In Eq. (5.7) of his book "Statistical Physics of Fields", M. Kardar proposes the identity
$$ \langle e^{\sum_i a_ix_i} \rangle =\exp{\left[\sum_{ij}\frac{a_ia_j}{2}\langle x_ix_j \rangle\right]}\tag{5.7} $$
where the expectation value is taken with respect to a set $\{x_i\}$ of Gaussian distributed variables. I attempted to prove this as follows:
$$ \begin{align} \langle e^{\sum_i a_ix_i} \rangle &\propto\prod_i\int dx_i\,e^{a_ix_i}\,e^{-b_ix_i^2}\\ &=\prod_i\int dx_i\,\exp{\left[-b_i\left(x_i^2-\frac{a_i}{b_i}\,x_i+\frac{a_i^2}{4b_i^2}\right)+\frac{a_i^2}{4b_i}\right]}\\ &=\prod_ie^{\frac{a_i^2}{4b_i}}\int dx_i \exp{\left[-b_i\left(x_i-\frac{a_i}{2b_i}\right)^2\right]}\\ &=\prod_i\sqrt{\frac{\pi}{b_i}}\,e^{\frac{a_i^2}{4b_i}}. \end{align} $$
The product $\prod_i\sqrt{\frac{\pi}{b_i}}$ will disappear once the distribution is properly normalized; ie.
$$ \langle e^{\sum_i a_ix_i} \rangle =\prod_i e^{\frac{a_i^2}{4b_i}} $$
On the other hand
$$ \begin{align} \langle x_i^2\rangle &\propto\prod_i\int dx_i\,x_i^2\,e^{-b_ix_i^2}\\ &=\prod_{j\ne i}\sqrt{\frac{\pi}{b_j}}\int dx_i\,x_i^2\,e^{-b_ix_i^2}\\ &=\frac{1}{2b_i}\prod_j\sqrt{\frac{\pi}{b_j}} \end{align} $$
and similarly
$$ \langle x_i^2\rangle =\frac{1}{2b_i}. $$
Combining these expressions:
$$ \langle e^{\sum_i a_ix_i} \rangle =\prod_i\exp{\left[\frac{a_i}{2}\,\langle x_i^2\rangle\right]} =\exp{\left[\sum_i\frac{a_i}{2}\,\langle x_i^2\rangle\right]} $$
which agrees with Kardar's result, modulo the cross-terms. By the even-odd symmetry of the integrand, I understand that these cross-terms $\langle x_i x_j\rangle$ will vanish for $i\ne j$, and that Kardar is therefore free to add them to his identity with impunity, but I don't understand the point of doing so. Is there a reason Kardar is including these terms, or have I made a mistake in my derivation?
