答案1
这似乎提供了不错的模拟。对于积分符号,您可以做得更好。
\documentclass{article}
\usepackage{mlmodern}
\usepackage[nolimits]{cmupint}
\usepackage{amsmath}
\usepackage{microtype}
\DisableLigatures[f]{encoding=\encodingdefault,family=\familydefault}
\DeclareSymbolFont{slanted}{\encodingdefault}{\familydefault}{m}{sl}
\DeclareMathSymbol{g}{\mathalpha}{slanted}{`g}
\DeclareFontFamily{U}{cmupint}{}
\DeclareFontShape{U}{cmupint}{m}{n}{ <-> s*[0.8] cmupint}{}
\begin{document}
$g_n$ is the coefficient of $x^n$ in the expansion of
\[
\frac{x^{n+1}}{\{y_1(x+a)\}^2}\exp\Bigl[-\int_0^x up(u)\,du\Bigr].
\]
It may happen that $g_n=0$ in which case $y_2(x)$ denotes
a logarithmic term.
\textbf{The Point at Infinity.}\quad In many problems
\end{document}
答案2
我的旧版本
\documentclass[12pt]{article}
\usepackage{mathspec}
\defaultfontfeatures{Mapping=tex-text}
\setallmainfonts{Old Standard}
\DeclareSymbolFont{mathptmxlargesymbols}{OMX}{ztmcm}{m}{n}
\DeclareMathSymbol{\upsumop}{\mathop}{mathptmxlargesymbols}{"50}
\DeclareMathSymbol{\upprodop}{\mathop}{mathptmxlargesymbols}{"51}
\begin{document}
, for there coordinates, Laplace's equation in the form
\[\frac{\partial^2\psi}{\partial \rho^2} +\frac{1}{\varrho}\frac{\partial\psi}{\partial\varrho}+\frac{1}{\varrho^2}\frac{\partial^2\psi}{\partial\phi^2}+\frac{\partial^2\psi }{\partial z^2}=0.\]
\end{document}