How can I show that these two integrals are equal? How can I show that there is an equality
$$
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2-y^2}dxdy=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-x^2}dx\right)?
$$
 A: 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.
A: 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.
A: $$\int_{ - \infty }^\infty  {\int_{ - \infty }^\infty  {e^{ - x^2  - y^2 } dx} dy}  = \int_{ - \infty }^\infty  {e^{ - y^2 } \bigg(\int_{ - \infty }^\infty  {e^{ - x^2 } dx} \bigg)dy}  = \bigg(\int_{ - \infty }^\infty  {e^{ - x^2 } dx} \bigg)\bigg(\int_{ - \infty }^\infty  {e^{ - y^2 } dy} \bigg).$$
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.
