Consider an elliptic function $f(z)$, we know how to compute the integral $$ \int_0^1 f(z)dz, $$ by expanding $f(z)$ as a sum of $\zeta(z)$ and its derivatives, where the coefficients of the expansion are Laurent coefficients of $f(z)$ at its poles.
Question: is there a similar trick to compute $$ \int_0^1 \frac{ \vartheta_4'(z, \tau) }{ \vartheta_4(z, \tau) }f(z)dz ? $$ The $\vartheta_4$ factor slightly breaks the ellipticity $$ \frac{ \vartheta'_4(z + \tau|\tau) }{ \vartheta_4(z + \tau |\tau) } = \frac{ \vartheta'_4(z|\tau) }{ \vartheta_4(z|\tau) } + 2\pi i \ . $$