# What is the functional derivative of an integral with d'Alembertian Operator?

I want to take the functional derivative of an integral with a d'Alembertian Operator:

$$\frac{\delta }{\delta F(x)} \int d^4y\,G(x) \partial_\mu \partial^\mu F(y)$$

I believe this is related to the product rule (or integration by parts) and tried the following:

$$\partial_\mu \partial^\mu (F\cdot G) =\partial_\mu \left( F \partial^\mu G + G \partial^\mu F \right )= 2 \partial_\mu G\, \partial^\mu F +F\partial_\mu \partial^\mu G +G\partial_\mu \partial^\mu F$$ which implies: $$\int d^4y\, G\,\partial_\mu \partial^\mu F = \int d^4y\, \partial_\mu \partial^\mu (F\cdot G) - 2\int d^4y\, \partial_\mu F \,\partial^\mu G -\int d^4y\,F\, \partial_\mu \partial^\mu G$$ And although I know that:

$$\frac{\delta}{\delta F(x)} \int d^4y\,F(y)h(y) = h(x)$$ and $$\frac{\delta}{\delta F(x)} \int d^4y \partial_\mu F(y)V(y)^\mu = - \partial_\mu V^\mu(x)$$ which can help me with the second term.

I still don't know what to make of the term: $$\int d^4y\, \partial_\mu \partial^\mu (F\cdot G)$$

You can use the identity $$G \left(\partial_\mu \partial^\mu F\right) = \partial_\mu \left( G \partial^\mu F\right) - \partial^\mu \left( \left(\partial_\mu G\right) F\right) + \left(\partial_\mu \partial^\mu G\right) F \, . \quad (*)$$ Then you have $$\int \text{d}^4x \, G \left(\partial_\mu \partial^\mu F\right) = \int \text{d}^4x \, \left(\partial_\mu \partial^\mu G\right) F \, ,$$ because the first two terms on the right-hand side of (*) are surface terms that do not contribute to the integral. The functional derivative is then easy $$\frac{\delta}{\delta F(x)} \int \text{d}^4x \, G \partial_\mu \partial^\mu F = \partial_\mu \partial^\mu G(x) \, .$$