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We consider the following function.

$$u(x) = \int_{0}^{x} \frac{dt}{\sqrt{1 - t^4}}$$

$u(x)$ is defined on $[-1, 1]$. Since $u'(x) = \frac{1}{\sqrt{1 - x^4}} > 0$ on $(-1, 1)$, $u(x)$ is strctly increasing on $[-1, 1]$. Hence there exists the inverse function of $u(x)$. We denote the inverse function of $u(x)$ by $s(u)$. We call $s(u)$ lemniscate sine(see this question).

We denote $u(1) = \int_{0}^{1} \frac{dx}{\sqrt{1 - x^4}}$ by $\omega$. $s(u)$ is defined on $[-\omega, \omega]$.

We define a function $c(u)$ by $c(u) = s(\omega - u)$ and call it lemniscate cosine. $c(u)$ is defined on $[0, 2\omega]$.

By this question, we get the following formulae.

$$s(-u) = -s(u), c(-u) = c(u)$$

$$c(u) = \sqrt{\frac{1 - s^2(u)}{1 + s^2(u)}}$$

$$s(u+v) = \frac{s(u)c(v) + s(v)c(u)}{1 - s(u)s(v)c(u)c(v)}$$

$$c(u+v) = \frac{c(u)c(v) - s(u)s(v)}{1 + s(u)s(v)c(u)c(v)}$$

By the above formulae, we get:

$s(u + \omega) = c(u)\space\space\space c(u + \omega) = -s(u)$

$s(u + 2\omega) = -s(u)\space\space\space c(u + 2\omega) = -c(u)$

$s(u + 3\omega) = -c(u)\space\space\space c(u + 3\omega) = s(u)$

$s(u + 4\omega) = s(u)\space\space\space c(u + 4\omega) = c(u)$

Since $s(u + \omega) = c(u)$, the domain of $s(u)$ can be extended to $[0, 2\omega]$. Since $s(-u) = -s(u)$, the domain of $s(u)$ can be extended to $[-2\omega, 2\omega]$. Since $s(u + 4\omega) = s(u)$,the domain of $s(u)$ can be extended to $(-\infty, \infty)$. Similarly the domain of $c(u)$ can be extended to $(-\infty, \infty)$.

We would like to define $s(u)$ and $c(u)$ as complex valued functions of a complex variable $u$.

By this question, it is natural to define $s(iu) = is(u)$.

By the formula $c(u) = \sqrt{\frac{1 - s^2(u)}{1 + s^2(u)}}$, $c(iu) = \sqrt{\frac{1 + s^2(u)}{1 - s^2(u)}} = \frac{1}{c(u)}$.

By the addition formula for $s(u)$, we define $$s(t+iv) = \frac{s(t)c(v)^{-1} + is(v)c(t)}{1 - is(t)s(v)c(t)c(v)^{-1}} = \frac{s(t) + ic(t)s(v)c(v)}{c(v) - ic(t)s(t)s(v)}$$

By the addition formula for $c(u)$, we define $$c(t+iv) = \frac{c(t)c(v)^{-1} - is(t)s(v)}{1 + is(t)s(v)c(t)c(v)^{-1}} = \frac{c(t) - is(t)s(v)c(v)}{c(v) + ic(t)s(t)s(v)}$$

My question How do we prove the following assertions?

(1) $s(u)$ and $c(u)$ are meromorphic functions on $\mathbb{C}$.

(2) $s(u)$ and $c(u)$ are doubly periodic. Namely,

$s(u + 4\omega) = s(u), s(u + i4\omega) = s(u)$

$c(u + 4\omega) = c(u), c(u + i4\omega) = c(u)$


Zeros of $s(u)$ are $u = 2m\omega + 2ni\omega$, where $(m, n) \in \mathbb{Z}^2$

Zeros of $c(u)$ are $u = (2m+1)\omega + 2ni\omega$, where $(m, n) \in \mathbb{Z}^2$

Poles of $s(u)$ are $u = (2m+1)\omega + (2n+1)i\omega$, where $(m, n) \in \mathbb{Z}^2$

Poles of $c(u)$ are $u = 2m\omega + (2n+1)i\omega$, where $(m, n) \in \mathbb{Z}^2$

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Generally speaking, $$\int_0^1\sqrt[m]{1-x^n}=\int_0^1\sqrt[m]{1-x^m}=\frac{\left(\frac1m\right)!\ \left(\frac1n\right)!}{\left(\frac1m+\frac1n\right)!}$$ In this case, $m=-2$ and $n=4$ , with $\left(-\frac12\right)!=\Gamma\left(\frac12\right)=\sqrt\pi$ , so $u(1)$ becomes $\displaystyle\sqrt\pi\cdot\frac{\left(+\frac14\right)!}{\left(-\frac14\right)!}‌​$ . See also Gauss's constant. – Lucian Nov 18 '13 at 21:06

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