# Multiplying two integrals becomes a double integral?

I ran by this in a textbook:

\begin{align*} I^2 & =\left(\int_{-\infty}^\infty e^{-x^2/2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2/2}dy\right)\\ & =\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2/2}e^{-y^2/2}dxdy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)/2}dxdy \end{align*}

How come this is valid? I assume it's not generally true because it doesn't seem like you can just go around mashing integrals together. Does it only work because they are independent variables? And therefore constants with respect to each other?

• Yes, that's right. Nov 25 '19 at 21:35
• you are not changing something, the first expression is exactly the same than the last one Nov 25 '19 at 21:38

Yes:\begin{align}\int_a^b\int_c^df(x)g(y)\,\mathrm dx\,\mathrm dy&=\int_a^bg(y)\int_c^df(x)\,\mathrm dx\,\mathrm dy\\&=\left(\int_a^bg(y)\,\mathrm dy\right)\left(\int_c^df(x)\,\mathrm dx\right).\end{align}