Casimir forces due to scalar field using Path integrals I have just started learning QFT. I have just completed scalar fields, which I learnt in using Canonical Quantisation and Path integrals. I did calculation of Casimir force between two metal plates using just free scalar field theory (using the vacuum energy). However, I am not able to find a way to do this thing using Path integrals and propagators. The partition function for the case of free scalar field (i.e KG field) turns out to be,
$$ Z[J] = \text{exp}\bigg(i\int \mathrm d^4x \;\mathrm d^4x'J(x')\Delta_F(x-x') J(x) \bigg) \qquad \qquad (1) $$
which after setting the $Z[J=0] =1$. I wish to know, how to approach my problem from here.
PS : I have not learnt vector or spinor fields yet. Most of the references or notes that I checked either assumed a prior knowledge of that or did not say how to quantise scalar fields.
EDIT : This is the integral to begin with right
$$ Z[J] = \frac{1}{Z_0} \int [d \phi] \text{exp}\bigg(-i\int d^4x \bigg[ \frac{1}{2}\phi (\Box  + m^2 -  i\epsilon)\phi - \phi J\bigg]\bigg) $$
All I did was to introduce $\phi \rightarrow \phi + \phi_0 $ and demand that
$$ (\Box + m^2 - i\epsilon)\phi_0 = J(x) $$ and $\Delta_F(x-x') $ is the Green's function involved in solving this equation.
Then I obtain the equation (1).
 A: What I would do is to calculate the effective action from integrating at one loop the propagator in a space with boundaries. The result is quite simple, schematically of the form $\mathrm{Tr}\log \Delta$ where $\Delta(x_1,x_2)$ is the propagator in position space. Indeed, the free action is quadratic in the field, $ S\sim -\frac{1}{2}\phi(\partial^2-m^2)\phi$, which makes the path integral Gaussian and hence explicitly calculable. 
Since $\Delta$ will depend on the geometry of your space, say the distance $L$ between two parallel planes, you will get that the vacuum energy depends on such a separation too. Taking minus the derivative w.r.t L gives the force. Of course, you need to have calculated what the propagator in such a non trivial space is, by solving e.g. the Klein- Gordon equations with boundary conditions at $y=L$ and $y=0$, $y$ being the coordinates orthogonal to the plates . This is not the usual Feynman propagator because of the non trivial boundary conditions (far away from the boundary you should recover Feynman). Note also that th effective action will be UV divergent (therefore you will need to regularize the trace above)  but its derivative w.r.t to $L$, the force, is finite and calculable.
