We consider Brownian motion as a process
$$ V(0)=0, \qquad \overline{V(x)}=0, \qquad \overline{(V(x)-V(y))^2}= 2 |x-y|. $$
I do not know how to prove the statement:
$$ \overline{\log\left(\int \mathrm{d}x \, \exp\left(-\frac{x^2}{2} -V(x) \right) \right)} = \overline{\log\left(\int \mathrm{d}x \, \exp\left(-\frac{x^2}{2} - V(x+a) \right) \right)},$$
where $a$ is an arbitrary number
I had the following attempt:
$ V(x+a) = V(x) + V(x+a) - V(x)$ and $V(x+a) - V(x)$ is distributed independently of $x$. $V(x)$ and $V(x+a)-V(x)$ are both independent. The idea was to replace $V(x+a)-V(x)$ by the variable $u$, which is independent of $x$ and distributed the same as $V(x+a)-V(x)$.
The logarithm gives $-u$ and averaging gives $0$. But this substitution for $u$ seems unfair to me, and I don't know what to do.
Thank you for your help in advance!
Let me use the probabilist's notation that most users here (including myself) are more familiar with.
Write $\tilde{V}(x) = V(x + a) - V(a)$. Then
\begin{align*} \mathbb{E}\left[ \log \left( \int_{\mathbb{R}} \mathrm{d}x \, e^{-\frac{1}{2}x^2 - V(x+a)} \right) \right] &= \mathbb{E}\left[ \log \left( \int_{\mathbb{R}} \mathrm{d}x \, e^{-\frac{1}{2}x^2 - \tilde{V}(x) - V(a)} \right) \right] \\ &= \mathbb{E}\left[ \log \left( \int_{\mathbb{R}} \mathrm{d}x \, e^{-\frac{1}{2}x^2 - \tilde{V}(x)} \right) - V(a) \right] \\ &= \mathbb{E}\left[ \log \left( \int_{\mathbb{R}} \mathrm{d}x \, e^{-\frac{1}{2}x^2 - V(x)} \right) \right]. \end{align*}
In the last step, we utilized the fact that (1) $\tilde{V}$ and $V$ have the same law as stochastic processes, and (2) $\mathbb{E}[V(a)] = 0$ by the assumption.