When doing nested integration, which of these is the correct notation for the order of integtation limits?

Question

If we wish to do a nested integration over a function of multiple variables, is there a (or, what is the) correct order to write the limits on the integral symbol?

For example, to integrate the function $f(x,y)$ first with respect to $x$ over the range $(x=-1$ to $x=1)$, and then secondly with respect to $y$ over the range $(y=-2$ to $y=2)$, which of the following is the correct notation:

Option 1: $$ \int_{-2}^{2} \int_{-1}^{1} f(x,y)\; dx\;dy $$

Option 2: $$ \int_{-1}^{1} \int_{-2}^{2} f(x,y)\; dx\;dy $$

Are the integration symbols "nested", starting with the inner ones and working outwards (as in option 1)? Or, do they maintain the order as the differential operators (as in option 2, i.e the $x$ one comes first, then the $y$ one).

Answer 1

Option 1 is the only convention I know (except the one kimchi wrote in the comments, but this is non of your two options and only used by physicists, as far as I know). When you write $$ \int_{-2}^{2} \int_{-1}^{1} f(x,y)\; dx\;dy $$ you can imagine brackets like in $$ \int_{-2}^{2}\left( \int_{-1}^{1} f(x,y)\; dx\right)\;dy $$ which are usually not written.

Answer 2

Short Answer: Integrals are nested; the standard, natural understanding is "option 1".

There are two common ways of thinking of the integral. The first is to think of integration as an uncountable summation procedure (i.e. as a Riemann sum, taken to the continuum limit); the second is to think of integration as a linear operator which acts on spaces of functions.

Integration as Summation

Heuristically, an integral (specifically, a Riemann integral) represents an area, which is obtained by adding up the areas of an infinite number of infinitesimally narrow rectangles. You might read the integral $$ \int_{a}^{b} f(x) \,\mathrm{d}x $$ as "add up the areas of a collection of infinitesimal rectangles of with $\mathrm{d}x$ and height $f(x)$, where $x$ ranges from $a$ to $b$". For any fixed $x$, the expression $f(x)\,\mathrm{d}x$ is the area of one rectangle, and the symbols $\int_{a}^{b}$ are just telling us which values of $x$ to consider.

An iterated integral in two variables resents a sum of infinitesimal volumes, rather than areas. Each volume is a prism of infinitesimal height (or thickness, if you prefer), with a base whose area is that enclosed between the $xy$-plane on the bottom, and the graph of $f$ on the top. In the following picture, the "height" is the infinitesimal $\mathrm{d}x$, while the area of the base is given by the area under the graph of $f(x,y)$, where $x$ is fixed and $y$ is allowed to vary.

To get the total volume, we have to add up all of these infinitesimal volumes, which is another integral:

$$ \int_{c}^{d} \underbrace{\left( \int_{a}^{b} f(x,y)\,\mathrm{d}y \right)}_{ \text{"base area"}} \,\underbrace{\mathrm{d}x}_{\text{"height"}}.$$

The order in which to write the integrals "makes sense" here, as we first need to compute the area of the base of an infinitesimally narrow region (the integral with respect to $y$ above), then add up all of those little regions (the integral with respect to $x$). Note that I have used parentheses here in order to highlight the fact that, in this context, the inner integral—including the $\mathrm{d}y$ term!—is a single value, representing an area.

Because there is actually no ambiguity here, the parentheses are almost always omitted. Indeed, you may think of $\int_{a}^{b}$ and $\mathrm{d}y$ as left- and right-delimiters, respectively. That is, $\int_{a}^{b}$ plays the same role as a left bracket or left parenthesis, while $\mathrm{d}y$ plays the same role as a right bracket or right parenthesis.

Again, note that the above is a heuristic, though it can be made rigorous.

Integration as an Operator

Motivated by kimchi lover's comment, another way of thinking of the integral is as an operator which eats a function and spits out a number. This is often how physicists think of things (though it shows up in functional analysis, as well), and it motivates the notation

$$ \int_{a}^{b} \mathrm{d}x\, f(x). $$

Here, the entire collection $\int_{a}^{b} \mathrm{d}x$ is one symbol, which denotes the integration of the function $f$ over the interval $[a,b]$ with respect to the variable $x$. This may be thought of as an analog to the differential notation

$$ D f(x) \qquad\text{or}\qquad \frac{\mathrm{d}}{\mathrm{d}x} f(x); $$

either $D$ or $\frac{\mathrm{d}}{\mathrm{d}x}$ is notation for an operator which eats functions. Technically, $D$ spits out another function (the derivative of $f$), so the indefinite integral gives a more appropriate analog:

$$ \int\mathrm{d}x f(x). $$

In any event, the notation

$$ \int_{c}^{d} \mathrm{d}x \int_{a}^{b}\mathrm{d}y\, f(x,y) = \int_{c}^{d} \mathrm{d}x \left( \int_{a}^{b}\mathrm{d}y\, f(x,y) \right) $$

represents the composition of two such operators: first, integrate with respect to $y$, then integrate with respect to $x$. Again, the order of integration is completely unambiguous, so the parentheses are typically omitted (as is done on the left-hand side).

Addendum

If there is any worry that the order of integration is going to be ambiguous, a little bit of extra notation goes a long way. While standard notation(s) outlined above are completely unambiguous to most (all?) working mathematicians, it is sometimes helpful to indicate the variable of integration in the limits of the integral. For example, instead of writing

$$ \int_{c}^{d} \int_{a}^{b} f(x,y)\,\mathrm{d}y \,\mathrm{d}x, $$

one could write

$$ \int_{x=c}^{d} \int_{y=a}^{b} f(x,y)\,\mathrm{d}y \,\mathrm{d}x. $$

This little bit of extra notation should resolve any lingering doubts.