\int {d^D k \over (2 \pi)^D} \phi_k(t) e^{i k.x} .\end{aligned}$$ The fact that $\phi(x,t)$ is real implies $\phi_k = \phi_{-k}^*$ and similarly for $\xi_k$. Because of these relations it is more convenient to work with $\phi_k^R = {\rm Re}(\phi_k)$ and $\phi_k^I = {\rm Im}(\phi_k)$, which obey the Langevin Eq. (\[Lang\_k\]) with $\xi_k^{R,I}$: $$\begin{aligned}
 \langle \xi_k(t)^{R,I}\rangle = 0, \quad 
\langle \xi_k^{R,I} (t)\xi_k^{R,I}(t') \rangle = {\Omega \over 2} 
\delta(t-t').\end{aligned}$$ The Fokker-Planck equation for the modes now is $$\begin{aligned}
&& \eta \partial_t P_k = -{\cal H}_k P_k \\
&& {\cal H}_k = - {\delta \over \delta \phi_k^{R,I}} 
\left[ {\Omega  \over 4 \eta} {\delta \over \delta \phi_k^{R,I}}  
+ \left(k^2 + m^2 \right) \phi_k^{R,I}  \right].
\label{FPk}\end{aligned}$$ which holds, as indicated, both for the real and imaginary components of $\phi_k$. The solution of the Fokker-Planck equation then follows as $$\begin{aligned}
P = \Pi_{k=0}^\infty \left[ P_k(\phi_k^R) P_k(\phi_k^I) \right]\end{aligned}$$ The time independent solution is clearly $$\begin{aligned}
P_k(\phi_k^R) P_k(\phi_k^I) \propto \exp \left[ -\beta (k^2 + m^2) \left( 
{\phi_k^R}^2 + {\phi_k^I}^2 \right) \right]\end{aligned}$$ which leads to $$\begin{aligned}
P_0 =&& {\cal N}\exp \left[ -\beta \int_0^\infty  {d^D k \over (2 \pi)^D}~
 (k^2 + m^2) \left( 
{\phi_k^R}^2 + {\phi_k^I}^2 \right) \right]  \\
 =&& {\cal N}\exp \left[ -\beta \int_{-\infty}^{+\infty}
{d^D k \over (2 \pi)^D}~ 