Integral of $e^{-xy}$ is a Dirac Delta

Question

I was looking at papers about the SYK model page 33 (equation 112), in which they write

$$\int\mathscr{D}\Sigma\,\mathscr{D}G~\exp\left\{-\frac{N}{2}\int\limits_{\left[0,\beta\right]^2} d\tau~d\tau^{\prime}~\Sigma(\tau,\tau^{\prime})\left(G(\tau,\tau^{\prime})~-~\sum_{i=1}^N\psi_i(\tau)\psi_i(\tau^{\prime})\right)\right\}~=~1.\tag{112}$$

This would follow from $$\int d\sigma\,dg\ e^{-\sigma g}=1.\tag{*}$$

Wolfram alpha confirms this as a Cauchy principle value.

How would this be proved?

Answer 1

OP asks about heuristic manipulations of a functional integral in Ref. 1. This question (v5) seems to belong more to Phys.SE than Math.SE. Functional integration is a huge topic. In this answer, we will only discuss the underlying ordinary integral.

1. Since the Grassmann-odd variables $\psi_i=\psi_i^{\ast}$ are real supernumbers, the constraint $$G(\tau,\tau^{\prime})~-~\sum_{i=1}^N\psi_i(\tau)\psi_i(\tau^{\prime})$$ is an $\color{red}{\text{imaginary}}$ Grassmann-even supernumber. (Recall that complex conjugation reverses the order of supernumbers.)

2. OP's underlying integral (*) in eq. (112) is therefore$^1$ $$\int_{\mathbb{R}} \mathrm{d}\sigma~ e^{-\sigma g}~=~2\pi \delta(\color{red}{i}g), \qquad \color{red}{i}g~\in~\mathbb{R}. $$

References:

1. G. Sarosi, AdS2 holography and the SYK model, arXiv:1711.08482.

--

$^1$ Ref. 1 uses a so-called Euclidean formulation, cf. footnote 36.