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Let's say I need to solve this functional integral: $$ Z=\int \mathcal{D}x(\lambda)F[x(\lambda)] $$ Then, I want to change the integration to $\mathcal{D}v(\lambda) $. Where $v$ is a shorthand for $v^\mu =\frac{dx^\mu}{d\lambda}$. The reason for this could be, for example, that in the functional $F$ there is no $x$ (just derivatives of $x$) and (for example) integrals of the form: $$\int_{\lambda_i}^{\lambda_f}\frac{dx^\mu}{d\lambda}\frac{dx_\mu}{d\lambda} d\lambda$$ So my guess is that $$Z=\int \mathcal{D}v(\lambda) \det(\frac{\delta x(\lambda)}{\delta v(\lambda)}) G[v(\lambda)]$$ where $$G[v(\lambda)]=F[x(\lambda)]|_{\frac{dx^\mu}{d\lambda}=v^\mu(\lambda)}$$

I wonder if the jacobian would be something relevant, or perhaps something kind of trivial, that I can take out in a constant N $$Z=N\int \mathcal{D}v(\lambda) G[v(\lambda)]$$

Are my steps correct? What would be the functional determinant?

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  • $\begingroup$ I don't believe this is possible in the general case. Your path integrals don't mean anything without specifying boundary conditions. And switching from $x$ to $v$ is not a regular change of variables when there's boundary conditions. At least, if you discretize the path integral such that it becomes a finite product of ordinary integrals, you can't do the $x \rightarrow v$ change of variables in these ordinary integrals such that it preserves the boundary conditions. $\endgroup$ – Prof. Legolasov May 9 '17 at 0:56

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