Why the functional integral of a functional derivative is zero?

I'm reading Quantum Field Theory and Critical Phenomena, 4th ed., by Zinn-Justin and on page 154 I came across the statement that the functional integral of a functional derivative is zero, i.e. $$\int [d\phi ]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} = 0$$ for any functional $F[\phi ]$.

I would be most thankful if you could provide a mathematical proof for this identity.

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Try discretizing spacetime. –  Michael Brown Aug 17 '13 at 8:06
@Michael Brown I tried and it doesn't work. –  user22208 Aug 17 '13 at 8:26
The functional integral of a total derivative vanishes, which results from the variation of fixed boundary conditions. –  soliton Aug 17 '13 at 8:27
@soliton Could you please write some equations to better understand your statement? I tried a simple case, $F[\phi] = \int_{-\infty}^{\infty} C(x)\phi (x)dx$. Then $\int [d\phi]\frac{\delta F}{\delta\phi (x)} = C(x)\int [d\phi] \neq 0$ –  user22208 Aug 17 '13 at 8:54
Sorry for late reply. As Trimok says, $F(-\infty) - F(\infty)$ makes sense. –  soliton Aug 18 '13 at 11:29

If the functional derivative

$$\tag{1} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}$$

exists (wrt. to a certain choice of boundary conditions), it obeys infinitesimally

$$\tag{2}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\int_M \!dx\sum_{\alpha\in J} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x).$$

OP's functional integral formula

$$\tag{3} \int [d\phi ]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} ~=~ 0$$

is really a shorthand for infinitely many integrations

$$\tag{4} \left[\prod_{y\in M,\beta\in J} \int d\phi^{\beta}(y) \right]\frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)} ~=~ 0.$$

Before we can proceed, the functional integral measure in (4) must be given a mathematical definition. The precise definition depends on context and method. Needless to say that a general mathematically rigorous definition of functional integrals is a well-known open problem in mathematics. For instance, one may try to construct the functional integral as an appropriate continuum limit of a discretized space-time $M$, as Michael Brown suggests in a comment.

Let us use DeWitt's condensed notation, where all indices (both continuous and discrete indices) are lumped together as

$$\tag{5} i~=~(\alpha,x)~\in~ I~:=~ J\times M,$$

and fields are written as

$$\tag{6} \phi^i ~:=~ \phi^{\alpha}(x)~,\qquad i~\in~ I.$$

We now discretize space-time $M$. The discretization means that we think of $I$ as a finite index-set. In other words, we now only have finitely many variables $\phi^i$, $i\in I$, in the theory. The functional derivative (1) [times$^1$ the volume $\Delta x$ of a single cell of the discretization] is replaced by a partial derivative

$$\tag{1'} \frac{\partial F[\phi]}{\partial\phi^{i}}.$$

An infinitesimally variation is given by the standard formula from calculus in several variables

$$\tag{2'}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\sum_{i\in I} \frac{\partial F[\phi]}{\partial\phi^{i}}\delta\phi^{i}.$$

Finally, OP's functional integral formula (4) becomes

$$\tag{4'} \left[\prod_{j\in I} \int d\phi^{j} \right] \frac{\partial F[\phi]}{\partial\phi^{i}}~=~ 0.$$

Equation (4') follows from the fact that an integral of a total derivative vanishes if the boundary contributions are zero.

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$^1$ Concerning dimensions of functional derivatives versus partial derivatives, see also this Phys.SE post.

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+1. Yes, this is exactly what I meant. Never knew that was called DeWitt notation though. –  Michael Brown Aug 17 '13 at 13:39
The expression : $[d\phi(x)] \frac{\delta F}{\delta \phi(x)}$ could be interpreted as a formal $dF(\phi)$ : $$\int [d\phi(x)] \frac{\delta F}{\delta \phi(x)} \sim \int \frac {\partial F}{\partial \phi_i} d\phi_i \sim \int dF(\phi) =F(+\infty) - F(-\infty)$$
So the left hand side of the expression is zero only for identical boudary conditions, for instance $F(-\infty) = F(+\infty)$
For instance, the function $F(\phi) = e^{- \frac{1}{2}\int dx~\phi^2(x)} =\Pi_x ~(e^{- \frac{1}{2} \phi^2(x)}$), is a valid function, because, at positive and negative infinite $\phi$, we have $F(\phi) = 0$
Your function $F = \int dx ~ C(x) \phi(x)$ is not valid because it takes different values at negative and positive infinite values of $\phi$. Moreover, the values of $F$ are infinite, for infinite $\phi$, so it is difficult to give a sense to $F(-\infty) - F(+\infty)$