Consider a meromorphic function $f(\mathfrak{a}_1, \mathfrak{a}_2)$, such that $$ \begin{align} f(\mathfrak{a}_1, \mathfrak{a}_2) = f (\mathfrak{a}_1 + 1, \mathfrak{a}_2) = f(\mathfrak{a}_1 + \tau, \mathfrak{a}_2) = f(\mathfrak{a}_1, \mathfrak{a}_2 + 1) = f(\mathfrak{a}_1, \mathfrak{a}_2+ \tau) \ . \end{align} $$ So essentially $f$ is elliptic w.r.t. both $\mathfrak{a}_{1,2}$.
My question is: consider the integral $$ I(\mathfrak{a}_1) \equiv \int_0^1 d \mathfrak{a}_2 f(\mathfrak{a}_1, \mathfrak{a}_2) \ . $$ Question: do we still have ellipticity $I(\mathfrak{a}_1 + 1) = I (\mathfrak{a}_1 + \tau) = I(\mathfrak{a}_1)$?
I once thought that the answer is of course positive, but recently I encounter an example where it is not.
Consider $$ f(\mathfrak{a}_1, \mathfrak{a}_2) \equiv \frac{\vartheta_1(2\mathfrak{a}_2)^2}{ \prod_\pm \vartheta_4(\pm 2\mathfrak{a}_2 - \mathfrak{a}_1)} \frac{1}{\vartheta_4(\mathfrak{a}_1)^4} \frac{\vartheta_1(2\mathfrak{a}_1)^2}{\prod_\pm \vartheta_4(\pm2\mathfrak{\mathfrak{b}} + \mathfrak{a}_1)} \ . $$ The function is elliptic w.r.t. both $\mathfrak{a}_{1,2}$.
Now we can integrate over $\mathfrak{a}_2$ from $0 \to 1$, using the formula (by first expanding the elliptic integrand in Weierstrass-$\zeta$ function and then integrate) $$ \int_0^1 d \mathfrak{a}_2 \prod_\pm \frac{\vartheta_1(2\mathfrak{a}_2)^2}{ \vartheta_4(\pm 2\mathfrak{a}_2 - \mathfrak{a}_1)} = \frac{1}{ \pi\eta(\tau)^3 }\frac{ \vartheta'_4(\mathfrak{a}_1) }{ \vartheta_1(2\mathfrak{a}_1) } \ . $$
Now, due to the presence of $\vartheta'_4(\mathfrak{a}_1)$, it is easy to see that $$ \int_0^1 f(\mathfrak{a}_1, \mathfrak{a}_2) d \mathfrak{a}_2 $$ is no longer elliptic in $\mathfrak{a}_1$ (it is still invariant under the unit shift, but not so for $\tau$-shift).
Why does the integration destroy the periodicity ?
