Is the following logic of simplifying the complicated expression correct? Assume 
I have three variables $s, \omega,\gamma$ and define a function $G(.)$. 

Next consider the following function

\begin{align}
\prod_{{i&gt;1}}\left(\int_{-\infty}^{+\infty}\Bigg(\frac{1}{1+s  G(\omega_i)\gamma_{i}^{-1} }\Bigg)\ d\omega_i\right)\\\nonumber
\end{align}

Notie that $\omega_i$ and $\gamma_i$ are both indexed by $i$.

Is it true that I can re-write the above as

\begin{align}
\prod_{{i&gt;1}}h(\gamma_i,s) \\\nonumber
\end{align}

where 

$$h(\gamma_i,s)= \int_{-\infty}^{+\infty}\Bigg(\frac{1}{1+s  G(\omega_i)\gamma_{i}^{-1} }\biggl) \ d\omega_i$$

It is the indexing under the $\prod$ that is confusing me, since both the integrand has index $i$ and the product outside too, is my logic correct?
Maybe it should be made clear what $\gamma_i$ and $\omega_i$ are, e.g. they are different possible values of $\gamma$ and $\omega$ (or components maybe)? In any case what you suggest here is simply writing the complicated-looking integral expression as a function with two parameters $s$ and $\gamma_i$. There are two possible explanations: either $\omega_i$'s are arbitrary dummy variables and the author wanted to emphasize that these are different integrals, or $\forall i$: $\gamma_i$ and $\omega_i$ are related, for instance componentwise. Depending on the context, it might disable you to first add the terms and then integrate.
Maybe it should be made clear what $\gamma_i$ and $\omega_i$ are, e.g. they are different possible values of $\gamma$ and $\omega$ (or components maybe)? In any case what you suggest here is simply writing the complicated-looking integral expression as a function with two parameters $s$ and $\gamma_i$. There are two possible explanations: either $\omega_i$'s are arbitrary dummy variables and the author wanted to emphasize that these are different integrals, or $\forall i$: $\gamma_i$ and $\omega_i$ are related, for instance componentwise. Depending on the context, it might disable you to first add the terms and then integrate.