cv.complex variables – The loss of double periodicity (ellipticity)

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
$$f(mathfrak{a}_1, mathfrak{a}_2) equiv frac{vartheta_1(2mathfrak{a}_2)^2}{ prod_pm vartheta_4(pm 2mathfrak{a}_2 – mathfrak{a}_1)} frac{1}{vartheta_4(mathfrak{a}_1)^4} frac{vartheta_1(2mathfrak{a}_1)^2}{prod_pm vartheta_4(pm2mathfrak{mathfrak{b}} + mathfrak{a}_1)} .$$
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
$$int_0^1 d mathfrak{a}_2 prod_pm frac{vartheta_1(2mathfrak{a}_2)^2}{ vartheta_4(pm 2mathfrak{a}_2 – mathfrak{a}_1)} = frac{1}{ pieta(tau)^3 }frac{ vartheta’_4(mathfrak{a}_1) }{ vartheta_1(2mathfrak{a}_1) } .$$

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 ?