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I've seen several times on this site people writing integrals like $ \int \! dxf(x) $. This seems confusing to me, especially in an iterated integral or next to a long integrand and it's nonstandard compared to my own personal experience. I'm wondering what are the reasons people use this notation swap and if it has broader support than I'm aware of.

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    $\begingroup$ fairly common in physics $\endgroup$ – Will Jagy Dec 16 '13 at 18:10
  • $\begingroup$ I've migrated this question to the main site, since it's a question about mathematical notation rather than about the website math.stackexchange.com. Remember, the meta site is not for meta mathematics, but for discussing the math.stackexchange.com site itself. $\endgroup$ – Alex Becker Dec 16 '13 at 18:12
  • $\begingroup$ Just a different convention. $\endgroup$ – copper.hat Dec 16 '13 at 18:19
  • $\begingroup$ It's allowed. So why not? $\endgroup$ – Don Larynx Dec 17 '13 at 4:29
  • $\begingroup$ It is also much closer to the notation for a sum. How often do you write a sigma with boundaries, but only reveal at the end of your expression which variables are global variables and which variable is your summation index? And yes, it is typically used in physics. $\endgroup$ – Phira Dec 24 '13 at 22:51
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When you have multiple variables, it's often not clear what bounds belong to which variable. Putting the differential at the beginning helps the reader immediately see what the appropriate integration limits for each of the variables are.

I'll add that this notation is often used by physicists, where triple integrals or even infinite dimensional integrals are often used. As an example, here is a relatively short expression taken from an old physics paper:

$$\sigma_t(E_s,E_p,T)=\int_0^T\frac{dt}{T}\int_{E_{smin}(E'_s)}^{E_s}dE'_s\int_{E_p}^{E_{pmax}(E'_p)}dE'_p\ I_e(E_s, E'_s,t)\sigma(E'_s,E'_p)I_e(E'_p, E_p,T-t)$$

Now try and write this expression with the differentials at the end. How do you know which belongs to what bounds? arguably, in this case the variables were helpfully named, but this is far from always being the case.

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    $\begingroup$ Multiple variables implies at least a double integral. How is $$\iint \operatorname{f}(r,\theta) \, \operatorname{d}\!r \, \operatorname{d}\!\theta$$ any harder to read than $$\iint \operatorname{d}\!r \, \operatorname{d}\!\theta \, \operatorname{f}(r,\theta)$$ $\endgroup$ – Fly by Night Dec 16 '13 at 18:16
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    $\begingroup$ @FlybyNight - I'll post a real example later on, you'll see why :) $\endgroup$ – nbubis Dec 16 '13 at 18:17
  • $\begingroup$ @FlybyNight Not always. It is very common to integrate over just one of many variables e.g. $\int f(x_1,\ldots,x_n)dx_5$ $\endgroup$ – icurays1 Dec 16 '13 at 18:18
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    $\begingroup$ My reading of your edited expression is as a product of separate integrals rather than as one multiple integral but I understand now how people use the notation differently so one just has to be careful with any expression they find "in the wild". I find also an easy way of fixing the problem of ambiguity on the limits of integration is to write $\int_{x=a}^{x=b} f dx$ or more succinctly $\int_{x=a}^b f dx$ whenever ambiguity may arise. $\endgroup$ – R R Dec 16 '13 at 19:01
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    $\begingroup$ @RR - true, but conventions will be conventions :) $\endgroup$ – nbubis Dec 16 '13 at 19:07
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I suspect that it's because with the "dx" at the end, you don't know what's the var of integration until you get there. Imagine that the integrand is some complicated expression involving $x$ and $t$; it's nice to know whether you're integrating with respect to $x$ or $t$ as you read the integrand. In multiple integrals, it's also nice to know in which order you're integrating so that you can see whether there's a constant (wrt the variable of integration) that can be dragged outside the integral.

All that said, I still find the notation a bit confusing at times, esp. since I tend to think of the $\int$ and the $dx$ as being like "open" and "close" parentheses, and hence used for grouping.

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  • $\begingroup$ I agree with your last comment on thinking of dx as being the closing of the operator. Incidentally I didn't realize this question had essentially been asked already, but an answer in a different thread likened the 'grouping' of $\int dx()$ as analogous to the grouping of $\frac{d}{dx}()$ as an operator which does make some sense to me. $\endgroup$ – R R Dec 16 '13 at 18:54

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