Should I \mathrm the d in my integrals?

When writing an integral, it seems like something should be done to separate the "d", as in \int f(x) dx, so as not to confuse it with a variable. I've seen it left as-is, bolded, and straightened. Even among those options there are several ways to accomplish each task; e.g., I could do a \mathrm or a \operatorname. What is the preferred method of dealing with the "d"?

\documentclass{article}
\usepackage{amsmath}
\newcommand*\diff{\mathop{}\!\mathrm{d}}
\newcommand*\Diff{\mathop{}\!\mathrm{d^#1}}
\begin{document}

\begin{align*}
\biggl(\int_{-\infty}^\infty e^{-x^2}\diff x\biggr)^2
&= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\diff x\diff y \\
&= \int_0^{2\pi}\int_0^\infty e^{-r^2}r\diff r\diff\theta                  \\
&= \int_0^{2\pi}\biggl(-{e^{-r^2}\over2}\bigg\vert_{r=0}^{r=\infty}\,\biggr)\diff\theta\\
&= \pi                                          \tag*{q.e.d.}\\
\end{align*}
%
$V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{|\mathbf{x}-\mathbf{y}|}\,\rho(\mathbf{y})\,\Diff3\mathbf{y}$

\end{document} • Herbert, are you recommending what's on the rhs? It looks odd to me, especially when I have an inline $dy/dx$. – Jim Hefferon Jun 20 '12 at 13:07
• IMO it makes sense to add a small explanation of why this solution was chosen, rather than just providing uncommented source code. – Marco Jun 20 '12 at 13:09
• if i'm not mistaken, the upright "d" is an iso standard. but it's not common practice in the u.s. (and perhaps elsewhere). certainly knuth uses -- intentionally -- an italic "d" as can be inferred from the italic correction "d" is given in the cmmi fonts, namely none. what i find peculiar in @Herbert's example is the italic "d" on the left side while upright is used on the right. in my opinion, whichever is chosen should be used consistently. – barbara beeton Jun 20 '12 at 13:19
• @JimHefferon: in inline mode I use only \mathrm{d} – user2478 Jun 21 '12 at 6:39
• Thank you, but (1) I think it should be \newcommand*\Diff{\mathop{}\!\mathrm{d}^#1}, because the n in my example should not be in roman. (2) I still find the distance between the d^n and the x too large. The x looks like a regular variable. I think I would rather write \newcommand*\Diff{\mathop{}\!\mathrm{d}^{#1}\!{#2}}, although it is a tad too close. – Jost Nov 5 '13 at 13:16

I found a TUGboat article some years ago which seems to deal with the spacing around the differential operator in the correct way (at least to me).

Example

\documentclass{article}

\makeatletter
\providecommand*{\dif}%
{\@ifnextchar^{\DIfF}{\DIfF^{}}}
\def\DIfF^#1{%
\mathop{\mathrm{\mathstrut d}}%
\nolimits^{#1}\gobblespace
}
\def\gobblespace{%
\futurelet\diffarg\opspace}
\def\opspace{%
\let\DiffSpace\!%
\ifx\diffarg(%
\let\DiffSpace\relax
\else
\ifx\diffarg$% \let\DiffSpace\relax \else \ifx\diffarg\{% \let\DiffSpace\relax \fi\fi\fi\DiffSpace} \makeatother \begin{document} \[ \int x \dif x$

\end{document}


Update

As pointed out by Enrico Gregorio and implemented by Herbert Voß, the following will do:

\documentclass{article}

\newcommand*\dif{\mathop{}\!\mathrm{d}}

\begin{document}

$\int x \dif x$

\end{document}

• Claudio Beccari later discovered that \newcommand\dif{\mathop{}\!\mathrm{d}} does the same with much less effort. – egreg Jan 28 '13 at 16:37
• According to tug.org/pipermail/texhax/2009-August/013018.html, the following by Morten Høgholm is an improved version of the large code chunk I posted: \newcommand*\dif{ \mathop{}\nobreak \mskip-\thinmuskip\nobreak \mathrm{d} } what is best of Morten's code and the code posted by @egreg ? – Svend Tveskæg Jan 28 '13 at 17:03
• It's just the same, with two redundant \nobreak that do exactly nothing, because a line break is not possible in a math formula after a mathop atom or after \mskip glue. – egreg Jan 28 '13 at 17:07
• Besides the fact that there is apparently a much shorter equivalent for the code: I believe \ifx\diffarg\[% should be \ifx\diffarg[%, – clemens Jan 28 '13 at 23:28
• @SvendTveskæg \mathop{} provides the thin space at the left when preceded by an ordinary symbol or a closing delimiter; the “d” after it inserts another thin space that's removed with \!. – egreg Sep 8 '14 at 9:19

I usually do this (which I've shamefully stolen from Niel de Beaudrap and modified):

\makeatletter \renewcommand\d{\ensuremath{%
\;\mathrm{d}#1\@ifnextchar\d{\!}{}}}
\makeatother


It renders nicely, especially with multiple integrals: • I find this wrong under many respects. The \; space is too much. The definition proposed by Herbert is certainly better. – egreg Jun 20 '12 at 13:08
• @egreg: I'm curious if there are other reasons aside from \; (perhaps you can replace it with \:, or with \mathop{}\! as in Herbet's solution) why you find the definition "wrong". As someone who is regularly doing all sorts of ad-hoc fooling around with spacing to try and better suggest logical groupings of symbols in my math typesetting, I'm interested in other people's notions of best practises. – Niel de Beaudrap Jun 20 '12 at 13:35
• I like this definition of \d (taking care of subsequent differentials). Just IMO: (1) \ensuremath is completely wrong here, (2) the space is indeed to large and \mathop{}\! gives some nice-looking result. – yo' Jun 20 '12 at 13:42
• @NieldeBeaudrap The \ensuremath is completely useless (your code didn't have it); the \; spacing is too much (\, is correct) and testing whether another \d follows should be omitted once a thin space instead of the thick space is used. – egreg Jun 20 '12 at 13:43
• @tohecz If \mathop{}\!d is used, then the spacing for subsequent differentials will be automatically added. – egreg Jun 20 '12 at 13:44

Have a look at

http://ctan.sharelatex.com/tex-archive/macros/latex/contrib/physics/physics.pdf

2.5 Derivatives

I use it and i am very happy with this package.

EDIT:

\documentclass{article}
\usepackage{amsmath}
\usepackage{physics}
\usepackage{amssymb}

\begin{document}

\begin{align}
\left(\int\limits_{-\infty}^\infty e^{-x^2} \dd{x} \right)^2
&=\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty {e^{-(x^2+y^2)}}\dd{x}\dd{y} \\
&=\int\limits_{0}^{2\pi} \int\limits_{0}^\infty e^{-r^2}r \dd{r}\dd{\theta} \\
&=\int\limits_{0}^{2\pi} {\left(\left.-\frac{e^{-r^2}}2\right|_{r=0}^{r=\infty}\right)}\dd{\theta}  \\
&=\pi
\end{align}

\begin{equation}
V(x)=-\int\limits_{\mathbb R^3} \frac G{|x-y|}\rho(y) \dd{y}
\end{equation}

\end{document} • Can you extend the answer a bit? Maybe give a minimal example and a screenshot? – Johannes_B Apr 15 '15 at 21:24
• Note that we can italicise the d's using \usepackage[italicdiff]{physics} – Mateen Ulhaq Apr 9 '17 at 5:22

Presumable you are trying to both save on typing, and to exert some consistent notation throughout you article (good idea).

If you are making a macro for infinitesimals, you might as well make a marco for a derivative and an integral with limits. Avoid single letter macros e.g. \d because they are often already defined.

\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}

\newcommand \dd  { \,\textrm d{#1}                       }   % infintesimal
\newcommand \de  { \frac{\mathrm d{#1}}{\mathrm d{#2}}   }   % first order derivative
\newcommand \intl{ \int\limits_{#1}^{#2}{#3}\dd{#4}      }   % integral with limits

\begin{document}

$$\dd x=-\dd u$$
$$y'=\de yx$$
$$\intl0\infty{f(t)}t$$

\begin{align*}
\left(\intl{-\infty}\infty{e^{-x^2}}x\right)^2
&=\intl{-\infty}\infty{\intl{-\infty}\infty{e^{-(x^2+y^2)}}x}y                      \\
&=\intl0{2\pi}{\left(\left.-\frac{e^{-r^2}}2\right|_{r=0}^{r=\infty}\right)}\theta  \\
&=\pi
\end{align*}

$$V(x)=-\intl{\mathbb R^3}{}{\frac G{|x-y|}\rho(y)}{^3}y$$

\end{document} 