homework and exercises - Calculating the generating functional for the free scalar field explicitly by completing the square

I'm trying to reproduce the calculation resulting in equation (3.12) in the following script:

http://www.desy.de/~jlouis/Vorlesungen/QFTII11/QFTIIscript.pdf

The only difference is that in my notes, the Feynman propagator was defined with an extra minus sign. Let $A_x = (\Box_x + m^2)$, then

$$A_xD(x-y) = -i\delta(x-y).$$

The generating functional is given by:

$$Z[J] = \int D \phi \exp{i\int d^4x(-\frac{1}{2}\phi A \phi +J\phi)}$$

Let $I = \int d^4x(-\frac{1}{2}\phi A \phi +J\phi)$ and change variables

$$\phi(x) = \phi'(x) + \phi_0 (x) \equiv \phi'(x) + i\int d^4y D(x-y)J(y)$$

so that $D\phi'=D\phi$ and

$$A_x\phi_0(x) = J(x)$$

Then

$$I = \int d^4x \left( -\frac{1}{2} \left[ \phi'A\phi' + \phi'A\phi_0 + \phi_0A\phi' + \phi_0A\phi_0 \right] +J\phi' + J\phi_0\right)$$

this simplifies to

$$I = \int d^4x \left( -\frac{1}{2} \phi'A\phi' -\frac{1}{2} \phi_0A\phi' +\frac{1}{2} J\phi' + \frac{1}{2} J\phi_0\right)$$

Now, we just want to keep the $-\frac{1}{2} \phi'A\phi'$ and $\frac{1}{2} J\phi_0= \frac{i}{2} \int d^4y J(x)D(x-y)J(y)$ parts. The first one is independent of $J$ and provides the normalization, and the second, upon multiplication by $i$ and exponentiation, leads to the correct answer:

$$Z[J] = N e^{-(1/2)\iint d^4x d^4y J(x)D(x-y)J(y)}.$$

However, I absolutely can't see why the other two parts of the expression for $I$ need to cancel. That is, why the following holds:

$$-\frac{1}{2} \phi_0(x) A\phi'(x) +\frac{1}{2} J(x)\phi'(x) = 0?$$

Answer

The cancellation occurs by integrating the term $-\frac12 \phi_0(x)A\phi'(x)$ twice by parts.

Ignoring surface terms and the $-\frac12$ coefficient, this is:

$$\begin{aligned} \int d^4x ~\phi_0(x)A\phi'(x) &= \int d^4x ~\phi_0(x)\partial^2\phi'(x) + \phi_0(x)m^2\phi'(x)\\ &= -\int d^4x ~\partial\phi_0(x)\partial\phi'(x) + \int d^4x ~\phi_0(x)m^2\phi'(x)\\ &= \int d^4x ~\partial^2\phi_0(x)\phi'(x) + \int d^4x ~\phi_0(x)m^2\phi'(x)\\ &= \int d^4x ~A\phi_0(x)\phi'(x)\\ &= \int d^4x J(x)\phi'(x) \end{aligned}$$