# A Question about Path Integral Measure

I want to do the following path integral.

$$\mathcal{Z}=\int\mathcal{D}x e^{iS[\dot{x}]}$$

The action only denpends on $$\dot{x}$$. For some reason, I want to replace the integral measure $$\mathcal{D}x$$ by $$\mathcal{D}\dot{x}$$.

So I have

$$\mathcal{Z}=\int\mathcal{D}\dot{x}\mathrm{Det}\left(\frac{\delta x}{\delta\dot{x}}\right)e^{iS[\dot{x}]}.$$

The variable $$x$$ is related with $$\dot{x}$$ via the linear transformation

$$x(t)=\int_{0}^{t}\dot{x}(s)ds,$$

which implies

$$\mathrm{Det}\left(\frac{\delta x}{\delta\dot{x}}\right)\equiv 1.$$

Am I correct in the above derivation?

1. For the corresponding problem with discretized time, the Jacobian determinant of the coordinate transformation $$(x^0,x^1,\ldots, x^N)\qquad\longrightarrow \qquad (x^0,v^{1/2},\ldots, v^{N-1/2}),$$ where $$v^{j+1/2}~:=~\frac{x^{j+1}-x^j}{\Delta t} ,$$ would be $$\det=(\Delta t)^{-N}$$, not unity.
2. For continuum time, the velocity is $$v(t)~=~\frac{dx(t)}{dt}~=~\int \!dt^{\prime} x(t^{\prime}) \frac{d}{dt}\delta(t\!-\!t^{\prime}).$$ Whether the functional determinant $${\rm Det}\frac{\delta v(t)}{\delta x(t^{\prime})}$$ is unity (or not) depends on regularization scheme and boundary conditions. However, see also this related Phys.SE post.