Should I \mathrm the d in my integrals?

Question

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"?

Answer 1

\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}

Answer 2

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}

Answer 3

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:

Answer 4

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[3]{y}
\end{equation}

\end{document}

Answer 5

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}