1

This question already has an answer here:

I want to create a math operator that is an integral with a square (\square) stack onto it such that the integral is still formated the way that the operator \int is. How do I do this?

marked as duplicate by egreg, Mensch, Stefan Pinnow, TeXnician, Circumscribe Jan 24 at 18:28

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3
\documentclass{article}
\usepackage{esint}
\begin{document}

\[ \sqint\limits_0^1 x\,\mathrm dx\ne\int\limits_0^1 x\,\mathrm dx \]           
\[ \scriptstyle\sqint_0^1 x\,\mathrm dx\ne\int_0^1 x\,\mathrm dx \]     
\[ \scriptscriptstyle\sqint_0^1 x\,\mathrm dx\ne\int_0^1 x\,\mathrm dx \]    

\[ \sqiint\limits_0^1 x\,\mathrm dx\ne\iint\limits_0^1 x\,\mathrm dx \]     
\[ \scriptstyle\sqiint_0^1 x\,\mathrm dx\ne\iint_0^1 x\,\mathrm dx \]    
\[ \scriptscriptstyle\sqiint_0^1 x\,\mathrm dx\ne\iint_0^1 x\,\mathrm dx\]
\end{document}

enter image description here

1

Adapting Heiko Oberdiek's code in https://tex.stackexchange.com/a/171445/4427 is almost straightforward

\documentclass[a5paper]{article}

\usepackage{graphicx}

\usepackage{amsmath}
\usepackage{amssymb}


\makeatletter
\let\DOTSI\relax % amsmath support for \dots
\newcommand*{\sqint}{%
  \DOTSI
  \mathop{%
    \mathpalette\@LetterOnInt{\square}%
  }%
  \mkern-\thinmuskip % thin space is inserted between two \mathop
  \int
}
\newcommand*{\@LetterOnInt}[2]{%
  \sbox0{$#1\int\m@th$}%
  \sbox2{$%
    \ifx#1\displaystyle
      \textstyle
    \else
      \scriptscriptstyle
    \fi
    #2%
  \m@th$}%
  \dimen@=.4\dimexpr\ht0+\dp0\relax
  \ifdim\dimexpr\ht2+\dp2\relax>\dimen@
    \sbox2{\resizebox*{!}{\dimen@}{\unhcopy2}}%
  \fi
  \dimen@=\wd0 %
  \ifdim\wd2>\dimen@
    \dimen@=\wd2 %
  \fi
  \rlap{\hbox to \dimen@{\hfil
    $#1\vcenter{\copy2}\m@th$%
  \hfil}}%
  \ifdim\dimen@>\wd0 %
    \kern.5\dimexpr\dimen@-\wd0\relax
  \fi
}
\makeatother

\begin{document}

Take the regulated function $f$, we want to compare the Riemann integral,
$\int_a^b f$
to the regulated integral $\sqint_a^b f$ by taking
sequences of step
functions.

We will prove that
\[
  \sqint_a^b f = \int_a^b f
\]

\[
  \displaystyle      \int_a^b f \dots \sqint_a^b f \qquad
  \textstyle         \int_a^b f \dots \sqint_a^b f \qquad
  \scriptstyle       \int_a^b f \dots \sqint_a^b f \qquad
  \scriptscriptstyle \int_a^b f \dots \sqint_a^b f
\]

\end{document}

enter image description here

0

Tweaked for CM font.

\documentclass{article}
\usepackage{amssymb,graphicx}
\newcommand\sqint{\mathchoice
  {\raisebox{.75pt}{$\scriptscriptstyle\square$}\mkern-17.5mu}
  {\raisebox{.75pt}{$\scriptscriptstyle\square$}\mkern-14mu}
  {\scalebox{.8}{$\scriptscriptstyle\square$}\mkern-13mu}
  {\scalebox{.55}{$\scriptscriptstyle\square$}\mkern-11mu}
  \int}
\begin{document}
$\displaystyle\sqint_0^1 x\,dx\ne\int_0^1 x\,dx$\smallskip

$\sqint_0^1 x\,dx\ne\int_0^1 x\,dx$\smallskip

$\scriptstyle\sqint_0^1 x\,dx\ne\int_0^1 x\,dx$\smallskip

$\scriptscriptstyle\sqint_0^1 x\,dx\ne\int_0^1 x\,dx$
\end{document}

enter image description here

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