# On the computation of functionals in QFT

Using the Gaussian (path)-integral

$$\int \mathcal{D}\eta e^{i\int_{t_i}^{t_f} dt \eta(t) O(t) \eta(t)} = N [\operatorname{det} O(t)]^{-1/2}$$

my book claims that we can compute the following integral as

$$N e^{\frac{i}{\hbar}S[x_{cl}]}\int \mathcal{D} \eta \exp\left(\frac{i}{2\hbar} \int \int dt_1 dt_2 \eta(t_1)\frac{\delta^2 S[x_{cl}]}{\delta x_{cl}(t_1) \delta x_{cl}(t_2)}\eta(t_2)\right) =\frac{N}{\sqrt{\operatorname{det}\Big(\frac{1}{\hbar} \frac{\delta^2 S[x_{cl}]}{\delta x_{cl}(t_1) \delta x_{cl}(t_2)}}\Big)}e^{\frac{i}{\hbar}S[x_{cl}]}\tag{7.35}$$

using the above identity. What happened with one integration over $$t_2$$ in applying the formula? Why is this valid?

Hint: For a local action functional $$S[x]$$ the Hessian $$\frac{\delta^2 S[x]}{\delta x(t_1) \delta x(t_2)}$$ is proportional to a Dirac delta distribution $$\delta(t_1\!-\!t_2)$$. See e.g. eq. (7.37) in Ref. 1.