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I've been making an effort to type all function names and operators in roman font. For example $$\int \operatorname{f}(x) \, \operatorname{d}\!x$$

This was all well and good until I tried to write out the rule for integration by parts. The usual formula comes from the product rule. If $\operatorname{u}$ and $\operatorname{v}$ are functions of $x$ then $$\frac{\operatorname{d}}{\operatorname{d}\!x}(\operatorname{uv}) = \frac{\operatorname{du}}{\operatorname{d}\!x}\operatorname{v}+\operatorname{u}\frac{\operatorname{dv}}{\operatorname{d}\!x}$$

Indeed, we can even write the rule for integration by parts:

$$\int \operatorname{u}\frac{\operatorname{dv}}{\operatorname{d}\!x}\, \operatorname{d}\!x = \operatorname{uv}-\int\operatorname{v}\frac{\operatorname{du}}{\operatorname{d}\!x}\, \operatorname{d}\!x$$

The problem comes when we try to write this in the form that I know it, i.e. by "cancelling the $\operatorname{d}\!x$". Without using any roman letters at all, I need to write:

$$\int u \, dv = uv-\int v\, du$$ It's tempting to write the first integrand as $\operatorname{u} \, \operatorname{d}\!v$ and the second as $\operatorname{v} \, \operatorname{d}\!u$. However, both $u$ and $v$ and $\operatorname{u}$ and $\operatorname{v}$ appear, meaning that they are sometimes functions and sometimes variables.

Does this mean that the roman system is doomed, or that the short-hand version is nonsensical? I always thought it held as an expression in terms of differential forms.

Fly by Night
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    “I've been making an effort to type all function names and operators in roman font.” -- Why? – Michael Hoppe Oct 08 '13 at 16:21
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    Yes, why would you want to type all function names and operators in roman font? This goes against the convention that nearly everyone else uses, and if you do that you should have a good reason. – Stefan Smith Oct 08 '13 at 16:25
  • This is hardly an appropriate question for this forum? – copper.hat Oct 08 '13 at 16:38
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    I make no attempts to make $f(x)$ or the $d$ in $dx$ upright, and I generally use common convention, as I got my habits from imitating what I see. I can add no more to the discussion except to cite a ISO/31XI which I am quite sure you will find interesting. Their standard (good but not one I really care about) is only things like variables should be italicized, and operators (like $d$) should be upright. Also see ...typeset.... – J. W. Perry Oct 08 '13 at 16:41
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    @copper.hat For the first 10 minutes after reading this I thought the same. On afterthought, at worst it belongs in tex.SE. After poking around a bit though, this question of upright and italic conventions definitely fits the (notation) tag. More than that, after a bit of digging it is apparently a subject for discussion and debate, although I have no interest in participating in that debate as I usually just root for the winning team in these cases. – J. W. Perry Oct 08 '13 at 16:47
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    Only well-known function names (such as sine ro cosine) should be typeset in roman (and are so by TeX makros, for example). With "throw-away" variably named functions, please stick to the traditional italics typesetting – Hagen von Eitzen Oct 08 '13 at 16:47
  • @HagenvonEitzen Literature published by several British universities, including Cambridge, use Roman for all functions; known or unknown. That's where I got the habit from. – Fly by Night Oct 08 '13 at 16:57
  • @MichaelHoppe To be consistent. – Fly by Night Oct 08 '13 at 16:57
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    None of these comments relate to my question. If you disagree with the style guide that I'm using then that's fine. But my question was not "do you agree with my style guide"! – Fly by Night Oct 08 '13 at 16:59
  • @copper.hat The question relates to the mathematical validity of an alternative form of a well-known rule. I would say that that is a mathematical question. – Fly by Night Oct 08 '13 at 17:02
  • @FlybyNight I am not sure I disagreed with anything you said. In fact, I just verified your mention of the british universities conventions. See examples where some integration occurs (upright $\operatorname{d}$ often). It appears that they are trying to adhere to the ISO/31XI standards. I am concluding that you are trying to do the same. It turns out that those standards are quite legitimate but frequently ignored, and the reason is probably just convenience (or laziness). – J. W. Perry Oct 08 '13 at 17:05
  • @FlybyNight. “That's where I got the habit from.” “De gustibus non est disputandum.” Why? Either you have them or not. – Michael Hoppe Oct 08 '13 at 17:17
  • @FlybyNight: You are asking for an opinion on the 'Roman system'. Basically asking for fashion advice. Hardly mathematics. – copper.hat Oct 08 '13 at 17:19

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To answer your first question: Not at all, as far as I'm concerned!

In fact, it can be argued that your method, which would yield:

$$\int {\rm u \, dv} = {\rm uv} - \int {\rm v\,du}$$

draws attention to the fundamental difference between the natures of both $\rm d$s.

For, the $\rm d$ in ${\rm d}x$ is not to be separated from the $x$. It is just a notation that makes the $\rm d$ look like an operator, while it is not. We could just as well have used a symbol like $\mu_x$ or something else expressing "an infinitesimal quantity in the variable $x$".

On the other hand, the $\rm d$ in $\rm du$ is of a fundamentally different nature: it is, in the language of differential geometry, an operation ${\rm d}: \wedge^0 T^*\Bbb R\to \wedge^1 T^*\Bbb R$. As such, it takes functions (like $\rm u$) to so-called $1$-forms (roughly "things with respect to which you can integrate"). In the one-dimensional case, this proceeds via the well-known rule $ {\rm du} = \dfrac{\rm du}{{\rm d}x}{\rm d}x$ (where both expressions on the right-hand side cannot be decomposed in any way). Any course or book on differential geometry will explain how to generalise these concepts from $\Bbb R$ to sufficiently well-behaved manifolds.

The notation for the expression defining $\rm du$ is thus certainly very convenient and reminiscent of ordinary numbers. Nonetheless, $\rm du$ and ${\rm d}x$ are (or should be) conceptually distinct, and I think your notation does a nice job of reflecting this distinction.

The supposed defect of the notation is thus not due to the notation itself, but due to the obfuscation of the distinction between the operator $\rm d$ and the use of $\rm d$ in denoting the basis element ${\rm d}x$ of $T^*\Bbb R$ that is very ubiquitous in mathematical literature.

Lord_Farin
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