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?