Test data: Latex - Integrals 1

By syntax

% Define command "diff"
\newcommand\diff{\mathop{}\!\mathrm{d}}
\begin{gather}
\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 \end{gather} \tag{q.e.d.}

% Define command "diff"
\newcommand\diff{\mathop{}\!\mathrm{d}}
% Define command "Diff" with an argument
\newcommand\Diff[1]{\mathop{}\!\mathrm{d^#1}}

V(\mathbf{x}) = -\int_{\mathbf{R}^3}
   \frac{G}{|\mathbf{x}-\mathbf{y}|}\,\rho(\mathbf{y})\,\Diff3\mathbf{y}

By render

(∫−∞∞e−x2 ⁣dx)2=∫−∞∞∫−∞∞e−(x2+y2) ⁣dx ⁣dy=∫02π∫0∞e−r2r ⁣dr ⁣dθ=∫02π(−e−r22∣r=0r=∞ ) ⁣dθ=π(q.e.d.)% Define command "diff"
\newcommand\diff{\mathop{}\!\mathrm{d}}
\begin{gather}
\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 \end{gather} \tag{q.e.d.}

V(x)=−∫R3G∣x−y∣ ρ(y)  ⁣d3y% Define command "diff"
\newcommand\diff{\mathop{}\!\mathrm{d}}
% Define command "Diff" with an argument
\newcommand\Diff[1]{\mathop{}\!\mathrm{d^#1}}

V(\mathbf{x}) = -\int_{\mathbf{R}^3}
   \frac{G}{|\mathbf{x}-\mathbf{y}|}\,\rho(\mathbf{y})\,\Diff3\mathbf{y}

Source

tex.stackexchange - Should i \mathrm the d in my Integrals