Infinite product representations of theta functions
Let $f(z) = \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})$. Each of these two products converge absolutely and uniformly in any bounded domain of values of $z$. Hence $f(z)$ is analytic throughout the finite part of the $z$ plane. The zeros of $f(z)$ are simple zeros at the points $z=\frac{2n+1}{2}\pi \tau i + m\pi i$ where $m,n\in \mathbb{Z}$. So, the zeros of $f(z)$ coincide with the zeros of $\vartheta_4(z)$. Therefore, the function $\frac{\vartheta_4(z)}{f(z)}$ has no poles or zeros in the finite part of the plane. Also, it is easy to see that $f(z+\pi) = f(z)$ and $$ \begin{aligned} f(z+\pi \tau) &= \prod_{n=1}^\infty (1-q^{2n+1}e^{2iz})(1-q^{2n-3}e^{-2iz}) \\ &= f(z) \frac{1-q^{-1}e^{-2iz}}{1-qe^{2iz}} \\ &= -q^{-1}e^{-2iz} f(z) \end{aligned} $$ This is, $\frac{\vartheta_4(z)}{f(z)}$ is a doubly periodic function with periods $\pi,\pi\tau$ and has no poles and zeros. By section 20.12 of [2], it is simply a constant, say $G$. We have $$ \vartheta_4(z) = G \prod_{n=1}^{\infty} (1-q^{2n-1}e^{2iz})(1-q^{2n-1}e^{-2iz})\quad (1) $$ Incrementing $z$ by the half periods $\frac{\pi}{2}, \frac{\pi \tau}{2}$ and $\frac{\pi +\pi\tau}{2}$, we get the product representations for the other theta functions: $$ \begin{aligned} \vartheta_3(z) &= G \prod_{n=1}^{\infty} (1+q^{2n-1}e^{2iz})(1+q^{2n-1}e^{-2iz})\quad (2) \\ \vartheta_1(z) &= 2G q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n}e^{2iz})(1-q^{2n}e^{-2iz}) \quad (3) \\ \vartheta_2(z) &= 2G q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1+q^{2n}e^{2iz})(1+q^{2n}e^{-2iz}) \quad (4) \end{aligned} $$ Now, all that left is to find the constant $G$. From identity (3), it easy to see that $$ \vartheta_1′(0) = 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 $$ Now, using the identity $\vartheta_1′(0)=\vartheta_2(0)\vartheta_3(0)\vartheta_4(0)$, we get: $$ \begin{aligned} 2G q^{\frac{1}{4}}\prod_{n=1}^{\infty} (1-q^{2n})^2 &= \left\{2G q^{\frac{1}{4}}\prod_{n=1}^{\infty}(1+q^{2n})^2 \right\} \times \left\{G\prod_{n=1}^{\infty}(1+q^{2n-1})^2 \right\} \\ &\quad \times \left\{G \prod_{n=1}^{\infty}(1-q^{2n-1})^2 \right\} \\ \implies G &= \prod_{n=1}^\infty (1-q^{2n}) \end{aligned} $$ Note that the rearrangements are justified since all products converge absolutely. Finally, we have $$ \begin{aligned} \vartheta_1(z) &= 2 q^{\frac{1}{4}}\sin(z) \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n} \cos(2z)+ q^{4n}) \quad (5) \\ \vartheta_2(z) &= 2 q^{\frac{1}{4}}\cos(z) \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n} \cos(2z) + q^{4n}) \quad (6) \\ \vartheta_3(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1+2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (7) \\ \vartheta_4(z) &= \prod_{n=1}^{\infty} (1-q^{2n})(1-2q^{2n-1} \cos(2z) + q^{4n-2}) \quad (8) \end{aligned} $$
Derivatives of Ratios of Theta Functions
Consider the function $\phi(z) = \frac{\vartheta_1′(z)\vartheta_4(z) – \vartheta_1(z)\vartheta_4′(z)}{\vartheta_2(z)\vartheta_3(z)}$. Now, $\phi(z)$ is doubly periodic with periods $\pi$ and $\frac{\pi\tau}{2}$. Relative to these periods, the only possible poles of $\phi(z)$ are at points congruent to $\pi\over 2$. It is easy to verify that $\frac{\pi}{2}$ is a removable singularity. This means that $\phi(z)$ is a constant. Letting $z\to 0$, we see that $\phi(z)=\vartheta_4^2(0)$. It is therefore established that $$\frac{d}{dz}\left(\frac{\vartheta_1(z)}{\vartheta_4(z)} \right)=\vartheta_4^2(0) \frac{\vartheta_2(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (9)$$ Two other identities of this kind can be derived using the same technique: $$ \begin{aligned} \frac{d}{dz}\left(\frac{\vartheta_2(z)}{\vartheta_4(z)} \right) &=-\vartheta_3^2(0) \frac{\vartheta_1(z)\vartheta_3(z)}{\vartheta_4^2(z)} \quad (10) \\ \frac{d}{dz}\left(\frac{\vartheta_3(z)}{\vartheta_4(z)} \right) &=-\vartheta_2^2(0) \frac{\vartheta_1(z)\vartheta_2(z)}{\vartheta_4^2(z)} \quad (11) \end{aligned}$$ By writing $\xi = \frac{\vartheta_1(z)}{\vartheta_4(z)}$ and making use of identities (13) and (14) from part 1, we obtain: $$\left(\frac{d\xi}{dz}\right)^2 = (\vartheta_2^2(0) – \xi^2 \vartheta_3^2(0))(\vartheta_3^2(0)-\xi^2 \vartheta_2^2(0))$$ Using the change of variables $y = \frac{\vartheta_3(0)}{\vartheta_2(0)}\xi$ and $u = \vartheta_3^2(0) z$, the above differential equation can be written as: $$ \left(\frac{dy}{du}\right)^2 = (1-y^2)(1-k^2 y^2) $$ where $k = \frac{\vartheta_2^2(0)}{\vartheta_3^2(0)}$. This differential equation has the particular solution: $$y = \frac{\vartheta_3(0)}{\vartheta_2(0)} \frac{\vartheta_1(u \vartheta_3^{-2}(0))}{\vartheta_4(u \vartheta_3^{-2}(0))}$$ We can write $y$ as a function of $u$ and $k$ as $y = \text{sn}(u,k)$ or simply $\text{sn}(u)$. Clearly, $\text{sn}(u,k)$ has periods $2\pi \vartheta^2_3(0)$ and $\pi \tau \vartheta_3^2(0)$. It has two simple poles congruent to $\frac{1}{2}\pi\tau \vartheta_3^2(0)$ and $\pi \vartheta_3^2(0) + \frac{1}{2}\pi\tau \vartheta_3^2(0)$. Also, it is easy to see that $$\int_0^1 \frac{dy}{\sqrt{(1-y^2)(1-k^2 y^2)}} = \text{sn}^{-1}(1) = \frac{\pi}{2}\vartheta_3^2(0)$$ The integral on the left is denoted by complete elliptic integral of the first kind and is denoted by $K(k)$.
References
- Derek F. Lawden (1989). Elliptic Functions and Applications. Springer-Verlag New York
- E. T. Whittaker, G. N. Watson (1927). A Course of Modern Analysis. Cambridge University Press