\subsection{General case integral}
\label{sec:integrals:general}
-We will show that, for any $(a,b > 0) \in \Reals$,%
-\nomenclature[aR]{\Reals}{Real numbers}
+We will show that, for any $(a,b > 0) \in \Reals$,
\begin{equation}
I = \iInfInf{z}{\frac{1}{(a^2-z^2)^2 + b^2 z^2}} = \frac{\pi}{b a^2} \;.
\end{equation}
+%
+\nomenclature[so aR ]{\Reals}{Real numbers.}
First we note that $\abs{f(z)} \rightarrow 0$ like $\abs{z^{-4}}$ for
$\abs{z} \gg 1$, and that $f(z)$ is even, so