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Looking the picture below it is possible to observe the same lenght of the the size of two signs of the integral.

enter image description here

For the first integral (red box), the size is correct. But for the second integral (violet box) the size is very small for the integrand. I have adjusted myself the lenght of the first integral.

But, how is possible to add a command (or more commands with different sizes) into my source to have a correct size for the second integral?

I have seen this link Big integral sign but I hope alone to find the right length about.

Thanks a lot to everybody and any further suggestions are always welcome.

Here there is my MWE:

\documentclass[12pt]{book}
\usepackage[top=2.5cm,bottom=2.5cm,left=3cm,right=3cm,headsep=10pt,a4paper]{geometry} 
\usepackage{times} % Use the Times font for headings
\usepackage[lite]{mtpro2} 
\usepackage{classico}
\usepackage{microtype} % Slightly tweak font spacing for aesthetics
\usepackage[utf8]{inputenc} % Required for including letters with accents
\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
\usepackage{mathtools}


%%%%%%% New integral with mathdesign %%%%%%%
\makeatletter
\def\upintkern@{\mkern-7mu\mathchoice{\mkern-3.5mu}{}{}{}}
\def\upintdots@{\mathchoice{\mkern-4mu\@cdots\mkern-4mu}%
 {{\cdotp}\mkern1.5mu{\cdotp}\mkern1.5mu{\cdotp}}%
 {{\cdotp}\mkern1mu{\cdotp}\mkern1mu{\cdotp}}%
 {{\cdotp}\mkern1mu{\cdotp}\mkern1mu{\cdotp}}}
\newcommand{\upiint}{\DOTSI\protect\UpMultiIntegral{2}}
\newcommand{\upiiint}{\DOTSI\protect\UpMultiIntegral{3}}
\newcommand{\upiiiint}{\DOTSI\protect\UpMultiIntegral{4}}
\newcommand{\upidotsint}{\DOTSI\protect\UpMultiIntegral{0}}
\newcommand{\UpMultiIntegral}[1]{%
  \edef\ints@c{\noexpand\upintop
    \ifnum#1=\z@\noexpand\upintdots@\else\noexpand\upintkern@\fi
    \ifnum#1>\tw@\noexpand\upintop\noexpand\upintkern@\fi
    \ifnum#1>\thr@@\noexpand\upintop\noexpand\upintkern@\fi
    \noexpand\upintop
    \noexpand\ilimits@
  }
  \futurelet\@let@token\ints@a
}
\makeatother

\DeclareFontFamily{OMX}{mdbch}{}
\DeclareFontShape{OMX}{mdbch}{m}{n}{ <->s * [1]  mdbchr7v }{}%old 0.8
\DeclareFontShape{OMX}{mdbch}{b}{n}{ <->s * [1]  mdbchb7v }{}%old 0.8
\DeclareFontShape{OMX}{mdbch}{bx}{n}{<->ssub * mdbch/b/n}{}

\DeclareSymbolFont{uplargesymbols}{OMX}{mdbch}{m}{n}
\SetSymbolFont{uplargesymbols}{bold}{OMX}{mdbch}{b}{n}
\DeclareMathSymbol{\upintop}{\mathop}{uplargesymbols}{82}
\DeclareMathSymbol{\upointop}{\mathop}{uplargesymbols}{"48}

\DeclareFontEncoding{MDB}{}{}
\DeclareFontFamily{MDB}{mdbch}{}
\DeclareFontShape{MDB}{mdbch}{m}{n}{ <->s * [0.8]  mdbchrmb }{}
\DeclareFontShape{MDB}{mdbch}{b}{n}{ <->s * [0.8]  mdbchbmb }{}
\DeclareFontShape{MDB}{mdbch}{bx}{n}{<->ssub * mdbch/b/n}{}
\DeclareFontSubstitution{MDB}{cmr}{m}{n}
\DeclareSymbolFont{mathdesignB}{MDB}{mdbch}{m}{n}%
\SetSymbolFont{mathdesignB}{bold}{MDB}{mdbch}{b}{n}%
\DeclareMathSymbol{\upintclockwise}{\mathop}{mathdesignB}{128}
\DeclareMathSymbol{\upointclockwise}{\mathop}{mathdesignB}{130}
\DeclareMathSymbol{\upointctrclockwise}{\mathop}{mathdesignB}{132}
\DeclareMathSymbol{\upoiint}{\mathop}{mathdesignB}{134}
\DeclareMathSymbol{\upoiiint}{\mathop}{mathdesignB}{136}

\makeatletter
\newcommand{\upint}{\DOTSI\upintop\ilimits@}
\newcommand{\upoint}{\DOTSI\upointop\ilimits@}
\makeatother


\begin{document}
\begin{equation}
dx=\upint \frac{dy}{\log x}
\end{equation}

\begin{equation}
\upint\frac{dy}{\sqrt{\dfrac{(n(y))^2}{c^2}-1}}
\end{equation}

\end{document}
Sebastiano
  • 54,118
  • 3
    Integrals don't change size; adapt the function instead, for instance using ^{-1/2}. – egreg Feb 10 '18 at 11:32
  • I had tried it at my house but I do not like the style ^{-1/2}. I have tried also, $\tfrac, \sfrac, $ etc. If it is not possible, I will place the square root in the numerator . Thanks a lot. – Sebastiano Feb 10 '18 at 11:42
  • The question here, https://tex.stackexchange.com/questions/39181/big-integral-sign, may be of some interest. – Steven B. Segletes Feb 27 '18 at 14:35

0 Answers0