Is it possible to make the integral sign smaller? I found that in the Display Mode the integral sign seems automatically longer. But I am in a context where smaller integral sign would be preferred.



2 Answers 2


If you prefer to have any size of the integral sign (you can choose the size that you like), it is possible to use scalerel package with the option \scaleobj{<scale factor>}{...}.

enter image description here


\int^{b}_{a}\frac{1}{x}\,dx, \quad 
  • Thank you really much! This is what I was talking about I guess :).
    – Yes
    Aug 25, 2019 at 5:19
  • @Megadeth I am happy for you, I am very glad to help you. Please, can you check instead Bernard's answer? Because it is better than mine for many reasons. Thank you very much.
    – Sebastiano
    Aug 25, 2019 at 11:34
  • Hi, easy! I was not making a comparison. I guess my comment was just a shout of joy. I appreciate every feedback in general.
    – Yes
    Aug 25, 2019 at 12:51
  • @Megadeth Hi, :-) my code is very generic and although you like it and there are many positive and negative aspects. I hope you change the green v to Bernard because I think it is fairer and more correct :-)
    – Sebastiano
    Aug 25, 2019 at 14:26

The nccmath package provides a medium-size integral sign , with the \medint\int command. There are also medium-sized fractions (\mfrac), the more general medmath{...} command. and medsize environments. You obtain formulæ with a size of ~80 % \displaystyle, avoiding the discrepancy with \textstyle .

The following code provides a demo:

\usepackage{mathtools, nccmath}


Different kinds of integrals: 

    \int_{0}^{1}\frac{\mathrm d x}{1 + x^2} & = \frac\pi 4 %
     &\texttt{\textbackslash medint: \quad} \medint\int_{0}^{1}\frac{\mathrm d x}{1 + x^2} & = \frac\pi 4 \\[2ex]
    \begin{tabular}{ >{\ttfamily}r}
    \textbackslash medint\\ +\textbackslash mfrac\enspace\end{tabular} \quad \medint\int_{0}^{1}\mfrac{\mathrm d x}{1 + x^2} & = \mfrac\pi 4
       & \texttt{\textbackslash textstyle: }\quad\textstyle{\int_{0}^{1}}\frac{\mathrm d x}{1 + x^2} & = \frac\pi 4


enter image description here

  • Yes, that was a private conversation.
    – Bernard
    Aug 25, 2019 at 11:58

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