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How can I show that there is an equality $$ \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2-y^2}dxdy=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-x^2}dx\right)? $$

Anna Lytics
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  • You can change the variable of integration in one of your integrals on the right-hand side from x to, e.g., y. Then $e^{-y^2}$ is a constant with respect to x. – gary Jun 29 '11 at 16:41
  • Are you asking for the formal manipulations to go from one side to the other side? Or are you asking for the mathematical justification for the steps of the manipulations? – SteveH Jun 29 '11 at 16:54
  • Why did no one mention Fubini's theorem? – Rasmus Jun 29 '11 at 17:54

3 Answers3

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You can separate them as follows

$$ \begin{align*} \iint e^{-x^2 - y^2} \mathrm{d}x \mathrm{d}y & = \iint e^{-x^2} e^{-y^2} \mathrm{d}x \mathrm{d}y\\ & = \int \left ( e^{-y^2} \int e^{-x^2} \mathrm{d}x \right ) \mathrm{d}y\\ &= \left ( \int e^{-x^2} \mathrm{d}x \right ) \left( \int e^{-y^2} \mathrm{d}y \right ) \end{align*} $$

Just in case, I will mention that whenever you have a double integral of the form

$$\int_{a}^{b} \int_{c}^{d} f(x) g(y) \mathrm{d}y \mathrm{d}x$$

you can separate it as a product of two integrals

$$ \int_{a}^{b} \int_{c}^{d} f(x) g(y) \mathrm{d}y \mathrm{d}x = \left ( \int _{a}^{b} f(x) \mathrm{d}x \right ) \left ( \int _{c}^{d} g(y) \mathrm{d}y \right ) $$

in the same way as before.

Nana
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Adrián Barquero
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Hint for the most reasonable way: 1) $e^{a + b} = e^a e^b$ 2) Remember that the x and the y are just 'dummy' variables

I should also point you to an old answer by Ross, which I can only imagine is the cause of this question.

Community
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davidlowryduda
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$$\int_{ - \infty }^\infty {\int_{ - \infty }^\infty {e^{ - x^2 - y^2 } dx} dy} = \int_{ - \infty }^\infty {e^{ - y^2 } \bigg(\int_{ - \infty }^\infty {e^{ - x^2 } dx} \bigg)dy} = \bigg(\int_{ - \infty }^\infty {e^{ - x^2 } dx} \bigg)\bigg(\int_{ - \infty }^\infty {e^{ - y^2 } dy} \bigg).$$

EDIT: It may be worth noting that there is a distinction between an iterated integral and a double integral. However, for any nonnegative measurable function $f(x,y)$ on $\mathbb{R}^2$, it holds $$ \int_{\mathbb{R}\times \mathbb{R}} {f(x,y)dxdy} = \int_{ - \infty }^\infty {\bigg(\int_{ - \infty }^\infty {f(x,y)dx} \bigg)dy} = \int_{ - \infty }^\infty {\bigg(\int_{ - \infty }^\infty {f(x,y)} dy\bigg)dx} . $$ The first integral is a double integral, the last two are iterated integrals.

Shai Covo
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