# $\int f(x) dx$ is appearing as $\int dx f(x)$. Why? [duplicate]

This question already has an answer here:

A few of us over on MITx have noticed that $\int f(x) dx$ is appearing as $\int dx f(x)$.

It's not the maths of it that worries me. It's just I recently read a justification (analytical?) of the second form somewhere but can't recall it or where I saw it.

Can anyone give me a reference?

-

## marked as duplicate by Grigory M, amWhyMay 13 '14 at 15:13

One "justification" I've heard is that if you consider integration as a function taking, say, integrable functions to real number, then writing $\int \mathrm{d}x f$ sort of looks like "applying the function $\int \mathrm{d}x$ to $f$". I don't like the argument myself, but it's one I've heard at least. –  gspr Jun 21 '12 at 13:48
You can also think of integrating as a kind of sum over these infinitesimal products $f(x)dx$. So then interchanging the factors seems natural, i.e. then both orders are regarded as equivalent. Especially if you have many different variables and dimensions it is prudent to see at the beginning with respect to which you are integration. So you don't first have to match all integral signs with their respective $d$s. –  canaaerus Jun 21 '12 at 13:53
The second form sometimes makes it easier for the reader to match variables of integrations with their limits. Compare $$\int_0^1\int_{-\infty}^{\infty}\int_{-\eta}^{\eta}\int_{0}^{|t|} \Big\{\text{some long and complicated formula here}\Big\}\,ds\,dt\,d\zeta\,d\eta$$ and $$\int_0^1 d\eta\int_{-\infty}^{\infty}d\zeta\int_{-\eta}^{\eta}dt\int_{0}^{|t|} ds\,\Big\{\text{some long and complicated formula here}\Big\}$$