# Series Expansion of JacobiZeta in Both Arguments

I am looking for a series expansion of the JacobiZeta function at the following argument values:

$$JacobiZeta[~ArcSin[1+a \epsilon]~,~1-b\epsilon~]_{\epsilon\approx 0}=?$$

where $a$ and $b$ are complex parameters and $\epsilon$ is going to zero. The main problem is due to the same power of $\epsilon$ appearing in both arguments at the respective places, since it causes conventional series for one argument not to terminate due to balancing of $\epsilon$ terms from the other argument after insertion.

Any help or suggestion how to proceed is appreciated!

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What do you mean that one argument does not terminate? –  Igor Rivin Jul 13 '13 at 14:05
By "conventional series for one argument" I mean series expansions around a specific value of one single argument, not both. But I actually obtained a solution for above problem an hour ago, will post it as an answer soon. –  Kagaratsch Jul 13 '13 at 15:47

## migrated from mathoverflow.netJul 13 '13 at 17:00

This question came from our site for professional mathematicians.

Basically, the question asks for the behaviour of the Jacobi Zeta function for $c\to0$

$$\mathcal{Z}(\arcsin(1+ac)|1-bc)_{c\approx0}=?$$

where $a$ and $b$ are parameters. Consider the definition:

$$\mathcal{Z}(z,m)=\mathbb{E}(z|m)-\frac{\mathbb{E}(m)}{\mathbb{K}(m)}F(z|m)$$

where $\mathbb{E}(z,m)$ is the incomplete elliptic integral of the first kind, $\mathbb{E}(m)=\mathbb{E}(\frac{\pi}{2}|m)$ is the complete version, $F(z|m)$ is the incomplete elliptic integral of the second kind and $\mathbb{K}(m)=F(\frac{\pi}{2}|m)$ is again the complete version. Series expansions for $\mathbb{E}(m)$ and $\mathbb{K}(m)$ around $m\approx 1$ are readily available in the literature. Similarly, to get the leading contribution one can just straightforwardly evaluate

$$\mathbb{E}(\arcsin(1)|1)=1.$$

So the main difficulty lies with the evaluation of $F(z|m)$. Consider the following expansion of $F(z|m)$ which can be found on the Wolfram functions website:

$$F(z|m)=\sum_{k=0}^\infty\frac{(-1)^k{\left(\frac{1}{2}\right)_k}^2}{(k!)^2(m-1)^{-k}}\left(\ln(\sec(z)+\tan(z))+\frac{1}{2}\csc(z)\sum_{j=1}^k\frac{(-1)^j(j-1)!\tan^{2j}(z)}{{\left(\frac{1}{2}\right)_j}}\right)$$

where $\left(a\right)_n=\Gamma(a+n)/\Gamma(a)$ is the Pochhammer function. This expansion, untruncated describes the exact function and is valid for $|Re(z)|\leq\pi/2$. Let us first consider the $\ln$ term in the big parenthesis. For $z=\arcsin(1+ac)$ it behaves as

$$\log(\sec(z)+\tan(z))\approx\ln\left(-i\sqrt{2/ac}\right)+O(c)$$

while the factor $(m-1)^k$ in front goes as $(m-1)^k=(-bc)^k$. Therefore, even though the $\ln$ term diverges logarithmically, it gets suppressed by $(-bc)^k$ for any value of $k$ other than zero when we take $c\to 0$. Therefore, the $\ln$ term contributes only for $k=0$ and for the rest of the $k$ values we can throw it away.

At $k>0$ the second term in the big parenthesis kicks in. Here we always have the suppressing factor $(m-1)^k=(-bc)^k$ in front. For the functions of $z$ involved we observe:

$$\csc(z)\approx 1+O(c)$$ $$\tan^2(z)\approx -\frac{1}{2ac}-\frac{3}{4}+O(c)$$

Therefore, we realize that only terms where $j=k$ survive $c\to0$, since then we get a finite contribution from $(m-1)^k\tan^{2k}(z)$. Keeping only $j=k$ summands in the second sum we can perform the first infinite sum over $k$ to find:

$$\sum_{k=1}^\infty\frac{\left(\frac{1}{2}\right)_k}{k!~2 k}\left(\frac{b}{2a}\right)^k=\ln\left(\frac{4\sqrt{a}}{2\sqrt{a}+\sqrt{4a-2b}}\right)$$

Combining the leading divergence and this correction we arrive at:

$$F(\arcsin(1+ac)|1-bc)_{c\approx0}=\ln\left(-i\frac{4\sqrt{2}}{\sqrt{c}(2\sqrt{a}+\sqrt{4a-2b})}\right)+O(c)$$

With this the concrete behaviour of $\mathcal{Z}(\arcsin(1+ac)|1-bc)_{c\approx0}$ follows straightforwardly.

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