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I would like to write a big integral in a align environment but bigint seems to be useless. Could anyone suggest? Here is a MWE containing the long expression:

\documentclass[a4paper,10pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{bigints}

\begin{document}

\begin{align}
P(r, \omega) \propto \bigint_{h \in I} 
{\left(  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \right)^{(1-h)/2}}
{\exp \left(  -\frac{L \omega}{2}  \left(  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \right)^{(1-h)/2} \right)}\\\nonumber
\left( \left( \frac{r}{L} \right)^{3-D(h)} + c \left( \frac{\eta}{L} \right)^{3-D(h)} \right)
dh.
\end{align}


\end{document}          
  • @StevenB.Segletes He said that it was not professional but with me it doesn't work at all. Do you have any suggestions? Thanks – Math Student May 15 '14 at 11:51
  • 1
    I tried your code in a basic document and it worked fine. Please edit your question so that it includes a complete example that reproduces the problem (from \begin{document} to \end{document}). Without this it will be difficult for anyone to help you. – Ian Thompson May 15 '14 at 12:07
  • @IanThompson Thanks. I have just edited the code. Did you get a big integral?? – Math Student May 15 '14 at 12:14
  • 1
    Yes. I get a big integral with your example code as well. – Ian Thompson May 15 '14 at 12:17
4

Here's a solution that loads the bigints package and employs that package's \bigintsss macro to create a slightly-enlarged integral symbol. If \bigintsss is too large for your taste, try \bigintssss; \bigintss creates a symbol that's definitely too large for the present example.

In the present case, I would also recommend that you place the limit-of-integration term below rather than to the right of the integral symbol. This helps fit the first line inside the text block. I've also moved the \nonumber instruction, and I've used square brackets instead of round parentheses for the three pairs of outer "fences".

By the way, I have encountered no problems using the bigints package inside an align environment. I have found, though, that bigints doesn't seem to be compatible with the lmodern package; I have no idea as to why these two packages aren't compatible.

enter image description here

\documentclass[a4paper,10pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{bigints}
\begin{document} 
\begin{align}
P(r, \omega) 
&\propto \bigintsss\limits_{\mkern-18mu h \in I}  % use '\mukern-18mu' to shift limit to the left
\biggl[  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \,\biggr]^{(1-h)/2}
\exp \biggl[  -\frac{L \omega}{2}  \left(  \left(  \frac{r}{L} \right)^2 +  
    c \left( \frac{\eta}{L}  \right)^2 \right)^{(1-h)/2} \,\biggr] \nonumber\\
&\qquad\times\biggl[ \left( \frac{r}{L} \right)^{3-D(h)} + 
    c \left( \frac{\eta}{L} \right)^{3-D(h)}\, \biggr] \,dh.
\end{align}
\end{document}

Addendum: A potential visual improvement may result from aligning the two opening square brackets vertically, as is done in the following example. This example also uses \bigintssss instead of \bigintsss.

enter image description here

\begin{align}
P(r, \omega) 
\propto \bigintssss\limits_{\mkern-18mu h \in I} % using "\bigintssss" now
&\biggl[  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \,\biggr]^{(1-h)/2}
\exp \biggl[  -\frac{L \omega}{2}  \left(  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \right)^{(1-h)/2} \,\biggr] \nonumber\\
{}\times{}&\biggl[ \left( \frac{r}{L} \right)^{3-D(h)} + c \left( \frac{\eta}{L} \right)^{3-D(h)}\, \biggr] \,dh.
\end{align}
  • Consider using the alignment character & after the \bigint limits, which avoids using the (arbitrary) \qquad for spacing. – Werner May 15 '14 at 19:46
  • @Werner - Thanks for this suggestion. I've posted a second example that incorporates the idea, along with \bigintssss instead of \bigintsss. – Mico May 15 '14 at 19:55
0

Probably bigint doesn't work with align. Here is an alternative way to increase the size of your integral:

\documentclass[a4paper,10pt]{report}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{scalerel}
\usepackage{bigints}


\def\stretchint#1{\vcenter{\hbox{\stretchto[220]{\displaystyle\int}{#1}}}}
\def\scaleint#1{\vcenter{\hbox{\scaleto[3ex]{\displaystyle\int}{#1}}}}
\def\bs{\!\!}

\begin{document}

\begin{align}
P(r, \omega) \propto {\scaleint{9ex}}_{\bs {h \in I} }
{\left(  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \right)^{(1-h)/2}}
{\exp \left(  -\frac{L \omega}{2}  \left(  \left(  \frac{r}{L} \right)^2 + c \left( \frac{\eta}{L}  \right)^2 \right)^{(1-h)/2} \right)}\\\nonumber
\left( \left( \frac{r}{L} \right)^{3-D(h)} + c \left( \frac{\eta}{L} \right)^{3-D(h)} \right)
dh.
\end{align}

\end{document}          

And it works!

  • 1
    I don't think your claim that "Probably bigint doesn't work with align" is correct; see my answer for successful uses of \bigintsss and \bigintssss in an align environment. You mentioned earlier that you want the integral symbol(s) to look "professional". IMNSHO, the "tall" integral symbol that's produced by your \scaleint macro is rather skinny and also too upright relative to the built-in slant of TeX's math italics (which is picked up the ordinary "large" integral sign and is preserved in the integrals produced by the macros of the bigints package). – Mico May 15 '14 at 20:26

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