Why is $(\int dx)^2 = \int dx \int dy$ 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.
 A: 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
$$
G(a)^2 = \int_{-\infty}^\infty \mathrm dx \int_{-\infty}^\infty \mathrm dy\  f(x,a)f(y,a).
$$
A: If you write it in terms of Riemann's integral definition, it boils down to:
$$(\sum_i a_i)(\sum_j b_j)=\sum_{ij}a_ib_j$$
