Editing Jacobi elliptic functions (section)

==Continued fractions==

Assuming real numbers <math>a,p</math> with <math>0<a<p</math> and the [[Nome (mathematics)|nome]] <math>q=e^{\pi i \tau}</math>, <math>\operatorname{Im}(\tau)>0</math> with [[Theta function|elliptic modulus]] <math display="inline">k(\tau)=\sqrt{1-k'(\tau)^2}=(\vartheta_{10}(0;\tau)/\vartheta_{00}(0;\tau))^2</math>. If <math>K[\tau]=K(k(\tau))</math>, where <math>K(x)=\pi/2\cdot {}_2F_1(1/2,1/2;1;x^2)</math> is the [[complete elliptic integral of the first kind]], then holds the following [[Continued fraction|continued fraction expansion]]<ref name="bagis-evaluations">N.Bagis.(2020)."Evaluations of series related to Jacobi elliptic functions". preprint https://www.researchgate.net/publication/331370071_Evaluations_of_Series_Related_to_Jacobi_Elliptic_Functions</ref>
:<math>
\begin{align}
&\frac{\textrm{dn}\left((p/2-a)\tau K\left[\frac{p\tau}{2}\right];k\left(\frac{p\tau}{2}\right)\right)}{\sqrt{k'\left(\frac{p\tau}{2}\right)}} = \frac{\sum^\infty_{n=-\infty}q^{p/2 n^2+(p/2-a)n}}{\sum^\infty_{n=-\infty}(-1)^nq^{p/2 n^2+(p/2-a)n}}\\[4pt]
={}&-1+\frac{2}{1-{}} \, \frac{q^a+q^{p-a}}{1-q^p+{}} \, \frac{(q^a+q^{2p-a})(q^{a+p}+q^{p-a})}{1-q^{3p}+{}} \, \frac{q^p(q^a+q^{3p-a})(q^{a+2p}+q^{p-a})}{1-q^{5p}+{}} \, \frac{q^{2p}(q^a+q^{4p-a})(q^{a+3p}+q^{p-a})}{1-q^{7p}+{}}\cdots
\end{align}
</math>
Known continued fractions involving <math>\textrm{sn}(t),\textrm{cn}(t)</math> and <math>\textrm{dn}(t)</math> with elliptic modulus <math>k</math> are

For <math>z\in\Complex</math>, <math>|k|<1</math>:<ref name="wall-analytic-theory">H.S. Wall. (1948). "Analytic Theory of Continued Fractions", Van Nostrand, New York.</ref> pg. 374
:<math>\int^{\infty}_{0}\textrm{sn}(t)e^{-tz}\, \mathrm dt=\frac{1}{1^2(1+k^2)+z^2-{}} \, \frac{1\cdot 2^2\cdot 3k^2}{3^2(1+k^2)+z^2-{}} \, \frac{3\cdot 4^2\cdot 5k^2}{5^2(1+k^2)+z^2-{}}\cdots</math>

For <math>z \in \Complex\setminus\{0\}</math>, <math>|k|<1</math>:<ref name="wall-analytic-theory"/> pg. 375
:<math>\int^{\infty}_{0}\textrm{sn}^2(t)e^{-t z}\,\mathrm dt=\frac{2z^{-1}}{2^2(1+k^2)+z^2-{}} \, \frac{2\cdot 3^2\cdot 4k^2}{4^2(1+k^2)+z^2-{}} \, \frac{4\cdot 5^2\cdot 6k^2}{6^2(1+k^2)+z^2-{}}\cdots</math>

For <math>z\in\Complex\setminus\{0\}</math>, <math>|k|<1</math>:<ref name="perron">Perron, O. (1957). "Die Lehre von den Kettenbruchen", Band II, B.G. Teubner, Stuttgart.</ref> pg. 220
:<math>\int^\infty_0 \textrm{cn}(t)e^{-t z}\, \mathrm dt=\frac{1}{z+{}} \, \frac{1^2}{z+{}} \, \frac{2^2k^2}{z+{}}\, \frac{3^2}{z+{}} \, \frac{4^2k^2}{z+{}} \, \frac{5^2}{z+{}} \cdots</math>

For <math>z\in\Complex\setminus\{0\}</math>, <math>|k|<1</math>:<ref name="wall-analytic-theory"/> pg. 374
:<math>\int^\infty_0\textrm{dn}(t)e^{-t z}\, \mathrm dt=\frac{1}{z+{}} \, \frac{1^2k^2}{z+{}} \, \frac{2^2}{z+{}} \, \frac{3^2k^2}{z+{}} \, \frac{4^2}{z+{}} \, \frac{5^2k^2}{z+{}} \cdots</math>

For <math>z\in\Complex</math>, <math>|k|<1</math>:<ref name="wall-analytic-theory"/> pg. 375
:<math>\int^{\infty}_{0}\frac{\textrm{sn}(t)\textrm{cn}(t)}{\textrm{dn}(t)}e^{-tz}\, \mathrm dt=\frac{1}{2\cdot 1^2(2-k^2)+z^2-{}} \, \frac{1\cdot 2^2\cdot 3k^4}{2\cdot 3^2(2-k^2)+z^2-{}} \, \frac{3\cdot 4^2\cdot  5k^4}{2\cdot 5^2(2-k^2)+z^2-{}}\cdots</math>