Changing the size of an operator according to its argument?

Question

I'm trying to use a Laplace transform symbol as an operator, so far it looks good, but I want it to be of the same size as the argument that it's working with, so if the argument has one row, the Laplace transform would be of the same hight as one row, and if the argument is a fraction with 2 rows, enlarge the symbol of the operator.

class test2(Scene):
def construct(self):
    my_template = TexTemplate()
    my_template.add_to_preamble(
        r"""
        \usepackage{mathrsfs}
        \let\RSFSmathscr\mathscr  % save the meaning of \mathscr
        \usepackage[scr]{rsfso}
        \let\RSFSOmathscr\mathscr % save the new meaning of \mathscr
        \let\mathscr\RSFSmathscr  % restore the previous meaning
    \newcommand{\Laplace}{\RSFSOmathscr{L}}
    """)
tex = MathTex(r"\Laplace \left \{ \frac{1}{s^2 + 1} \right \}", tex_template=my_template)

and this is the result I'm getting

I would also like to add the arguments t and s, similar to wolfram alpha do it, here is an image of that.

in other words, the final result should look something like this:

I was thinking of maybe something like this?:

            \usepackage{mathrsfs}
            \let\RSFSmathscr\mathscr
            \usepackage[scr]{rsfso}
            \let\RSFSOmathscr\mathscr
            \let\mathscr\RSFSmathscr
        \newcommand{\Laplace}{\RSFSOmathscr{L} _t \left\{ somehow_argument_here  \right \} (s) }

Answer 1

You can do this by putting the ℒ and its arguments in boxes, and resizing them. Since you want to be able to add subscripts, I made that an optional argument.

\documentclass{article}
\tracinglostchars=2
\usepackage[paperwidth=10cm]{geometry} % To format the MWE for Tex.SX
\usepackage{iftex} % For \iftutex
\usepackage{settobox} % For \settoboxheight
\usepackage{graphicx} % For \resizebox
\iftutex
  \usepackage{newcomputermodern} % For example
%% By default, \mathscr duplicates \mathcal.  You wanted a symbol like
%% Boondoxo, which is available as a stylistic set in the STIX fonts.
  \setmathfont{STIX Two Math}[
    Scale=MatchUppercase,
    StylisticSet=1,
    range={scr, bfscr}]
\else
  \usepackage{lmodern} % For example
  \usepackage{amsmath}
  \usepackage[scr=boondoxo]{mathalpha}
\fi
\newlength{\LaplaceHeight}
\newbox{\LaplaceArg}
\newcommand\Laplace[2][]{%
  \savebox{\LaplaceArg}{(#2)}%
  \settoboxheight{\LaplaceHeight}{\LaplaceArg}%
  \mathop{\resizebox{!}{\LaplaceHeight}{$\mathscr{L}$}}\nolimits_{#1} \usebox{\LaplaceArg}%
}
\begin{document}
[ \Laplace{\sin t}(s) ]
[ \Laplace[t]{\left{ \frac{1}{t+1} \right} }(s) ]
\end{document}

This version matches only the height and not the depth. To match your illustration more closely, but drop below the baseline, you can instead resize to the argument’s total height and lower by its depth:

\documentclass{article}
\tracinglostchars=2
\usepackage[paperwidth=10cm]{geometry} % To format the MWE for Tex.SX
\usepackage{iftex} % For \iftutex
\usepackage{settobox} % For \settoboxtotalheight, \settoboxdepth
\usepackage{graphicx} % For \resizebox
\iftutex
  \usepackage{newcomputermodern} % For example
%% By default, \mathscr duplicates \mathcal.  You wanted a symbol like
%% Boondoxo, which is available as a stylistic set in the STIX fonts.
  \setmathfont{STIX Two Math}[
    Scale=MatchUppercase,
    StylisticSet=1,
    range={scr, bfscr}]
\else
  \usepackage{lmodern} % For example
  \usepackage{amsmath}
  \usepackage[scr=boondoxo]{mathalpha}
\fi
\newlength{\LaplaceHeight}
\newlength{\LaplaceDepth}
\newbox{\LaplaceArg}
\newcommand\Laplace[2][]{%
  \savebox{\LaplaceArg}{(#2)}%
  \settoboxtotalheight{\LaplaceHeight}{\LaplaceArg}%
  \settoboxdepth{\LaplaceDepth}{\LaplaceArg}%
  \mathop{\raisebox{-\LaplaceDepth}{\resizebox{!}{\LaplaceHeight}{$\mathscr{L}$}}}\nolimits_{#1} \usebox{\LaplaceArg}%
}
\begin{document}
[ \Laplace{\sin t}(s) ]
[ \Laplace[t]{\lef

As several have mentioned in the comments, scaling the ℒ makes it look too heavy. You could possibly scale it only along the vertical axis, but that doesn’t look good either.

Here’s an alternative you might like better, which defines a normal-size \laplace and a double-sized \Laplace, scaling up the latter from a lighter weight of the font family. It requires LuaLaTeX or XeLaTeX.

\documentclass{article}
\tracinglostchars=2
\usepackage{newcomputermodern} % For example
\setmathfont{KPMath-Regular}[ Scale=MatchUppercase, range={scr,bfscr}]
\setmathfontface\bigmath{KPMath-Light}[Scale=2.0]
\newcommand\laplace{\mathop{\mathscr{L}}\nolimits}
\newcommand\Laplace{\mathop{\bigmath{ℒ}}\nolimits}
\begin{document}
[ \laplace{\sin t}(s) ]
[ \Laplace_t {\left{ \frac{1}{t+1} \right} }(s) ]
\end{document}

Answer 2

Looks like the following code may get to your goal:

\documentclass{article}
\usepackage[scr]{rsfso}
\usepackage{amsmath}
\newcommand{\Laplace}[1]{\mathop{}\!\mathscr{L}_t\!\left\{#1\right\}\!}
\begin{document}
You can write
\[
\Laplace{\sin t}(s)
\quad\text{and}\quad
\Laplace{\frac{1}{t+1}}(s).
\]
Let's leave ``$(s)$'' out of the definition, so you can write
\[
\Laplace{\exp(-t)}(s)=\frac{1}{s+1}
\quad\text{and}\quad
\Laplace{\exp(-t)}(x)=\frac{1}{x+1}.
\]
You can even write some savvy convolution:
\[
\Laplace{(f\ast g)(t)}(s)=\Laplace{f(t)}(s)\Laplace{g(t)}(s),
\]
where $\ast$~is the convolution symbol defined as
\[
(f\ast g)(t)=\int f(x) g(t-x) \, dx.
\]
\end{document}

The \mathop{}\! at the very start takes care of spacing (when you write product of Laplace transforms, it adds a tiny amount of space to separate the terms). If you don’t like it, you can remove it. The other two \!’s cancel the extra spaces around \left\{...\right\}.

Answer 3

The \scalerel macro takes its first argument and scales it to the vertical footprint of the 2nd argument, working in math mode. Adjustments needed were to exclude the _t from the scale calculation, and to have a minimum size of the scale-to footprint, in this case the size of a cap X.

\documentclass{article}
\usepackage{mathrsfs}
  \let\RSFSmathscr\mathscr
  \usepackage[scr]{rsfso}
  \let\RSFSOmathscr\mathscr
  \let\mathscr\RSFSmathscr
\newcommand{\Laplace}[1]{\scalerel{\RSFSOmathscr{L}
  \smash{_t}}{\vphantom{X}#1} (s) }
\usepackage{scalerel}
\begin{document}
\[
\Laplace{\sin t}
\]
\[
\Laplace{\left\{\frac{1}{t^2+1}\right\}}
\]
\end{document}

SUPPLEMENT

The OP asks, I think, if the size of the Laplace operator can be made smaller in the second case, perhaps a fraction of the height of the argument, rather than the full height of the argument. Here, I make it .6 times the height of the argument, with a minimum size corresponding to the size of capital X.

\documentclass{article}
\usepackage{mathrsfs}
  \let\RSFSmathscr\mathscr
  \usepackage[scr]{rsfso}
  \let\RSFSOmathscr\mathscr
  \let\mathscr\RSFSmathscr
\newcommand{\Laplace}[1]{\ThisStyle{
  \sbox0{$\SavedStyle#1$}
  \ht0=.6\ht0
  \dp0=.6\dp0
  \scalerel{\RSFSOmathscr{L}
  \smash{_t}}{\vphantom{X\box0}}#1 (s)}}
\usepackage{scalerel}
\begin{document}
\[
\Laplace{\sin t}
\]
\[
\Laplace{\left\{\frac{1}{t^2+1}\right\}}
\]
\end{document}