Gaussian functional integral I am trying to understand how a functional integral is carried out. We have
$$F = \frac{1}{2}\sigma \int d^2x|\vec{\nabla} h|^2$$
where $h(\vec{x})$ is a scalar function of the position vector $\vec{x}$. We want to compute the functional integral
$$\langle (h(\vec{y})-h(\vec{0}))^2\rangle=\frac{\int Dh(\vec{x})|h(\vec{y})-h(\vec{0})|^2e^{-F/k_BT}}{\int Dh(\vec{x})e^{-F/k_BT}}$$
It is said that this integral can be computed in Fourier space to yield
$$\langle (h(\vec{y})-h(\vec{0}))^2\rangle=\frac{2k_BT}{\sigma}\int \frac{d^2q}{(2\pi)^2}\frac{1}{q^2}(1-e^{i\vec{q}\cdot\vec{y}})$$
Could you help me figure our how this integral is carried out?
 A: Trying to replicate the steps in the text the first they do is say that 
$\int d^4x (\partial_{\mu}\phi(x_i))^2-m^2\phi(x_i)=\frac{1}{V}\sum_n(|k_n|^2-m^2)|\phi(k_n)|^2$, simply writing the function as a fourier series should be enough but even so at first I wasn't able until it clicked that this equality is the plancherel/parseval identity for derivatives in a multidimensional case, so proving it was copy-pasting
$$
F
=\int d^2x|\partial_{\mu}\phi(x_i)|^2=\int d^2x(\partial_{\mu}(\frac{1}{V}\sum_n e^{-ik_nx_i}\phi(k_n))\partial_{\mu}(\overline{\frac{1}{V}\sum_m e^{-ik_nx_i}\phi(k_m)})
\\=\int d^2x\frac{1}{V^2}\sum_n\sum_m e^{i(k_m-k_n)x_i}k_n\overline{k_m}\phi(k_n)\overline{\phi(k_m)}
=\frac{1}{V}\sum_n\sum_m (\int \frac{d^2x}{V}e^{i(k_m-k_n)x_i})k_n\overline{k_m}\phi(k_n)\overline{\phi(k_m)}
\\=\frac{1}{V}\sum_n\sum_m \delta(k_m-k_n)k_n\overline{k_m}\phi(k_n)\overline{\phi(k_m)}
=\frac{1}{V}\sum_n|k_n|^2|\phi(k_n)|^2
$$
So using norm notation we have shown that
$$
||\partial_{\mu}\phi(x)||\equiv\int d^2x|\partial_{\mu}\phi(x_i)|^2
=\frac{1}{V}\sum_n|k_n|^2|\phi(k_n)|^2\equiv||k\phi(k)||
$$

Then they say that $\prod_{i}\int d\phi(x_i)=\prod_{k_n}\int dRe(k_n)dIm(k_n)$ and argue is true because is a unitary transformation and it kinds of makes sense intuitively but I'm confused about how to actually calculate this, in any case I think without decomposing it in real and imaginary parts this should be $\prod_{i}\int d\phi(x_i)=\prod_{k_n}\int d\phi(k_n)$
Or using path integral notation
$$\int D\phi(x_i)\equiv\prod_{i}\int d\phi(x_i)=\prod_{k_n}\int d\phi(k_n)\equiv\int D\phi(k_n)$$

Next step would be to calculate the denominator, here we only need to know that $\int_{-\infty}^{\infty} e^{-a x^2}\,dx = \sqrt{\pi \over a}$, they decompose the norm in real and imaginary parts but I fail to see why this is necessary so boldly don't do so
$$
\int D\phi \exp(-\frac{\sigma}{2k_{B}T}F)
=(\prod_{k_n}\int d\phi(k_n))\exp(-\frac{\sigma}{k_{B}TV}\sum_n|k_n|^2|\phi(k_n)|^2)
\\=\prod_{k_n}(\int d\phi(k_n)\exp(-\frac{\sigma|k_n|^2}{k_{B}TV}|\phi(k_n)|^2))
=\prod_{k_n}\sqrt{\frac{\pi k_{B}TV}{\sigma|k_n|^2}}
$$
Or using the notation for determinants
$$
\int D\phi \exp(-\frac{\sigma}{2k_{B}T}F)\equiv\det{\frac{\sigma}{2k_{B}T}F}
$$

Now we need to calculate the numerator and like in the previous two steps again I don't use imaginary and real parts, I got lost in the analysis of how they decide which terms contribute and which don't so I simply copy-pasted to our case without much thinking.
$$
\int D\phi \langle\phi(x_i),\phi(x_j)\rangle\exp(-\frac{\sigma}{2k_{B}T}F)
\\=\int D\phi (\frac{1}{V}\sum_n e^{-ik_nx_i}\phi(k_n))(\overline{\frac{1}{V}\sum_m e^{-ik_nx_j}\phi(k_m)})\exp(-\frac{\sigma}{2k_{B}T}F)
\\=\int D\phi (\frac{1}{V^2}\sum_n\sum_m e^{i(k_mx_j-k_nx_i)}\phi(k_n)\overline{\phi(k_m)})\exp(-\frac{\sigma}{2k_{B}T}F)
=(\prod_{k_l}\int d\phi(k_l))(\frac{1}{V^2}\sum_n\sum_m e^{i(k_mx_j-k_nx_i)}\phi(k_n)\overline{\phi(k_m)})\exp(-\frac{\sigma}{k_{B}TV^2}\sum_l|k_l|^2|\phi(k_l)|^2)
=_{somehow}(\prod_{k_l}\sqrt{\frac{\pi k_{B}TV}{\sigma|k_l|^2}}
)\frac{1}{V^2}\sum_n e^{ik_n(x_j-x_i)}\frac{k_{B}TV}{|k_n|^2}
\\=(\int D\phi \exp(-\frac{\sigma}{2k_{B}T}F))(\frac{k_{B}T}{\sigma})\frac{1}{V}\sum_n e^{ik_n(x_j-x_i)}\frac{1}{|k_n|^2}
\\=(\int D\phi \exp(-\frac{\sigma}{2k_{B}T}F))(\frac{k_{B}T}{\sigma})\int \frac{d^2 q}{(2\pi)^2}e^{iq(x_j-x_i)}\frac{1}{q^2}
$$
Or writing it as a fraction
$$
\frac{\int D\phi \langle\phi(x_i),\phi(x_j)\rangle\exp(-\frac{\sigma}{2k_{B}T}F)
}{\int D\phi \exp(-\frac{\sigma}{2k_{B}T}F)}
=\frac{k_{B}T}{\sigma}\int \frac{d^2 q}{(2\pi)^2}e^{iq(x_j-x_i)}\frac{1}{q^2}
$$

Finally 
$$
\langle (h(\vec{y})-h(\vec{0}))^2\rangle
=\langle 0|(h(\vec{y})-h(\vec{0}))^2|0\rangle
=\langle 0|h(\vec{y})^2|0\rangle-2\langle 0|h(\vec{y})h(\vec{0})|0\rangle+\langle 0|h(\vec{0})^2|0\rangle
\\=\frac{k_{B}T}{\sigma}\int \frac{d^2 q}{(2\pi)^2}e^{iq(\vec{y}-\vec{y})}\frac{1}{q^2}-2\frac{k_{B}T}{\sigma}\int \frac{d^2 q}{(2\pi)^2}e^{iq(\vec{y}-\vec{0})}\frac{1}{q^2}+\frac{k_{B}T}{\sigma}\int \frac{d^2 q}{(2\pi)^2}e^{iq(\vec{0}-\vec{0})}\frac{1}{q^2}
\\=\frac{2k_{B}T}{\sigma}\int \frac{d^2 q}{(2\pi)^2}(1-e^{iq\vec{y}})\frac{1}{q^2}
$$
