Delta functional representation in response field formalism

Question

A general way of obtaining a field-theoretical description of Langevin dynamics is via the Martin-Siggia-Rose (MSR) response fields. This is essentially just representing the identity - up to some Jacobian - in terms of a path integral that takes into account the dynamical equation of interest (as explained is section 5.1 (eq 32) of this article (title: Field theoretic methods, by Tauber)):

\begin{align} 1 &= \int \mathcal{D} [S] \prod_{x,t} \delta \left( \partial_t S(x,t) - F[S] (x,t) - \zeta (x,t) \right) \\ &= \int \mathcal{D}[S] \mathcal{D}[i\tilde{S}] \exp\left[ -\int_{x,t} \tilde{S}(x,t) \left( \partial_t S(x,t) - F (x,t) - \zeta (x,t) \right) \right], \end{align} where $F$ and $\zeta$ are the deterministic force and noise, respectively. This identity can then be used to compute averages of various quantities w.r.t. to the noise realizations.

I have two questions regarding this:

1. in going from the first line to the second line, I would expect to get something like $\int \mathcal{D}[\tilde{S}] e^{-i\int_{x,t} \tilde{S} (\partial_t S - F - \zeta)} $ in similarity with the 1d delta function $\delta(x) = \int \frac{dk}{2\pi} e^{ikx}$. But it seems that here we are using a different representation, $\delta(x) = \int \frac{d \, ik}{2\pi}e^{-kx}$.
   I have seen it stated* that in order to ensure the convergence of the path integral, the integration needs to be performed along the imaginary axis, hence the $\mathcal{D}[i\tilde{S}]$ measure in the second line. I don't quite understand this statement; in particular, how is it that if the path integral has convergence issues if we integrate over real fields with complex weights (i.e., $\int \mathcal{D}[\tilde{S}] e^{i\int \tilde{S}(\ldots)}$), it suddenly becomes well-defined by integrating along the imaginary axis via a change of variables (i.e., $\int \mathcal{D}[i\tilde{S}] e^{-\int \tilde{S}(\ldots)}$) ?

2. this probably has a very simple answer, but does it matter if in the above formula I switch $\tilde{S}$ with $-\tilde{S}$ (such that we have $e^{\int \tilde{S}(\ldots)}$ instead of $e^{-\int \tilde{S}(\ldots)}$ for the weights)?

*Page 134 of "Critical Dynamics - A Field Theory Approach to Equilibrium and Non-Equilibrium Scaling Behavior"

Answer 1

As written above, yes, the integral needs to be taken along the imaginary axis. The reason for writing it this way is not because it improves convergence with respect to your second representation, but rather because when finding saddle-points, one often finds that (in your original representation) $\tilde S$ is real [1]. So it is convenient just to omit the $i$ by shoving it into the measure.

Regarding your second question, no, it does not matter.

[1]. This is not a contradiction, of course, because saddle-points can and do occur off the original integration contour.

Note: the above representation without a Jacobian is correct only for Ito discretization.

Answer 2

OP's questions seem to boil down to the fact that $$\delta(x) ~=~ \delta(\pm x) ~=~\int_{\mathbb{R}} \frac{dk}{2\pi} e^{\pm ikx}~\stackrel{\kappa=ik}{=}~\int_{i\mathbb{R}} \frac{d\kappa}{2\pi i} e^{\pm \kappa x}.$$