Let $X$ be a compact Riemann surface, $u$ a meromorphic function on $X$ with divisor supported on a set of points $\{P\_1, ..., P\_n\}$, and $f$ a meromorphic function on $X$ such that $f$ has no pole or zeros on $\{P\_1, ..., P\_n\}$. Let $d\log(u)$ be the logarithmic differential form of $u$. Do we have the following equality in $\mathbf{C}^{\times}$:
$$\exp(\int\_{div(f)} d\log(u)) = \prod\_{i=1}^n f(P\_i)^{\text{ord}\_{P\_i}(u)}\text{ ?}$$
Here, $\text{div}(f)$ is the divisor of $f$, and the integral is well-defined (independant of the choice of the path) up to $2\pi i \mathbf{Z}$.


This identity is true if for all $i$ we have $f(P\_i)=1$, as it relates to the analytic Abel-Jacobi description of the generalized Jacobian of $X$ with respect to the divisor $(P\_1)+...+(P\_n)$.



Extended Abel-Jacobi theorem
This is true. Given two points $a,b\in X$ with $u(a)$ and $u(b)\neq 0$, one can choose a path from $a$ to $b$ and a determination of $\log u$ along that path, so that $\exp\int^b\_a d\log u=u(b)/u(a)$. Thus, if $ \operatorname{div}(f)=\sum n\_jQ\_j$, the left-hand side is $\prod u(Q\_j)^{n\_j}$. 


Recall that if $f,g$ are two meromorphic functions on $X$, of order $n$ and $m$ at a point $P$,
 the *tame symbol* $(f,g)\_{P}$ is $(-1)^{nm}[g^{n}/f^m](P)$. 
 Thus, if $\operatorname{div}(u)=\sum m\_iP\_i$, we have $(f,u)\_{P\_i}= f(P\_i)^{-m\_i}$, $(f,u)\_{Q\_j}= u(Q\_j)^{n\_j}$, and $(f,u)\_{P}=1$ at the other points. Now the product formula $\prod (f,u)\_P=1$ gives your equality. 


