7
$\begingroup$

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)? $

  • 0
    Why did no one mention [Fubini's theorem](http://en.wikipedia.org/wiki/Fubini%27s_theorem)?2011-06-29

3 Answers 3

5

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.

  • 0
    No problem :)2011-06-29
1

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.

1

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

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.

  • 0
    Of course, the situation is much simpler when $f$ is of the form $f(x,y)=g(x)h(y)$.2011-06-29