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A special path integral

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May be $f(\vec{x}), \vec{g}(\vec{x})$ an arbitrary functions dependent on the coordinates $\vec{x}=(x,y,z)^T$. Defining the following function dependent on a 3-dimensional curve $\vec{\gamma(t)}$ parametrized by $t \in [0,1]$:

$S := \int_{R^3}\int_{R^3} (f(\vec{x})exp(\int_0^1 \vec{g}(\vec{\gamma(t)}) \vec{\gamma'(t)}dt)f(\vec{x'}))d^3x d^3x'$.

It was defined $\vec{x}=\vec{\gamma(0)}, \vec{x'}=\vec{\gamma(1)}$ and it was integrated over all possible endpoints of the curve $\vec{\gamma(t)}$. How I can compute

$I := \int Sd[\vec{\gamma(t)}]$ ($[\vec{\gamma(t)}]$ denotes that every possible path with endpoints $\vec{x}, \vec{x'}$ that is connected must be integrated up)?

This Integration is similar to Feynman's path-integral over all possible paths in Quantum field theory. The exponential I can expand into a taylor series, but how can I evaluate the products $\int \vec{\gamma'(t_1)} \otimes ... \otimes \vec{\gamma'(t_n)}d[\vec{\gamma(t)}]$ where every $\vec{\gamma(t_i)},i \in \{ 1,...,n \}$ has the same endpoints (there occur both open and closed paths)?

Every help would be greatly appreciated.

This post imported from StackExchange Physics at 2015-03-10 13:00 (UTC), posted by SE-user kryomaxim
asked Mar 9, 2015 in Theoretical Physics by kryomaxim (65 points) [ no revision ]
Are you aware that we, in general, don't really compute the path integral?

This post imported from StackExchange Physics at 2015-03-10 13:00 (UTC), posted by SE-user ACuriousMind
I am asking about integrating over a dyadic product of Tangent vectors.

This post imported from StackExchange Physics at 2015-03-10 13:00 (UTC), posted by SE-user kryomaxim

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