Coefficient matrix of quadratic Lagrangian I've been studying path integrals from Weinbergs QToF vol 1. He says that when the $\mathcal{L_0}$ is quadratic in fields we can always write free term $I_0$ in the generalized quadratic form 
$$I_0[\psi]=-\frac{1}{2}\int d^4xd^4y\; D_{x,x'}\;\psi(x)\psi(x'),$$ $$I_0[\psi]=\int d^4x\: \mathcal{L_0}(\psi(x),\partial_\mu\psi(x)).$$
On p.398 he says that when considering real massive vector field we have the unperturbed Lagrangian 
$$\mathcal{L_0}=-\frac{1}{4}(\partial_\mu A_\nu-\partial_\nu A_\mu)(\partial^\mu A^\nu-\partial^\nu A^\mu)-\frac{1}{2}m^2A_\mu A^\mu,$$
and we can write the coefficient matrix $D_{x,x'}$ as 
$$D_{\rho x,\sigma y}=\left[\eta_{\rho\sigma}\frac{\partial^2}{\partial x^\mu \partial y_\mu}-\frac{\partial^2}{\partial x^\sigma \partial y^\rho}+m^2\eta_{\rho\sigma}\right]\delta^4(x-y).$$
However, I don't see it how to get the coefficient matrix straight from the Lagrangian. For example I know that $\partial_\mu=\frac{\partial}{\partial x^\mu}$, but I don't know how to interpret the first term in the matrix since it has the lowercase index in $y$. From my understanding we should be able to get the Lagrangian back when we substitute $D$ back in to the $I_0$. Any help regarding on how to interpret the coefficient matrix or how to get it from the Lagrangian is appreciated.
 A: First expand the product in $\mathcal{L}_0$:
$$\mathcal{L}_0=-\frac{1}{2}\left(\partial_{\mu}A_{\nu}\partial^{\mu}A^{\nu}-\partial_{\nu}A_{\mu}\partial^{\mu}A^{\nu}\right)-\frac{1}{2}m^2A_{\mu}A^\mu.$$
Now in the term $\partial_\mu A_\nu\partial^\mu A^\nu$, the vector fields $A$ have the same index, hence the metric tensor in the first term of $D_{\rho x,\sigma y}$. Also, the partial derivatives in this term have the same index, hence so do the partial derivatives in the first term of $D_{\rho x,\sigma y}$.
Specifically, (in what follows, $\partial_\mu\equiv\frac{\partial}{\partial x^{\mu}}$ and $\partial'_\mu\equiv\frac{\partial}{\partial y^{\mu}}$):
$$\eta_{\rho\sigma}\partial_\mu\partial'^\mu A^{\rho}(x)A^\sigma(y)=\partial_\mu A_\sigma(x)\partial'^\mu A^\sigma(y)$$
which after multiplying with the $\delta$-distribution and integrating over $y$ gives $\partial_\mu A_\nu(x)\partial^\mu A^\nu(x)$. This explains the first term of $D_{\rho x,\sigma y}$.
The second term follows from
$$\partial_\sigma\partial'_\rho A^\rho(x)A^\sigma(y)=\partial_\sigma A^\rho(x)\partial'_\rho A^\sigma(y),$$
which gives the term $-\partial_{\nu}A_{\mu}\partial^{\mu}A^{\nu}$ after multiplying with the $\delta$-distribution and integrating over $y$. The third term is trivial.
Combining all this, it follows that indeed $$I_0[A_\mu]\equiv \int d^4x\: \mathcal{L_0}(A_\mu(x),\partial_\mu A_\nu(x)) = -\frac{1}{2}\int d^4xd^4y\; D_{\rho x,\sigma y}\;A^\rho(x)A^\sigma(y).$$
