\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[tracking]{microtype} \usepackage[sc,osf]{mathpazo}%With old-style figures and real smallcaps. \usepackage[euler-digits,small]{eulervm} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by mforbes at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} %must be before next line \usepackage{mathpazo,mathabx} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{kpfonts} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{newtxtext,newtxmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
fi letters in text. But Math looks ok.
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage[libertine]{newtxmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \usepackage{stix} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
fi letters in text. But Math looks ok.
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \usepackage{lmodern} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \usepackage{mathpazo} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \usepackage{txfonts} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Answer by Mico at Tex.stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \usepackage{XCharter} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Tex.Stackexchange
\documentclass{article} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \usepackage[charter]{mathdesign} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
Math Code fragment thanks to Tex.Stackexchange
\documentclass{article} \usepackage[math]{anttor} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
http://www.tug.dk/FontCatalogue/anttor/
\documentclass{article} \usepackage[condensed,math]{anttor} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
http://www.tug.dk/FontCatalogue/anttor/
\documentclass{article} \usepackage[light,math]{anttor} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
http://www.tug.dk/FontCatalogue/anttor/
\documentclass{article} \usepackage[utf8]{inputenc} \usepackage{arev} \usepackage[T1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
N/A did not compile.
http://www.tug.dk/FontCatalogue/anttor/
\documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage[lf]{Baskervaldx} % lining figures \usepackage[bigdelims,vvarbb]{newtxmath} % math italic letters from Nimbus Roman \usepackage[cal=boondoxo]{mathalfa} % mathcal from STIX, unslanted a bit \renewcommand*\oldstylenums[1]{\textosf{#1}} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
fi, but math looks ok.
http://www.tug.dk/FontCatalogue/anttor/
\documentclass{article} \usepackage{amsmath} %\usepackage{boisik} %causes problems \usepackage[OT1]{fontenc} \usepackage{ntheorem} \newtheorem{theorem}{Theorem} \usepackage{amsmath} \DeclareMathOperator{\Res}{Res} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \blindtext \pagestyle{empty} \begin{theorem}[Residue Theorem] Let $f$ be analytic in the region $G$ except for the isolated singularities $a_1,a_2,\dots,a_m$. If $\gamma$ is a closed rectifiable curve in $G$ which does not pass through any of the points $a_k$ and if $\gamma\approx 0$ in $G$, then \[ \frac{1}{2\pi i}\int_\gamma\! f = \sum_{k=1}^m n(\gamma;a_k)\Res(f;a_k)\,. \] \end{theorem} \end{document}
fi, but math looks ok.