Funcional Integral in QFT I am reading Peskin and Schroeder's chapter on functional methods. They propose the amplitude:
$$
\langle \phi_b(\vec{x})|e^{-iHT}| \phi_a(\vec{x})\rangle 
=
\int \mathcal{D}\phi\mathcal{D}\pi 
\exp \bigg[
 i\int_0^T \left(
 \pi \dot\phi - \frac{1}{2}\pi^2 - \frac{1}{2} (\nabla\phi)^2-V(\phi)
 \right) \bigg]
$$
They then said that we can complete the square and integrate over $\mathcal{D}\pi$ integral. How is this done?
I was able to complete the square: $\pi\dot\phi - \frac{1}{2} \pi^2=-\frac{1}{2}(\pi-\dot\phi)^2+\frac{1}{2}\dot\phi^2$ and rewrite the amplitude as:
$$
\langle \phi_b(\vec{x})|e^{-iHT}| \phi_a(\vec{x})\rangle 
=
\int \mathcal{D}\phi
\exp \bigg[
 i\int_0^T \left(
 \frac{1}{2} \partial_\mu\phi\partial^\mu\phi-V(\phi) 
 \right) \bigg]
\int \mathcal{D}\pi \bigg[ -\frac{1}{2}i \int^T_0 (\pi-\dot\phi^2) \bigg]
$$
I am only confused about the functional integral (mathematically), how is it done?
 A: You do the following
$$\int D\pi \exp[ -\frac12 i \int_0^T (\pi-\dot \phi)^2] =\int D\pi \exp[-\frac12 i\int_0^T \pi^2]= constant$$
where I've shifted $\pi\to \pi+\dot \phi$ which leaves the measure $D\pi$ invariant. The result is a $\phi$ independent multiplicative constant. It's not quite well-defined; you'll need to determine this constant by other means, if necessary.
A: well functional integral can be thought as vector integral
$$I=\int d\vec{v}e^{-\frac{1}{2}\vec{v}^tA\vec{v}}=\prod_{a}\int dv_ae^{-1/2u_a^2(\lambda_{a})}=\sqrt{\det(2\pi A^{-1})}$$
where $\lambda_a(u_a)$ is eigenvalues(vectors) of $A$ the above calculation is very easy gaussian vector integral. Now instead of discrete vector we have continuous "vectors"
$$I=\int dv(x)e^{-\frac{1}{2}\int dx\int dy\ v(x)A(x,y)v(y)}=\prod_{x}\int du(x)e^{-1/2u(x)^2(\lambda(x))}\propto\sqrt{\det( A^{-1})}$$
in your case,
$$I=\int dv(x)e^{-\frac{1}{2}\int dx\int dy\ v(x)A(x,y)v(y)+\int dxj(x)v(x)}\propto\sqrt{\det( A^{-1})}\int d v(x)e^{\frac{1}{2}\int d x\int dy  v(x)A^{-1}(x,y)v(y)}$$
so this last equation is automatically square completed version, just apply $\pi$ integral to your case in your case:
$A=i$ and$j=i\dot{\theta}$ so if we apply our formula we get, $$\text{constant}\times e^{i1/2\int dx\dot{\theta}^2}$$
