Stacking Symbols for a Math Operator [duplicate]












1
















This question already has an answer here:




  • Superimpose letter on integral symbol

    3 answers




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?










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


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















  • Related: subseteq + circ as a single symbol ("open subset")

    – Werner
    Jan 24 at 17:30











  • I took tex.stackexchange.com/a/171445/4427 and changed R into square, getting what probably is your desired symbol.

    – egreg
    Jan 24 at 17:39
















1
















This question already has an answer here:




  • Superimpose letter on integral symbol

    3 answers




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?










share|improve this question















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


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















  • Related: subseteq + circ as a single symbol ("open subset")

    – Werner
    Jan 24 at 17:30











  • I took tex.stackexchange.com/a/171445/4427 and changed R into square, getting what probably is your desired symbol.

    – egreg
    Jan 24 at 17:39














1












1








1









This question already has an answer here:




  • Superimpose letter on integral symbol

    3 answers




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?










share|improve this question

















This question already has an answer here:




  • Superimpose letter on integral symbol

    3 answers




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?





This question already has an answer here:




  • Superimpose letter on integral symbol

    3 answers








math-operators






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share|improve this question













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edited Jan 24 at 17:36









Steven B. Segletes

154k9194402




154k9194402










asked Jan 24 at 17:20









LinearGuyLinearGuy

61




61




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


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









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


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Related: subseteq + circ as a single symbol ("open subset")

    – Werner
    Jan 24 at 17:30











  • I took tex.stackexchange.com/a/171445/4427 and changed R into square, getting what probably is your desired symbol.

    – egreg
    Jan 24 at 17:39



















  • Related: subseteq + circ as a single symbol ("open subset")

    – Werner
    Jan 24 at 17:30











  • I took tex.stackexchange.com/a/171445/4427 and changed R into square, getting what probably is your desired symbol.

    – egreg
    Jan 24 at 17:39

















Related: subseteq + circ as a single symbol ("open subset")

– Werner
Jan 24 at 17:30





Related: subseteq + circ as a single symbol ("open subset")

– Werner
Jan 24 at 17:30













I took tex.stackexchange.com/a/171445/4427 and changed R into square, getting what probably is your desired symbol.

– egreg
Jan 24 at 17:39





I took tex.stackexchange.com/a/171445/4427 and changed R into square, getting what probably is your desired symbol.

– egreg
Jan 24 at 17:39










3 Answers
3






active

oldest

votes


















3














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

[ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
[ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
[ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]

[ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
[ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
[ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
end{document}


enter image description here






share|improve this answer































    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
    letDOTSIrelax % 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{$#1intm@th$}%
    sbox2{$%
    ifx#1displaystyle
    textstyle
    else
    scriptscriptstyle
    fi
    #2%
    m@th$}%
    dimen@=.4dimexprht0+dp0relax
    ifdimdimexprht2+dp2relax>dimen@
    sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
    fi
    dimen@=wd0 %
    ifdimwd2>dimen@
    dimen@=wd2 %
    fi
    rlap{hbox to dimen@{hfil
    $#1vcenter{copy2}m@th$%
    hfil}}%
    ifdimdimen@>wd0 %
    kern.5dimexprdimen@-wd0relax
    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






    share|improve this answer

































      0














      Tweaked for CM font.



      documentclass{article}
      usepackage{amssymb,graphicx}
      newcommandsqint{mathchoice
      {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
      {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
      {scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
      {scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
      int}
      begin{document}
      $displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

      $sqint_0^1 x,dxneint_0^1 x,dx$smallskip

      $scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

      $scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
      end{document}


      enter image description here






      share|improve this answer






























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3














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

        [ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
        [ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
        [ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]

        [ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
        [ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
        [ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
        end{document}


        enter image description here






        share|improve this answer




























          3














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

          [ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
          [ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
          [ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]

          [ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
          [ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
          [ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
          end{document}


          enter image description here






          share|improve this answer


























            3












            3








            3







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

            [ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
            [ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
            [ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]

            [ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
            [ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
            [ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
            end{document}


            enter image description here






            share|improve this answer













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

            [ sqintlimits_0^1 x,mathrm dxneintlimits_0^1 x,mathrm dx ]
            [ scriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]
            [ scriptscriptstylesqint_0^1 x,mathrm dxneint_0^1 x,mathrm dx ]

            [ sqiintlimits_0^1 x,mathrm dxneiintlimits_0^1 x,mathrm dx ]
            [ scriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx ]
            [ scriptscriptstylesqiint_0^1 x,mathrm dxneiint_0^1 x,mathrm dx]
            end{document}


            enter image description here







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Jan 24 at 17:49









            HerbertHerbert

            273k24412725




            273k24412725























                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
                letDOTSIrelax % 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{$#1intm@th$}%
                sbox2{$%
                ifx#1displaystyle
                textstyle
                else
                scriptscriptstyle
                fi
                #2%
                m@th$}%
                dimen@=.4dimexprht0+dp0relax
                ifdimdimexprht2+dp2relax>dimen@
                sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
                fi
                dimen@=wd0 %
                ifdimwd2>dimen@
                dimen@=wd2 %
                fi
                rlap{hbox to dimen@{hfil
                $#1vcenter{copy2}m@th$%
                hfil}}%
                ifdimdimen@>wd0 %
                kern.5dimexprdimen@-wd0relax
                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






                share|improve this answer






























                  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
                  letDOTSIrelax % 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{$#1intm@th$}%
                  sbox2{$%
                  ifx#1displaystyle
                  textstyle
                  else
                  scriptscriptstyle
                  fi
                  #2%
                  m@th$}%
                  dimen@=.4dimexprht0+dp0relax
                  ifdimdimexprht2+dp2relax>dimen@
                  sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
                  fi
                  dimen@=wd0 %
                  ifdimwd2>dimen@
                  dimen@=wd2 %
                  fi
                  rlap{hbox to dimen@{hfil
                  $#1vcenter{copy2}m@th$%
                  hfil}}%
                  ifdimdimen@>wd0 %
                  kern.5dimexprdimen@-wd0relax
                  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






                  share|improve this answer




























                    1












                    1








                    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
                    letDOTSIrelax % 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{$#1intm@th$}%
                    sbox2{$%
                    ifx#1displaystyle
                    textstyle
                    else
                    scriptscriptstyle
                    fi
                    #2%
                    m@th$}%
                    dimen@=.4dimexprht0+dp0relax
                    ifdimdimexprht2+dp2relax>dimen@
                    sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
                    fi
                    dimen@=wd0 %
                    ifdimwd2>dimen@
                    dimen@=wd2 %
                    fi
                    rlap{hbox to dimen@{hfil
                    $#1vcenter{copy2}m@th$%
                    hfil}}%
                    ifdimdimen@>wd0 %
                    kern.5dimexprdimen@-wd0relax
                    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






                    share|improve this answer















                    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
                    letDOTSIrelax % 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{$#1intm@th$}%
                    sbox2{$%
                    ifx#1displaystyle
                    textstyle
                    else
                    scriptscriptstyle
                    fi
                    #2%
                    m@th$}%
                    dimen@=.4dimexprht0+dp0relax
                    ifdimdimexprht2+dp2relax>dimen@
                    sbox2{resizebox*{!}{dimen@}{unhcopy2}}%
                    fi
                    dimen@=wd0 %
                    ifdimwd2>dimen@
                    dimen@=wd2 %
                    fi
                    rlap{hbox to dimen@{hfil
                    $#1vcenter{copy2}m@th$%
                    hfil}}%
                    ifdimdimen@>wd0 %
                    kern.5dimexprdimen@-wd0relax
                    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







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    answered Jan 24 at 17:45


























                    community wiki





                    egreg
























                        0














                        Tweaked for CM font.



                        documentclass{article}
                        usepackage{amssymb,graphicx}
                        newcommandsqint{mathchoice
                        {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
                        {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
                        {scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
                        {scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
                        int}
                        begin{document}
                        $displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                        $sqint_0^1 x,dxneint_0^1 x,dx$smallskip

                        $scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                        $scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
                        end{document}


                        enter image description here






                        share|improve this answer




























                          0














                          Tweaked for CM font.



                          documentclass{article}
                          usepackage{amssymb,graphicx}
                          newcommandsqint{mathchoice
                          {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
                          {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
                          {scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
                          {scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
                          int}
                          begin{document}
                          $displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                          $sqint_0^1 x,dxneint_0^1 x,dx$smallskip

                          $scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                          $scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
                          end{document}


                          enter image description here






                          share|improve this answer


























                            0












                            0








                            0







                            Tweaked for CM font.



                            documentclass{article}
                            usepackage{amssymb,graphicx}
                            newcommandsqint{mathchoice
                            {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
                            {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
                            {scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
                            {scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
                            int}
                            begin{document}
                            $displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                            $sqint_0^1 x,dxneint_0^1 x,dx$smallskip

                            $scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                            $scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
                            end{document}


                            enter image description here






                            share|improve this answer













                            Tweaked for CM font.



                            documentclass{article}
                            usepackage{amssymb,graphicx}
                            newcommandsqint{mathchoice
                            {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-17.5mu}
                            {raisebox{.75pt}{$scriptscriptstylesquare$}mkern-14mu}
                            {scalebox{.8}{$scriptscriptstylesquare$}mkern-13mu}
                            {scalebox{.55}{$scriptscriptstylesquare$}mkern-11mu}
                            int}
                            begin{document}
                            $displaystylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                            $sqint_0^1 x,dxneint_0^1 x,dx$smallskip

                            $scriptstylesqint_0^1 x,dxneint_0^1 x,dx$smallskip

                            $scriptscriptstylesqint_0^1 x,dxneint_0^1 x,dx$
                            end{document}


                            enter image description here







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Jan 24 at 17:36









                            Steven B. SegletesSteven B. Segletes

                            154k9194402




                            154k9194402















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