I am getting quite confused with Riemann bilinear relations.
Let $\Sigma$ be a compact Riemann surface of genus $g$, with a canonical homology basis $a_1,b_1,\dots,a_g,b_g$, with associated normalised holomorphic and anti-holomorphic one-forms $\omega_I,\bar \omega_I$, $I=1\dots,g$, such that $\oint_{a_I} \omega_J = \delta_{IJ}$.
Let further $\tau_{P,Q}$ be an abelian differential on $\Sigma$ with poles at $P,Q$ of residues $\pm1$, normalised so that its $a$-cycles periods are zero : $$\oint_{a_I} \tau_{PQ}=0.$$ It is called an abelian differential of the third kind and the normalisation guarantees that it is unique. Besides, it is closed on $\Sigma-\{P,Q\}$. For more details, see Bobenko or textbooks on Riemann surfaces like Farkas & Kra.
I've read in a few places the following statement: $$\iint_\Sigma \tau_{PQ} \wedge \bar \omega_J \propto \int_Q^P \Im \omega_J$$ up to some factors of $2,\pi,i$ and sign.
I can't get where the imaginary part comes from. Calling $f=\overline{\int^z\omega_J}$, it seems to me that the standard proof of the Riemann identities based on Stokes theorem, $\int_{\partial \Sigma} f \tau_{PQ} = \iint \bar \omega_J \wedge \tau_{PQ}$ will hold ($\partial \Sigma$ means the union of $a,b,a^{-1}$ and $b^{-1}$ cycles that constitute a canonical dissection of the surface; the standard 4g-gon) and because $\tau_{PQ}$ has zero a-periods this is evaluated to $$\iint_\Sigma \tau_{PQ} \wedge \bar \omega_J = \sum_{I=1}^g \oint_{a_I} \bar \omega_J \oint_{b_I} \omega_{P,Q}, $$which, using some reciprocity formula like $\oint_{b_i} \omega_{P,Q} = 2i\pi \int_Q^P \omega_i $ will yield $$\iint_\Sigma \tau_{PQ} \wedge \bar \omega_J = 2i\pi \int_Q^P \omega_J$$ without the imaginary part ?