Given the function
\begin{equation}
G(X, y)=-\int\frac{g(x, y)}{\frac{\partial}{\partial y}g(x, y)}f(x)dx
\end{equation}
where $f(x)$ is a probability density function, $X$ is a random variable and $y$ is a choice variable, and $g(x, y)$ is a continuous function of both arguments, and the expected value
\begin{equation}
\mathbb{E}[g(X, y)]=\int g(x,y)f(x)dx.
\end{equation}
$\textbf{Question}$: What is the product of these two functions
\begin{equation}
G(X, y)\cdot\mathbb{E}[g(X, y)]?
\end{equation}
Is it equal to
\begin{equation}
-\int \frac{[g(x, y)]^2}{\frac{\partial}{\partial y}g(x, y)}f(x)dx
\end{equation}
or
\begin{equation}
-\int \frac{[g(x, y)]^2}{\frac{\partial}{\partial y}g(x, y)}[f(x)]^2dx?
\end{equation}
Many thanks
You could write it as $$-\int\frac{g(x, y)}{\frac{\partial}{\partial y}g(x, y)}f(x)dx \int g(x,y)f(x)dx$$ or combine the integrals into a double integral (changing the name of one of the integration variables) $$ - \int \int \frac{g(x, y)}{\frac{\partial}{\partial y}g(x, y)}f(x) g(z,y) f(z)\; dx \; dz$$ But there is no reason why it should be any of your other two integrals.