Should I \mathrm the d in my integrals?

\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}

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}

Have a look at

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

2.5 Derivatives

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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}