Ellptic\Jacobi theta function and its residue integral

Question

The Ellptic\Jacobi theta function is given by \begin{align} \theta_1(\tau|z)&=\theta_1(q,y)=-iq^{\frac{1}{8}}y^{\frac{1}{2}}\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^{k-1}) \\ &= -i\sum_{n\in \mathbb{Z}}(-1)^n e^{2\pi i z(n+\frac{1}{2})} e^{\pi i \tau(n+\frac{1}{2})^2} \end{align} Where $q=e^{2\pi i \tau}$, and $y=e^{2\pi i z}$ and dedkind eta function is given as

$\eta(\tau)=\eta(q)=q^{\frac{1}{24}} \prod_{n=1}^{\infty}(1-q^n) $

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Now i want to evaluate following relation $\theta'_1(\tau|0)=\theta'_1(q,1)=2\pi \eta(q)^3$

The results is well known (i guess...) but i can not find any papers or textbooks contain some procedure of this.

Answer 1

Rewrite the product in your first equation as $-iq^{1/8}y^{1/2}(1-y^{-1})\prod_{k=1}^{\infty}(1-q^k)(1-yq^k)(1-y^{-1}q^k)$ (just tweak the third part of the product), divide out $(1-y^{-1})$ from the equation, then use L'Hôpital's rule (with respect to $z$ as it approaches $0$) to calculate the limit of the left-hand theta expression; the result you want follows immediately.

You can also substitute $y=e^{2i\pi/3}$ and $q^{1/3}\mapsto q$ to evaluate the Dedekind eta function as a theta function (since $(1-q^{k/3}e^{2i\pi/3})(1-q^{k/3}e^{-2i\pi/3})(1-q^{k/3})=1-q^k$).

As far as I know, I don't think this problem needs residue calculus to solve it, so the title may be misleading.

Answer 2

From Antonio DJC's comment i explicit compute

\begin{align} &\theta_1(q, y)= -i y^{\frac{1}{2}}q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n-1} \right) \\ &\phantom{\Theta(q,y)}= -i y^{\frac{1}{2}}q^{\frac{1}{8}}(1-y^{-1}) \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} \right), \\ &\frac{\theta_1(q,y)}{1-y^{-1}}= -i y^{\frac{1}{2}} q^{\frac{1}{8}} \prod_{n=1}^\infty \left(1-q^n\right)\left(1-y q^n\right) \left(1-y^{-1}q^{n} \right) \\ & \lim_{z \rightarrow 0} \frac{\theta_1(\tau|z)}{1-y^{-1}} =\lim_{z \rightarrow 0} \frac{\theta_1'(\tau|z)}{y^{-2} 2\pi i} = \frac{\theta_1'(q,1)}{2\pi i} = -i q^{\frac{1}{8}} \prod_{n=1}^{\infty} (1-q^n)^3 = -i \eta(q)^3 \end{align} where we used $\frac{df(y)}{dz} = \frac{df}{dz} \frac{dy}{dz} = \frac{df}{dz} 2\pi i$. And Dedkind eta function is given as $\begin{align} \eta(q)=q^{\frac{1}{24}} \prod_{n=1}^\infty (1-q^n) \end{align}$