 at 1:22
  • $\begingroup$ Oops, nvm. It's fubini's theorem. $\endgroup$ – Tac-Tics Jan 10 '17 at 7:29
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Fubini-Tonelli theorem implies that if $\int_{\mathbb{R}} dx \left(\int_{\mathbb{R}} dy |f(x,y)|\right) < \infty$ then $$\int_{\mathbb{R}^2} f(x,y)dxdy = \int_{\mathbb{R}} dx \left(\int_{\mathbb{R}} dy f(x,y)\right).$$ First, to establish the premise, note that $f(x,y) = e^{-(x^2+y^2)},$ is positive so $|f(x,y)| = f(x,y).$ So we just need to establish the finiteness of the RHS. The inner integral is $$ \int_{\mathbb{R}} dy e^{-(x^2+y^2)} = e^{-x^2}\int_{\mathbb{R}} dy e^{-y^2}. $$ By comparing the integrand with $e^{-|y|}$ which can be integrated to 2 by elementary means, we see that $$\int_{\mathbb{R}} e^{-y^2}dy = C < \infty$$ so that $$\int_{\mathbb{R}} dx \left(\int_{\mathbb{R}} dy e^{-(x^2+y^2)}\right) = C\int_{\mathbb{R}} dx e^{-x^2} = C^2 < \infty.$$

Thus the theorem gives $$ \int_{\mathbb{R}^2} dxdy e^{-(x^2+y^2)} = C^2 $$ where $$ C = \int_{\mathbb{R}} dx e^{-x^2}.$$

The two-dimensional integral can then be transformed to polar coordinates by the change of variables theorem, which (by Fubini again) is a doable iterated integral that comes out to $\pi.$

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The integral $\displaystyle \int_{-\infty}^\infty e^{-y^2} \,dy$ does not change as $x$ goes from $-\infty$ to $\infty$, so you have \begin{align} \int_{-\infty}^\infty e^{-x^2} \, dx \cdot \int_{-\infty}^\infty e^{-y^2} \, dy & = \int_{-\infty}^\infty e^{-x^2} \, dx \cdot \text{constant} \\[10pt] & = \int_{-\infty}^\infty \left( e^{-x^2}\cdot\text{constant} \right)\,dx \\[10pt] & = \int_{-\infty}^\infty \left( e^{-x^2} \int_{-\infty}^\infty e^{-y^2}\,dy \right)\,dx \end{align} The factor $e^{-x^2}$ does not change as $y$ goes from $-\infty$ to $\infty$, so you now have \begin{align} \int_{-\infty}^\infty \left( e^{-x^2} \int_{-\infty}^\infty e^{-y^2}\,dy \right) \, dx &= \int_{-\infty}^\infty \left( \text{constant} \cdot \int_{-\infty}^\infty e^{-y^2}\,dy \right) \,dx \\[10pt] & = \int_{-\infty}^\infty \left( \int_{-\infty}^\infty \text{constant} \cdot e^{-y^2}\,dy \right) \,dx \\[10pt] & = \int_{-\infty}^\infty \left( \int_{-\infty}^\infty e^{-x^2} \cdot e^{-y^2}\,dy \right) \,dx \end{align}

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  • $\begingroup$ The presentation is nice. +1 $\endgroup$ – Simply Beautiful Art Jan 11 '17 at 21:31
  • $\begingroup$ +1 Always enjoy reading a Michael Hardy answer $\endgroup$ – Ovi Apr 26 '18 at 9:30
  • $\begingroup$ @Ovi : Thank you. $\endgroup$ – Michael Hardy Apr 26 '18 at 17:15
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Sorry This is not a complete answer but i hope it can help you

if we integral of dx then $e^{-y^2}$ assume constant $$\displaystyle\int\limits_{y=-\infty}^{y=\infty}\int\limits_{x=-\infty}^{x=\infty} e^{-x^2} e^{-y^2} dx\thinspace dy=\Big(\int\limits_{y=-\infty}^{y=\infty} e^{-y^2} dy\Big)\Big(\displaystyle\int\limits_{x=-\infty}^{x=\infty} e^{-x^2} dx\Big)=\thinspace\int\limits_{y=-\infty}^{y=\infty} e^{-y^2}dy \displaystyle\int\limits_{x=-\infty}^{x=\infty} e^{-x^2} dx $$

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