# Path integral with double integration involving the free particle case

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.

• Strictly speaking,no it would not be correct. But usually physicists don’t care. The path integral is mathematically ill defined anyway. The physicists approach is - just do it and see what we get. If the final result makes sense, then the manipulation is probably correct. Mar 28 at 14:46
• Assuming $a>0$ the path integral over $y\in\mathbb{R}$ is divergent. Is this from a reference? Which page? Mar 28 at 15:20
• It is from journals.aps.org/prd/abstract/10.1103/PhysRevD.43.2572, equation (4.9). It is said that the integral can be easily calculated, but they do not show how. They talk about a contour in the complex plane but it is not clear which one to choose and why. Mar 28 at 16:08
• Am I reading something wrong in that paper? In equation (4.9), the action $I_2$ from equation (4.8) has (only) kinetic terms, it comes as $\frac{1}{8N}\int_0^1\mathrm{d}\tau (-\dot{X}^2+\dot{Y}^2)$. Apr 1 at 11:38

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)].