An exercise in a quantum mechanics book:
Show that
$$ G(a) = \int_{-\infty}^\infty e^{-\frac{1}{2}ax^2} = \sqrt\frac{2\pi}{a}$$
Solution
$$G(a)^2 = \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy\space e^{-\frac{a}{2}(x^2+y^2)} = \space ...$$
Why is $G(a)^2$ equal to an integral of $dx$ and $dy$?
Edit/Note: this is a Gausian integral. This solution is also used on this wikipedia article.
Let's look at this a bit more generally. We have a function of $a$ of the form $$ G(a) = \int_{-\infty}^\infty f(x,a)\,\mathrm dx. $$ Note that the "$x$" here as a integration variable is just bound inside the integral, so equivalently we can write this as $$ G(a) = \int_{-\infty}^\infty f(\tau,a)\,\mathrm d\tau, $$ or indeed any other symbol instead of $x$ or $\tau$. This is analogous to summation indices: $$ \sum_{k=1}^3 a_k = a_1 + a_2 + a_3 = \sum_{n=1}^3 a_n, $$ it doesn't matter if the index is called $k$ or $n$, as long as it's used consistently.
Now squaring $f(a)$ yields $$ G(a)^2 = \left(\int_{-\infty}^\infty f(\tau,a)\,\mathrm d\tau\right)\left(\int_{-\infty}^\infty f(\tau,a)\,\mathrm d\tau\right). $$ We can now rename one of the indices (back) to $x$ and the other $y$: $$ G(a)^2 = \left(\int_{-\infty}^\infty f(x,a)\,\mathrm dx\right)\left(\int_{-\infty}^\infty f(y,a)\,\mathrm dy\right). $$ To go from this to a double integral, we treat one as a constant factor and use linearity of the integral: $c\int \dots\,\mathrm dx = \int c(\dots)\,\mathrm dx$. This then yields $$ G(a)^2 = \int_{-\infty}^\infty \underbrace{\left(\int_{-\infty}^\infty f(y,a)\,\mathrm dy\right)}_c f(x,a)\,\mathrm dx. $$ Again $f(x,a)$ may now be treated as a constant factor to the $\mathrm dy$-integral and we get $$ G(a)^2 = \int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(x,a)f(y,a)\,\mathrm dy\right) \,\mathrm dx. $$ This is the double integral you encountered and in a notation used mostly by physicists that treats $\int\mathrm dx$ as an operator applied from the left, this is written as $$ G(a)^2 = \int_{-\infty}^\infty \mathrm dx \int_{-\infty}^\infty \mathrm dy\ f(x,a)f(y,a). $$