# 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"?

• See also What's the proper way to typeset a differential operator? and the comments at Top four LaTeX mistakes -- there appear to be some regional variances, and some attempts at standardization. Commented Jun 20, 2012 at 13:10
• Using \mathrm or not depends on the traditions in your field. A thin space before the "d" in integrals is certainly required, Herbert's solution shows how to get it automatically (but using a macro for getting the "d"). Commented Jun 20, 2012 at 13:10
• Related Question: new command for the dx of intergral. Commented Nov 17, 2012 at 9:42
• The standard in mathematics is not to use mathrm on the d: just use \, dx. In engineering and physics, they do things differently. Commented Nov 27, 2013 at 9:15
• If that's any indication, the amsmath manual does not \rm it's integral d's: mirrors.ctan.org/macros/latex/required/amsmath/amsldoc.pdf
– bers
Commented Mar 26, 2019 at 16:18

\documentclass{article}
\usepackage{amsmath}
\newcommand*\diff{\mathop{}\!\mathrm{d}}
\newcommand*\Diff[1]{\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 $dy/dx$. Commented Jun 20, 2012 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. Commented Jun 20, 2012 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. Commented Jun 20, 2012 at 13:19
• @JimHefferon: in inline mode I use only \mathrm{d}
– user2478
Commented Jun 21, 2012 at 6:39
• Thank you, but (1) I think it should be \newcommand*\Diff[1]{\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[2]{\mathop{}\!\mathrm{d}^{#1}\!{#2}}, although it is a tad too close.
– Jost
Commented Nov 5, 2013 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. Commented Jan 28, 2013 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 ? Commented Jan 28, 2013 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. Commented Jan 28, 2013 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[%, Commented Jan 28, 2013 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 \!. Commented Sep 8, 2014 at 9:19 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} $$V(x)=-\int\limits_{\mathbb R^3} \frac G{|x-y|}\rho(y) \dd[3]{y}$$ \end{document}  • Note that we can italicise the d's using \usepackage[italicdiff]{physics} Commented Apr 9, 2017 at 5:22 I usually do this (which I've shamefully stolen from Niel de Beaudrap and modified): \makeatletter \renewcommand\d[1]{\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. Commented Jun 20, 2012 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. Commented Jun 20, 2012 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' Commented Jun 20, 2012 at 13:42 • @tohecz If \mathop{}\!d is used, then the spacing for subsequent differentials will be automatically added. Commented Jun 20, 2012 at 13:44 • Why is the d on the left italic? – user47077 Commented Nov 25, 2014 at 15:55 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[1] { \,\textrm d{#1} } % infintesimal \newcommand \de[2] { \frac{\mathrm d{#1}}{\mathrm d{#2}} } % first order derivative \newcommand \intl[4]{ \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}  My answer is short, but it is surely an answer. So I post it here as an answer rather than a comment. In my opinion, the letter "d" should be regarded as an operator rather than a variable when it is used to express the differential in mathematics. So the simplest way to realize this is to define a command via "\DeclareMathOperator{\td}{d}" in the whole document or to employ "\operatorname{d}" locally and occationally. Moreover, the command "\td" comes from "\textup{d}" in AmS-LaTeX. See also the following comments at https://mathoverflow.net/questions/455136/how-to-prove-negativity-of-a-3-times3-determinant-whose-elements-involve-triga#comment1178363_455136 and https://mathoverflow.net/questions/455136/how-to-prove-negativity-of-a-3-times3-determinant-whose-elements-involve-triga#comment1178365_455136 • This is wrong, sorry. Try \td x to see what goes wrong (you got comments by LSpice telling the same). Commented Sep 24, 2023 at 7:46 • There should be no space between “d” (upright or not) and the variable, but you always get a thin space. Look in every book on your shelf. Commented Sep 24, 2023 at 13:11 • @egreg What's wrong? I used the command "\td" for many years and I didn't find any error. Please notice that there should and must be some spaces in front of and behind any standard operator name. Compare x\sin x, x\operatorname{sin}x, and x\operatorname{d}x. Is there any difference about the spaces in front of and behind these operators? Nothing, so they are all standard. Commented Sep 24, 2023 at 13:31 • One more word. In my opinion, we should do our best to use existed standard LaTeX commands as possible as we can, and we shouldn't manage to and try to construct our own specific LaTeX commands. By the way I have been using \operatorname{e}^x and \operatorname{i} to express the exponential function and the imaginary unit. Commented Sep 24, 2023 at 13:40 • You didn't look into your books, did you? The fact is that the differential “d” is not really a math operator like log or sin. Commented Sep 24, 2023 at 13:59 As user69453 already said, I'd recommend you to use the Physics package. Instead of using macros and dealing with spaces, you can just use \dd for the non-italic d. But, as barbara beeton stated, it's more important to write consistently than following a standard, which may not be used universally. Most of my college teachers (UPM, Spain) write the italic d for differentials. ### Usage Differential of x will be written as \dd{x}, while the nth differential of x would be \dd[n]{x}. ### Example \documentclass{article} \usepackage{amsmath} \usepackage{physics} \begin{document} \begin{align*} \biggl(\int_{-\infty}^\infty e^{-x^2}\dd{x}\biggr)^2 &= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\dd{x}\dd{y} \\ &= \int_0^{2\pi}\int_0^\infty e^{-r^2}r\dd{r}\dd{\theta} \\ &= \int_0^{2\pi}\biggl(-{e^{-r^2}\over2}\bigg\vert_{r=0}^{r=\infty}\,\biggr)\dd{\theta}\\ &= \pi \tag*{q.e.d.}\\ \end{align*} % \[ V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{|\mathbf{x}-\mathbf{y}|}\,\rho(\mathbf{y})\,\dd[3]{\mathbf{y}}

\end{document}


### Output

The main advantages of using this solution:

• When using brackets, it chooses the ideal separation depending on neighbours. If you write it without brackets, for example \dd x, there will be no separation.
• By default, it follows the ISO 80000-2:2019 standard. However, you can add a package option to set italic as default style: \usepackage[italicdiff]{physics}

• Sorry, but I disagree. The definition of \dd is uselessly complicated and doesn't work as it should. Commented Feb 2, 2021 at 20:37
• Thanks both, Werner and egreg. I agree with Werner, this should be a comment. However, I have little reputation for now and decided to make an answer as I can't comment. I thought it would be useful. Anyway, I apologise if it wasn't supposed to be like this. Should I delete the answer? Or maybe fix it? And, well... I didn't know that the definition of \dd was so complicated :( I don't want to start a chat here, so is there anyway you could explain me that out of here, egreg? Again, thank you! Commented Feb 2, 2021 at 21:40