TITLE: Why does combining $\int_{-\infty}^{\infty} e^{-x^2} dx\int_{-\infty}^{\infty} e^{-y^2} dy$ into a $dx\,dy$ integral work?
QUESTION [8 upvotes]: 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!

REPLY [3 votes]: 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}