TITLE: How can I show that these two integrals are equal?
QUESTION [7 upvotes]: How can I show that there is an equality
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2-y^2}dxdy=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-x^2}dx\right)?
$$

REPLY [5 votes]: 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.