Path integral with double integration involving the free particle case

Question

Suppose we have the path integral:

\begin{equation} Z=\int \mathcal{D}x\mathcal{D}y\,\exp\left[-\frac{a}{2}\int_0^1 dt\,\left(\,\dot{x}(t)^2-\,\dot{y}(t)^2\right)\right]. \end{equation}

The integration over $x$ is trivial: it is the well known result for a free particle:

\begin{equation} \int \mathcal{D}x\,\exp\left[-\frac{a}{2}\int_0^1 dt\,\dot{x}(t)^2\right]=\sqrt{\frac{a}{2\pi}}\exp\left[-\frac{a}{2}(x_f-x_i)^2\right] \end{equation}

where $x(1)=x_f$, $x(0)=x_i$.

What about the other integration? It has the "wrong" sign. One could think to apply the same formula with $a\to-a$, hence getting an imaginary unit from the negative square root. Would be this correct? How would you compute this path integral?

The integral comes from https://doi.org/10.1103/PhysRevD.43.2572, equation (4.9). One should be able to pick up a convergent contour in the complex plane where the integral should converge. It is however not clear which contour to choose and why.

Answer 1

Hint: For a Gaussian integral $$ \int_{e^{i\phi}\mathbb{R}} \mathrm{d}z ~e^{-\frac{1}{2}re^{i\theta}z^2}~\stackrel{z=e^{i\phi} x}{=}~e^{i\phi} \int_{\mathbb{R}} \mathrm{d}x ~e^{-\frac{1}{2}re^{i(\theta+2\phi)}x^2}~=~\sqrt{\frac{2\pi}{re^{i\theta}}} \tag{1}$$ to be convergent, $$\begin{align}0~\leq&~{\rm Re} (e^{i(\theta+2\phi)})\cr &\Updownarrow\cr \phi~ \in~& \left[-\frac{\pi}{4}- \frac{\theta}{2},\frac{\pi}{4}- \frac{\theta}{2} \right] +\pi\mathbb{Z}. \end{align} \tag{2}$$ Notice that as long as the integral is convergent, it doesn't depend on the contour angle $\phi$ [up to a subtlety with the double-valueness of the square-root in eq. (1)].