How to write a big integral in an " align" environment?

Question

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}          

Answer 1

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.

\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.

\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}

Answer 2

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!