User:Exotic formulae

Alternating symmetric formula
$$\frac{2^3}{\pi}=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4n^2-1}\right)^{(-1)^{n+1}}} {\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}$$

$$\frac{1}{J_{4}\left(e^{-k\pi}\right)}=\prod_{n=1}^{\infty}\frac{e^{nk\pi}+1}{e^{nk\pi}-1}$$

Inspired by your website

These are the formulas that I derived by visiting your website for the last fews months.

Definition of:

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots$$

$$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots$$

$$\lambda(s)=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}=1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\cdots$$

$$\beta(s)=\sum_{n=0}^{\infty}\frac{(-1)^{k}}{(2n+1)^s}=1-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\cdots$$

Where p is prime numbers

$$\phi=\frac{\sqrt{5}+1}{2}$$

See Mock theta function

$$\theta(q),f(q),v(q),w(q)\cdots(1)$$

$$\theta^'(q)=\sqrt{\frac{J_2(0,q)}{J_3(0,q)}}$$

$$\psi(q)=\frac{J_2(0,q^{\frac{1}{2}})}{2q^{\frac{1}{8}}}=\sum_{k=0}^{\infty}q^{\frac{k(k+1)}{2}}$$

$$\chi(q)=\prod_{k=1}^{\infty}\left(1+q^{2k-1}\right)$$

$$g_n-G_n\cdot$$

Refer to Ramanujan g_n and G_n functions

Derived Formulae

$$\prod_{k=1}^{\infty}\left(1-\frac{1}{e^{k\pi\sqrt{n}}}\right)= \frac{2^{\frac{1}{2}}g_n^2\psi(e^{-\pi{\sqrt{n}}})}{e^{\frac{\pi\sqrt{n}}{12}}}\cdots(1)$$

$$\prod_{k=1}^{\infty}\left(1-\frac{1}{e^{2k\pi\sqrt{n}}}\right)= \frac{2g_n^2G_n^2\psi(e^{-2\pi\sqrt{n}})}{e^{\frac{\pi\sqrt{n}}{6}}}\cdots(2)$$

$$\prod_{k=1}^{\infty}\left(1+\frac{1}{e^{k\pi\sqrt{n}}}\right)= \frac{\sqrt{2}{\cdot}G_n^2{\cdot}\psi(e^{-\pi\sqrt{n}})} {e^{\frac{\pi\sqrt{n}}{12}}\cdot{J_3(0,e^{-\pi\sqrt{n}})}}\cdots(4)$$

$$\psi(q^{-1})=\prod_{k=1}^{\infty}\left(\frac{q^k}{q^k-1}\right)^{(-1)^{k+1}}\cdots(5)$$

$$2J_1{'}(0,q^{-\frac{1}{2}}){\cdot}{J_1{'}(0,q^{-1})}=\left[J_2(0,q^{-\frac{1}{2}}) {\cdot}{J_4(0,q^{-1})}\right]^3\cdots(6)$$

$$\frac{J_3(0,e^{-\pi\sqrt{n}})}{J_4(0,e^{-\pi\sqrt{n}})}=\frac{G_n^2}{g_n^2}\cdots(7)$$

$$\prod_{k=1}^{\infty}\frac{e^{k\pi\sqrt{n}}+1}{e^{k\pi\sqrt{n}}-1}= \frac{G_n^2}{g_n^2\cdot{J_3(0,e^{-\pi\sqrt{n}})}}\cdots(8)$$

$$\frac{\psi^2(e^{-\pi\sqrt{n}})}{\psi(e^{-2\pi\sqrt{n}})}=J_3(0,e^{-\pi\sqrt{n}})\cdots(9)$$

$$\frac{q^{\frac{1}{24}}\psi(q^{-1})\left[J_1^{'}(0,q^{-\frac{1}{2}})\right]^{\frac{1}{3}}}{2^{\frac{1}{3}}}= \prod_{k=1}^{\infty}\left(1-\frac{1}{q^{2k}}\right)^2\cdots(10)$$

$$\frac{q^{\frac{1}{24}}\left[J_1^{'}(0,q^{-\frac{1}{2}})\right]^{\frac{1}{3}}}{2^{\frac{1}{3}}\psi(q^{-1})}= \prod_{k=1}^{\infty}\left(1-\frac{1}{q^{2k-1}}\right)^2\cdots(11)$$

Jacobi theta function

$$J_2(0,e^{-\pi})=J_4(0,e^{-\pi})=\left(\frac{\pi}{2}\right)^{\frac{1}{4}}\cdot \frac{1}{\Gamma\left(\frac{3}{4}\right)}\cdots(1)$$

$$J_2(0,e^{-\frac{\pi}{2}})=\frac{2^{\frac{5}{4}}}{e^{\frac{\pi}{6}}}\cdots(2){error!}$$

$$\frac{J_{1}^'(0,e^{-\pi})}{J_{2}(0,e^{-\pi})}=\left(\frac{\pi^2}{2} \right)^{\frac{1}{4}}\cdot\frac{1}{\Gamma^2\left(\frac{3}{4}\right)}\cdots(3)$$

$$J_{1}^'(0,e^{-\pi})=\frac{(\pi)^{\frac{3}{4}}}{2^{\frac{1}{2}}} \cdot\frac{1}{\Gamma^3\left(\frac{3}{4}\right)}\cdots(4)$$

$$J_{1}^'(0,e^{-2\pi})=\frac{(\pi)^{\frac{3}{4}}}{2^{\frac{13}{8}}} \cdot\frac{1}{\Gamma^3\left(\frac{3}{4}\right)}\cdots(4b)$$

$$J_1^{'}(e^{-\frac{\pi}{2}})=\frac{\pi^{\frac{3}{4}}}{2^{\frac{1}{8}}\cdot{\Gamma^3\left(\frac{3}{4}\right)}}\cdots(5)$$

$$J_{4}\left(0,e^{-2\pi}\right)= \left(\frac{\pi^2}{2}\right)^{\frac{1}{8}}\cdot\frac{1}{\Gamma\left(\frac{3}{4}\right)} \cdots(6)$$

$$\theta^{'}\left(e^{-\pi}\right)=\frac{1}{2^{\frac{1}{8}}}\cdots(7)$$

$$\psi(e^{-\pi})=\frac{\pi^{\frac{1}{4}}\cdot{e^{\frac{\pi}{8}}}} {2^{\frac{5}{8}}\cdot{\Gamma\left(\frac{3}{4}\right)}}\cdots(8)$$

$$\psi(e^{-2\pi})=\left(\frac{e^{\pi}\cdot{\pi}}{2^5} \right)^{\frac{1}{4}}\frac{1}{\Gamma\left(\frac{3}{4}\right)}\cdots(9)$$

$$\chi(e^{-\pi})=\frac{2^{\frac{1}{4}}}{e^{\frac{\pi}{24}}}\cdots(10)$$

$$2\theta(-e^{-\pi})-f(e^{-\pi})=\left(\frac{\pi^6}{8e^{\pi}} \right)^{\frac{1}{24}}\frac{1}{\Gamma\left(\frac{3}{4}\right)}\cdots(11)$$

$$2\theta(-e^{-2\pi})-f(e^{-2\pi})=\left(\frac{8\pi^3}{e^{\pi}} \right)^{\frac{1}{12}}\frac{1}{\Gamma\left(\frac{3}{4}\right)}\cdots(12)$$

$$v(e^{-\pi})+{e^{-\pi}}w(e^{-2\pi})=\frac{e^{\frac{\pi}{3}} \cdot{\pi^{\frac{1}{4}}}}{2^{\frac{13}{8}}} \cdot{\frac{1}{\Gamma\left(\frac{3}{4}\right)}}\cdots(13)$$

$$v(e^{\frac{-\pi}{2}})+e^{\frac{-\pi}{2}}w(e^{-\pi})=2^{\frac{1}{8}}\cdots(14)$$

Infinite product

$$\frac{8}{\pi}=\frac{\prod_{n=1}^{\infty}\left(1+\frac{1}{4n^2-1}\right)^{(-1)^{n+1}}} {\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}\cdots(0)$$

$$\pi=\frac{\prod_{n=1}^{\infty}\left[1+\frac{1}{4(3n)^2-1}\right]}{\sum_{n=1}^{\infty}\frac{1}{4^n} }\cdots(1)$$

$$\frac{\Gamma^2(\frac{1}{4})}{\pi^{\frac{3}{2}}}=\frac{\prod_{n=1} ^{\infty}\left[1+\frac{1}{(4n-1)^2-1}\right]}{\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}\cdots(2)$$

$$\frac{\Gamma^2(\frac{1}{4})}{\pi^{\frac{5}{2}}}={\sum_{n=1}^{\infty}\frac{1}{4n^2-1}}\cdot \cdots(3)$$

$$\left[\frac{\Gamma^(\frac{1}{4})}{2}\right]^2\cdot{\frac{1}{\sqrt{2\pi}}}=\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{7}{6}\cdot\frac{8}{9} \cdot\frac{11}{10}\cdot\frac{12}{13}=\prod_{k=2,4,6,...}^{\infty}\left(\frac{k+1}{k}\right)^{(-1)^{\frac{k}{2}+1}} \cdots(4)$$

$$\frac{\left(2\pi\right)^{\frac{3}{2}}}{\Gamma^2\left(\frac{1}{4}\right)}=\frac{4}{3}\cdot\frac{5}{6}\cdot\frac{8}{7}\cdot\frac{9}{10} \cdots=\prod_{n=2}^{\infty}\left(\frac{2n-1}{2n}\right)^{(-1)^{n}}\cdots(5)$$

$$\frac{4}{\pi}=\lim_{n\to\infty}(4n+1)\cdot\left(\frac{1}{2}\cdot \frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2n-1}{2n}\right)^2\cdots(6)$$

$${\pi}=\lim_{n\to\infty}(4n+3)\cdot\left(\frac{2}{3}\cdot \frac{4}{5}\cdot\frac{6}{7}\cdots\frac{2n}{2n+1}\right)^2\cdots(7)$$

$$\frac{\pi}{2}=\lim_{n\to\infty}\sqrt{\frac{4n+3}{4n+1}} \left(\frac{2^2}{2^2-1}\cdot\frac{4^2}{4^2-1}\cdot\frac{6^2} {6^2-1}\cdots\frac{4n^2}{4n^2-1}\right)\cdots(8)$$

$$2^{\frac{1}{4}}=\prod_{n=1}^{\infty}\frac{e^{(2n-1)\pi}+1}{e^ {(3n-1)\pi}}\cdot\frac{(e^{n\pi}+1)^3}{e^{2n\pi}+1}\cdots(9)$$

$$\pi=\lim_{n\to\infty}\left[\prod_{k=1}^{n}\left(1+\frac{1}{2k-1}\right)^2 -\prod_{k=1}^{n-1}\left(1+\frac{1}{2k-1}\right)^2\right]\cdots(10)$$

$$\frac{4}{\pi}=\lim_{n\to\infty}\left[\prod_{k=1}^{n}\left(1+\frac{1}{2k}\right)^2 -\prod_{k=1}^{n-1}\left(1+\frac{1}{2k}\right)^2\right]\cdots(11)$$

$$\frac{e^{\pi}}{2^3}=\prod_{n=1,3,5,...}^{\infty}\left(\frac{e^{n\pi}} {e^{n\pi}-1}\right)^{24}\cdots(12)$$

$$\frac{e^{\pi}}{2^6}=\prod_{n=1,3,5,...}^{\infty}\left(\frac{e^{n\pi}} {e^{n\pi}+1}\right)^{24}\cdots(13)$$

$$\frac{e^{2\pi}}{2^9}=\prod_{n=1,3,5,...}^{\infty} \left(\frac{e^{2n\pi}}{e^{2n\pi}-1}\right)^{24}\cdots(14)$$

$$2=\prod_{n=1,3,5,...}^{\infty} \left(\frac{e^{n\pi}+1}{e^{n\pi}-1}\right)^{8}\cdots(15)$$

$$\left(\frac{2}{\pi}\right)^{\frac{1}{4}}\cdot{\Gamma\left(\frac{3}{4}\right)}=\prod_{n=1}^{\infty} \left(\frac{e^{n\pi}+1}{e^{n\pi}-1}\right)\cdots(16)$$

$$2^{\frac{1}{8}}=\prod_{n=1,3,5,...}^{\infty} \left(\frac{e^{n\pi}+1}{e^{n\pi}-1}\right)\cdots(17)$$

$$\left(\frac{2}{\pi^2}\right)^{\frac{1}{8}}\cdot\Gamma\left(\frac{3}{4}\right) =\prod_{n=1}^{\infty}\frac{e^{2n\pi}+1}{e^{2n\pi}-1}\cdots(18)$$

$$2=\prod_{n=1}^{\infty}\left[1+\frac{1}{e^{(2n-1)\pi}-1}\right]^{20}\left[\left(1+\frac{1} {e^{n\pi}}\right)\left(1-\frac{1}{e^{n\pi}}\right)^2\right]^4\cdots(19)$$

$$2=\prod_{n=1}^{\infty}\left(1+\frac{1}{e^{n\pi}}\right) ^{20}\left(1-\frac{1}{e^{2n\pi}+1}\right)^{8}\left(1-\frac{1} {e^{(2n-1)\pi}}\right)^{4}\cdots(20)$$

$$2=\prod_{n=1}^{\infty}\left(1+\frac{1}{e^{n\pi}}\right) ^{14}\left(1-\frac{1}{e^{n\pi}}\right)^{8}\left(1+ \frac{1}{e^{(2n-1)\pi}-1}\right)^{10}\cdots(21)$$

$$e^{\pi}=\prod_{n=1}^{\infty}\left(1+\frac{1}{e^{n\pi}-1}\right) ^{30}\left(1+\frac{1}{e^{(2n-1)\pi}-1}\right)^{42} \left(1-\frac{1}{e^{2n\pi}}\right)^{54}\cdots(22)$$

$$\frac{e^{\frac{\pi}{24}}}{2^{\frac{1}{8}}} =\prod_{n=1}^{\infty}\left(1+\frac{1}{e^{n\pi}}\right)\cdots(23)$$

$$\frac{2^{\frac{1}{8}}}{e^{\frac{\pi}{24}}}=\prod_{n=1} ^{\infty}\left(1-\frac{1}{e^{(2n-1)\pi}}\right)\cdots(24)$$

$$\frac{2^{\frac{1}{4}}}{e^{\frac{\pi}{24}}}=\prod_{n=1}^{\infty} \left(1+\frac{1}{e^{(2n-1)\pi}}\right)\cdots(25)$$

$$\frac{2^{\frac{3}{8}}}{e^{\frac{\pi}{12}}}=\prod_{n=1}^ {\infty}\left(1-\frac{1}{e^{2(2n-1)\pi}}\right)=\prod_{n=1}^ {\infty}\left(1-\frac{1}{e^{2n\pi}+1}\right)\cdots(26)$$

$$\frac{e^{\frac{\pi}{12}}\cdot\pi^{\frac{1}{4}}} {2^{\frac{1}{2}}\cdot\Gamma\left(\frac{3}{4}\right)} =\prod_{n=1}^{\infty}\left(1-\frac{1}{e^{2n\pi}}\right)\cdots(27)$$

$$\frac{\pi^{\frac{1}{4}}\cdot{e^{\frac{\pi}{24}}}}{2^{\frac{3}{8}}\cdot{\Gamma\left(\frac{3}{4}\right)}}=\prod_{n=1}^{\infty} \left(1-\frac{1}{e^{n\pi}}\right)\cdots(28)$$

$$\frac{1}{2}=\prod_{n=1}^{\infty}\left(1+\frac{2}{n}\right)^{(-1)^n}\cdots(29)$$

$$\frac{5}{\pi}=\lim_{n\to\infty}\frac{5n^2+15n+11}{(4n+7)(n+1)^2} \left(\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}\cdots\frac{2n+1}{2n}\right)^2\cdots(30)$$

$$\frac{6}{\pi}=\lim_{n\to\infty}\frac{6n^2+18n+13}{(4n+7)(n+1)^2} \left(\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}\cdots\frac{2n+1}{2n}\right)^2\cdots(31)$$

$$\frac{1}{\pi}=\lim_{n\to\infty}\frac{n+2}{(4n+7)(n+1)} \left(\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{7}{6}\cdots\frac{2n+1}{2n}\right)^2\cdots(32)$$

$$\frac{\Gamma\left(\frac{1}{4}\right)^4}{2^5}=\lim_{n\to\infty}\left(8n+3\right) \left(\frac{4}{5}\cdot\frac{8}{9}\cdot\frac{12}{13}\cdots\frac{4n}{4n+1}\right)^4\cdots(33)$$

$$\frac{2^7\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}=\lim_{n\to\infty}\left(8n+5\right) \left(\frac{5}{6}\cdot\frac{9}{10}\cdot\frac{13}{14}\cdots\frac{4n+1}{4n+2}\right)^4\cdots(34)$$

$$4\pi^2=\lim_{n\to\infty}\left[(8n+5)(8n+3)\right] \left(\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots\frac{2n}{2n+1}\right)^4\cdots(35)$$

$$\frac{2}{\sqrt{\phi+2}}=\frac{5^2}{5^2-1}\cdot\frac{15^2}{15^2-1}\cdot\frac{25^2}{25^2-1} \cdot\frac{35^2}{35^2-1}\cdots=\prod_{n=0}^{\infty}\left[1+\frac{1}{25(2n+1)^2-1}\right]\cdots(36)$$

$$\frac{10}{{\phi}{\pi}\sqrt{\phi+2}}=\frac{5^2}{5^2-1}\cdot\frac{10^2-1}{10^2} \cdot\frac{15^2}{15^2-1}\cdot\frac{20^2-1}{20^2}\cdots=\prod_{n=1}^{\infty}\left(1+\frac{1}{25n^2-1}\right)^{(-1)^{n+1}}\cdots(37)$$

$$\frac{2\sqrt{3}}{\pi}=\frac{3^2}{3^2-1}\cdot\frac{6^2-1}{6^2}\cdot\frac{9^2}{9^2-1} \cdot\frac{12^2-1}{12^2}\cdots=\prod_{n=1}^{\infty}\left(1+\frac{1}{9n^2-1}\right)^{(-1)^{n+1}}\cdots(38)$$

$$\frac{\Gamma^4\left(\frac{1}{4}\right)}{8\pi^2}=\frac{3}{1}\cdot\frac{3}{5}\cdot \frac{7}{5}\cdot\frac{7}{9}\cdot\frac{11}{9}\cdot\frac{11}{13}\cdots=\prod_{n=1}^{\infty}\left(\frac{2n+1}{2n-1}\right)^{(-1)^{n+1}}\cdots(39)$$

$$\sqrt{2-\sqrt2}=\frac{2}{3}\cdot\frac{6}{5}\cdot\frac{10}{11}\cdots=\prod_{n=0}^{\infty}\left[1-\frac{(-1)^n}{4n+2+(-1)^n}\right]\cdots(40)$$

$$\left(\frac{\pi}{5}\right)^2=\frac{\sum_{n=0}^{\infty}\frac{(n!)^2}{\phi^{2(n+1)} \cdot(2n+1)!}}{\prod_{n=1}^{\infty}(1+\frac{1}{25n^2-1})^{(-1)^{n+1}}}\cdots(41)$$

$$\frac{\pi}{2\sqrt3}=\prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\right)^2 \left(\frac{6n-4}{6n-1}\right)\left(\frac{6n-2}{6n+1}\right)\cdots(42)$$

$$\frac{\pi}{8\left(\sqrt2-1\right)}=\prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\right)^2 \left(\frac{8n-5}{8n-1}\right)\left(\frac{8n-3}{8n+1}\right)\cdots(43)$$

$$\pi{\sqrt2}\cdot\frac{\sqrt{5-\sqrt5}}{5\left(\sqrt5-1\right)^2}=\prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\right)^2 \left(\frac{10n-6}{10n-1}\right)\left(\frac{10n-4}{10n+1}\right)\cdots(44)$$

$$\frac{\pi}{6\left(\sqrt3-1\right)^2}=\prod_{n=1}^{\infty}\left(\frac{2n}{2n-1}\right)^2 \left(\frac{12n-7}{12n-1}\right)\left(\frac{12n-5}{12n+1}\right)\cdots(45)$$

$$\left(\frac{2}{e^{\frac{\pi}{4}}}\right)^{\frac{1}{2}}= \prod_{n=1}^{\infty}\frac{1}{\left(1+e^{-n\pi}\right)\left(1+e^{-2n\pi}\right)}\cdots(46)$$

$$\left(\frac{2^5}{e^\pi}\right)^{\frac{1}{8}}=\prod_{n=1}^{\infty} \frac{1+e^{-n\pi}}{\left(1+e^{-2n\pi}\right)^2}\cdots(47)$$

$$\left(\frac{2^9}{e^{2\pi}}\right)^{\frac{1}{24}}= \prod_{n=1}^{\infty}\frac{1}{1+e^{-2n\pi}}\cdots(48)$$

$$\frac{1}{2^{\frac{1}{8}}}=\prod_{n=1}^{\infty} \frac{1+e^{-2n\pi}}{\left(1+e^{-n\pi}\right)^2}\cdots(49)$$

$$\frac{2}{e^{\frac{5\pi}{24}}}=\prod_{n=1}^{\infty}\frac{1+e^{-n\pi}}{(1+e^{-2n\pi})^3}\cdots(50)$$

$$\left(\frac{2}{e^{\frac{\pi}{6}}}\right)^{\frac{1}{4}}= \prod_{n=1}^{\infty}\frac{1+e^{-n\pi}}{1+e^{-2n\pi}}\cdots(51)$$

$$\left(\frac{2^{21}}{e^{5\pi}}\right)^{\frac{1}{24}}=\prod_{n=1}^{\infty}\frac{1}{ \left(1+e^{-n\pi}\right)\left(1+e^{-2n\pi}\right)^2}\cdots(52)$$

$$\frac{5\phi}{\pi}\left(\phi\sqrt2-\sqrt{5-\sqrt5}\right)^2=\frac{10^2}{10^2-1}\cdot\frac{20^2-1}{20^2} \cdot\frac{30^2}{30^2-1}\cdot\frac{40^2-1}{40^2}\cdots(53)$$

$$\frac{2}{\psi^{12}(e^{-\pi})}=\prod_{n=1}^{\infty}\left(\frac{e^{n\pi}}{e^{n\pi}-1}\right)^4\cdots(54)$$

$$\frac{e^{\pi}}{2^3\psi^{12}(e^{-\pi})}=\prod_{n=1}^{\infty}\left(\frac{e^{n\pi}}{e^{n\pi}-1}\right)^{12}\cdots(55)$$

$$\frac{e^{\pi}\psi^{12}(e^{-2\pi})}{\psi^{42}(e^{-\pi})}=\prod_{n=1}^{\infty}\left(\frac{e^{n\pi}}{e^{n\pi}-1}\right)^{30}\cdots(56)$$

$$\frac{5^4\psi^{26}(e^{-2\pi}) \psi^7(e^{-5\pi})}{\psi^{91}(e^{-\pi})\psi^2(e^{-10\pi})}= \prod_{n=1}^{\infty}\left(\frac{e^{n\pi}}{e^{n\pi}-1} \right)^{57}\left(\frac{e^{5n\pi}}{e^{5n\pi}-1}\right)^3\cdots(57)$$

$$\frac{3^3\psi^{14}(e^{-2\pi})\psi^7(e^{-3\pi})}{\psi^{49}(e^{-\pi})\psi^2(e^{-6\pi})}= \prod_{n=1}^{\infty}\left(\frac{e^{n\pi}}{e^{n\pi}-1}\right)^{27} \left(\frac{e^{3n\pi}}{e^{3n\pi}-1}\right)^3\cdots(58)$$

$$\left(\frac{e^{\pi}\cdot{\pi}}{2^5}\right)^{\frac{1}{4}} \cdot{\frac{1}{\Gamma\left(\frac{3}{4}\right)}}= \prod_{n=1}^{\infty}\left(\frac{e^{2n\pi}}{e^{2n\pi}-1}\right)^{(-1)^{n+1}}\cdots(59)$$

$$\frac{\pi^{\frac{1}{4}}\cdot{e^{\frac{\pi}{8}}}}{2^{\frac{5}{8}}\cdot{\Gamma\left(\frac{3}{4}\right)}} =\prod_{n=1}^{\infty} \left(\frac{e^{n\pi}}{e^{n\pi}-1}\right)^{(-1)^{n+1}}\cdots(60)$$

$$\frac{e^{\frac{\pi}{8}}}{2^{\frac{5}{8}}}=\prod_{n=1}^{\infty}\left(\frac{e^{n\pi}}{e^{n\pi}+1}\right)^{(-1)^{n+1}}\cdots(61)$$

$$\frac{\pi^{\frac{1}{4}}\cdot{e^{\frac{\pi}{6}}}}{2^{\frac{7}{8}}\cdot \Gamma\left(\frac{3}{4}\right)}= \prod_{n=1}^{\infty}\left(1-\frac{1}{e^{4n\pi}}\right)\cdots(62)$$

Infinite sum

$$\pi=72\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(e^{n\pi}-1)}+24\sum_{n=1}^{\infty} \frac{(-1)^n}{n(e^{2n\pi}-1)}\cdots(1)$$

$$\frac{1}{2}=\sum_{n=1}^{\infty}\frac{1}{e^{\frac{(2n-1)^2\pi}{4}}}+ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{e^{n^2\pi}}\cdots(2)$$

$$\frac{\pi}{4}=\sum_{n=0}^{\infty}\frac{(-1)^{n}\zeta(4n+2)}{(2n+1) \cdot{2^{2n+1}}}\cdots(3)$$

$$\frac{\pi}{4}=2\sum_{n=0}^{\infty}\frac{\zeta(8n+2)}{(4n+1) \cdot2^{4n+1}}-\sum_{n=0}^{\infty}\frac{\zeta(4n+2)}{(2n+1)\cdot2^{2n+1}}\cdots(4)$$

$$\frac{\pi}{4}=\sum_{n=0}^{\infty}\frac{\zeta(8n+2)}{(4n+1) \cdot2^{4n+1}}-\sum_{n=0}^{\infty}\frac{\zeta(8n+6)}{(4n+3)\cdot2^{4n+3}}\cdots(5)$$

$$arctan\left(\frac{1}{2}\right)=\sum_{k=0}^{\infty} \frac{(-1)^k\cdot2^{2k+1}\left[\lambda(4k+2)-1\right]}{2k+1}\cdots(6)$$

$$arctan\left(\frac{1}{2}\right)=2\sum_{n=0}^{\infty}\frac{2^{4n+1}\left[\lambda(8n+2)-1\right]}{4n+1} -\sum_{n=0}^{\infty}\frac{2^{2n+1}\left[\lambda(4n+2)-1\right]}{2n+1}\cdots(7)$$

$$arctan\left(\frac{1}{2}\right)=\sum_{n=0}^{\infty}\frac{2^{4n+1} \left[\lambda(8n+2)-1\right]}{4n+1}- \sum_{n=0}^{\infty}\frac{2^{4n+3}\left[\lambda(8n+6)-1\right]}{4n+3}\cdots(8)$$

$$\gamma=\sum_{n=1}^{\infty}\frac{1-\beta(2n)}{n}+\sum_{n=2}^{\infty} \frac{(-1)^n\lambda(n)}{2^{n-2}\cdot{n}}\cdots(9)$$

$$\frac{\pi^{\frac{1}{4}}}{\Gamma\left(\frac{3}{4}\right)}=2\sum_{n=1}^ {\infty}e^\frac{-(2n-1)^2\pi}{4}+4\sum_{n=1}^{\infty}e^{-(2n-1)^2\pi}\cdots(10)$$

$$2^{\frac{5}{4}}\sum_{n=1}^{\infty}e^{-\frac{(2n-1)^2\pi}{4}}=1+2\sum_{n=1}^{\infty}e^{-n^2\pi}\cdots(11)$$

$$2^{\frac{9}{8}}\sum_{n=1}^{\infty}e^{-\frac{(2n-1)^2\pi}{4}}=1+2\sum_{n=1}^{\infty}e^{-2n^2\pi}\cdots(12)$$

$$2^{\frac{1}{4}}\left(1+2\sum_{n=1}^{\infty}(-1)^{n}e^{-n^2\pi}\right)=1+2\sum_{n=1}^{\infty}e^{-n^2\pi}\cdots(13)$$

$$2^{\frac{1}{8}}\left(1+2\sum_{n=1}^{\infty}(-1)^{n}e^{-2n^2\pi}\right)=1+2\sum_{n=1}^{\infty}e^{-n^2\pi}\cdots(14)$$

$$\frac{\pi}{4}=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty} \frac{(-1)^n2^{2n}}{\left(2k^2+1\right)^{4n+3}} \left[\frac{(2k^2+1)^2}{4n+1}+\frac{2(2k^2+1)}{4n+2}+\frac{2}{4n+3}\right]\cdots(15)$$

$$\frac{\pi}{4}=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty} \frac{(-1)^n2^{2n}}{\left(k^2+k\right)^{4n+3}} \left[\frac{(k^2+k)^2}{4n+1}+\frac{2(k^2+k)}{4n+2}+\frac{2}{4n+3}\right]\cdots(16)$$

$$\pi=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot2^{2n+4}}{\left(1+\sqrt3\right)^{4n+3}} \left(\frac{2+\sqrt3}{4n+1}+\frac{1}{4n+3}\right)\cdots(17)$$

$$1+\frac{1}{e^\pi}+\frac{2}{e^{2\pi}}+\frac{3}{e^{3\pi}}+\frac{4}{e^{4\pi}}+\cdots= \frac{2^{\frac{3}{8}} \cdot{\Gamma\left(\frac{3}{4}\right)}}{\pi^{\frac{1}{4}}\cdot{e^{\frac{\pi}{24}}}}\cdots(18)$$

Beta function

$$ln\left[\frac{\Gamma^2(\frac{1}{4})}{4\sqrt{2\pi}}\right]= \sum_{n=1}^{\infty}\frac{1-\beta(n)}{n}\cdots(1)$$

$$ln\left[\frac{\Gamma^{4}(\frac{1}{4})}{16\pi^2}\right]=\sum_{k=1}^{\infty} \frac{1-\beta(2k)}{k}\cdots(2)$$

$$ln\sqrt\frac{\pi}{2}=\sum_{k=0}^{\infty}\frac{1-\beta(2k+1)}{2k+1}\cdots(3)$$

$$ln\left[\frac{4\pi^2}{\sqrt{2\pi}}\cdot\frac{1}{\Gamma^2\left(\frac{1}{4}\right)}\right]=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\left [1-\beta(n)\right]}{n}\cdots(4)$$

prime number

$$ln\sqrt{\frac{7}{6}}=\sum_{k=0}^{\infty}\frac{1}{2k+1}\sum_{p=2}^ {\infty}\frac{1}{p^{8k+4}}\cdots(1)$$

$$ln\sqrt{\frac{5}{2}}=\sum_{k=0}^{\infty}\frac{1}{2k+1}\sum_{p=2}^{\infty}\frac{1} {p^{4k+2}}\cdots(2)$$

Zeta function

$$ln\left(\frac{\pi}{sinh\pi}\right)=\sum_{n=1}^{\infty}\frac{(-1) ^k\zeta(2n)}{n}\cdots(1)$$

$$ln\sqrt{\frac{3}{2}}=\sum_{k=0}^{\infty}\frac{\zeta(6k+3)-1}{2k+1}\cdots(2)$$

$$ln\sqrt{\frac{sinh\pi}{\pi}}=\sum_{k=0}^{\infty}\frac{\zeta(4k+2)-1}{2k+1}\cdots(3)$$

$$ln\pi-\gamma=\sum_{k=2}^{\infty}\frac{\zeta(k)}{k\cdot2^{k-1}}\cdots(4)$$

$$ln2-\gamma=\sum_{k=1}^{\infty}\frac{\zeta(2k+1)}{(2k+1)2^{2k}}\cdots(5)$$

$$ln(2\pi)=\sum_{n=1}^{\infty}\frac{\left(2^{2n+1}+1\right)\zeta(2n) -2^{2n+1}}{2^{2n}\cdot{n}}\cdots(6)$$

$$ln\left(\frac{\pi}{2}\right)=\sum_{n=1}^{\infty}\frac{2^{2n}n}\cdots(7)$$

$$ln\left(\frac{\pi}{3}\right)=\sum_{n=1}^{\infty}\frac{6^{2n}n}\cdots(8)$$

$$ln\left(\frac{\pi\phi}{5}\right)=\sum_{n=1}^{\infty} \frac{10^{2n}n}\cdots(9)$$

$$ln\left(\frac{2\pi\phi}{5\sqrt{\phi+2}}\right)= \sum_{n=1}^{\infty}\frac{5^{2n}n}\cdots(10)$$

$$ln(2)=\sum_{n=1}^{\infty}\frac{\zeta(2n)-1}{n}\cdots(11)$$

$$ln\left(\frac{4\pi}{sinh\pi}\right)=\sum_{n=1}^{\infty}\frac{\zeta(4n)-1}{n}\cdots(12)$$

$${\gamma}=\lim_{k\to\infty}\left[\sum_{n=1}^k{\frac{\zeta(2n)}{n}-ln2k}\right]\cdots(13)$$

Eta function

$$ln\left[\frac{\Gamma^2\left(\frac{1}{4}\right)}{4\sqrt{2\pi}}\right]= \sum_{k=1}^{\infty}\frac{(-1)^{k+1}\eta(k)}{2^k\cdot{k}}\cdots(1)$$

$$ln\left(\frac{2}{\pi}\right)=\sum_{n=1}^{\infty}\frac{(-1)^n\eta(n)}{n}\cdots(2)$$

$$ln\sqrt2=\sum_{k=0}^{\infty}\frac{1-\eta(2k+1)}{2k+1}\cdots(3)$$

$$ln\left(\frac{\pi}{e}\right)=2\sum_{n=0}^{\infty}\frac{1-\eta(2n)}{2n+1}\cdots(4)$$

$$ln\left(\frac{8}{\pi^2}\right)=\sum_{n=1}^{\infty}\frac{\eta(2n)-1}{n}\cdots(5)$$

$$ln\left(\frac{2}{\pi}\right)=\sum_{n=1}^{\infty}\frac{\eta(n)-1}{n}\cdots(6)$$

$$ln\left(\frac{10}{{\pi}{\phi}\sqrt{\phi+2}}\right)=\sum_{n=1}^{\infty}\frac{\eta(2n)}{5^{2n}\cdot{n}}\cdots(7)$$

$$ln\left(\frac{4}{\pi}\right)=\sum_{n=1}^{\infty}\frac{\eta(2n)}{2^{2n}\cdot{n}}\cdots(8)$$

$$ln\left(\frac{2\sqrt{3}}{\pi}\right)=\sum_{n=1}^{\infty}\frac{\eta(2n)}{3^{2n}\cdot{n}}\cdots(9)$$

$$ln\left[\frac{\Gamma^2\left(\frac{1}{4}\right)}{2\pi{\sqrt2}}\right]=\sum_{n=0}^{\infty}\frac{\eta(2n+1)}{2^{2n+1}\cdot{(2n+1)}}\cdots(10)$$

Lambda function

$$ln\left(\frac{4}{\pi}\right)= \sum_{n=1}^{\infty}\frac{\lambda(2n)-1}{n}\cdots(1)$$

$$ln\left(\frac{1}{\pi}\right)+2-\gamma=2\sum_{n=2}^{\infty}\frac{\lambda(n)-1}{n}\cdots(2)$$

$${\gamma}=\lim_{k\to\infty}\left[\sum_{n=1}^k{\frac{\lambda(2n)}{m}-ln \left(\frac{4k}{\pi}\right)}\right]\cdots(3)$$

$$\gamma+ln\left(\frac{4}{\pi}\right)=2\sum_{n=2} ^{\infty}\frac{(-1)^n\lambda(n)}{n}\cdots(4)$$

$$ln\left[\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\right]+\gamma=4\sum_{n=2}^{\infty}\frac{(-1)^n\lambda(n)}{2^n\cdot{n}}\cdots(5)$$

$$ln\left(\frac{2}{\sqrt{\phi+2}}\right)=\sum_{n=1}^{\infty}\frac{\lambda(2n)}{5^{2n}\cdot{n}}\cdots(6)$$

$$ln\sqrt{2}=\sum_{n=1}^{\infty}\frac{\lambda(2n)}{2^{2n}\cdot{n}}\cdots(7)$$

$$ln\left[\sqrt{2(2-\sqrt2)}\right]=\sum_{n=1}^{\infty}\frac{\lambda(2n)}{4^{2n}\cdot{n}}\cdots(8)$$

$$ln\left(\frac{2}{\sqrt{3}}\right)=\sum_{n=1}^{\infty}\frac{\lambda(2n)}{3^{2n}\cdot{n}}\cdots(9)$$

$$ln\left[\frac{\Gamma^2\left(\frac{1}{4}\right)}{2\pi{\sqrt2}}\right]-\frac{\gamma}{2}=2\sum_{n=1}^{\infty}\frac{\lambda(2n+1)}{2^{2n+1}\cdot{(2n+1)}}\cdots(10)$$

$$ln\left(\phi^2\sqrt2-\phi{\sqrt{5-\sqrt5}}\right)=\sum_{n=1}^{\infty}\frac{\lambda(2n)}{n\cdot10^{2n}}\cdots(11)$$

$$\pi+ln\left(\frac{1}{2^3}\right)=24\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(e^{n\pi}-1)}\cdots(1)$$

$$2\pi+ln\left(\frac{1}{2^9}\right)=24\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n(e^{2n\pi}-1)}\cdots(2)$$

$$ln2=24\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(e^{n\pi}-1)}-8 \sum_{n=1}^{\infty}\frac{1}{n(e^{n\pi}-1)}\cdots(3)$$

$${4\pi}+ln\left(\frac{1}{2^{18}}\right)=24\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1} {e^{2k\pi}+1}+\frac{1}{e^{2k\pi}-1}\right)\cdots(4)$$

$$\frac{1}{8}ln2=\sum_{k=0}^{\infty}\frac{1}{2k+1}\left (\frac{1}{e^{(2k+1)\pi}+1}+\frac{1}{e^{(2k+1)\pi}-1}\right)\cdots(5)$$

$${\pi}+ln\left(\frac{1}{2^6}\right)=12\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left(\frac{1}{e^{n\pi}+1}+ \frac{1}{e^{n\pi}-1}\right)\cdots(6)$$

$$ln\frac{\pi^{\frac{1}{4}}}{\Gamma(\frac{3}{4})}=2\sum_{k=0}^{\infty} \frac{1}{(2k+1)(e^{\pi(2k+1)}+1)}\cdots(7)$$

$$ln\left[\left(\frac{2}{\pi}\right)^{\frac{1}{4}}\cdot{\Gamma\left(\frac{3}{4}\right)}\right]=2\sum_{k=0}^{\infty} \frac{1}{(2k+1)(e^{\pi(2k+1)}-1)}\cdots(8)$$

$$ln\left[\Gamma\left(\frac{3}{4}\right)\cdot\left(\frac{2}{\pi^2}\right) ^{\frac{1}{8}}\right]=\sum_{n=1}^{\infty}\frac{1}{2k-1} \left(\frac{1}{e^{(2n-1)\pi}-1}-\frac{1}{e^{(2n-1)\pi}+1}\right)\cdots(9)$$

$$ln\left[\left(\frac{2}{\pi^2}\right)^{\frac{1}{8}} \cdot{\Gamma\left(\frac{3}{4}\right)}\right] =2\sum_{n=0}^{\infty}\frac{1}{2n+1}\left(\frac{1}{e^{2\pi(2n+1)}-1}\right)\cdots(10)$$

$$ln3=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{3n}\left(\frac{84}{e^{n\pi}-1}- \frac{28}{e^{2n\pi}-1}-\frac{12}{e^{3n\pi}-1}+ \frac{4}{e^{6n\pi}-1}\right)-\sum_{n=1}^{\infty}\frac{1}{3n} \left(\frac{8}{e^{n\pi}-1}-\frac{8}{e^{3n\pi}-1}\right)\cdots(11)$$

$$ln5=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{4n}\left(\frac{156}{e^{n\pi}-1}- \frac{52}{e^{2n\pi}-1}-\frac{19}{e^{5n\pi}-1}+\frac{4}{e^{10n\pi}-1}\right) -\sum_{n=1}^{\infty}\frac{1}{4n}\left(\frac{8}{e^{n\pi}}-\frac{1}{e^{5n\pi}-1}\right)\cdots(12)$$

$$ln\left(\sqrt{\phi+2}-\phi\right)+\frac{2\pi}{5} =\sum_{n=1}^{\infty}\frac{e^{2n\pi}-e^{4n\pi}-e^{6n\pi}+e^{8n\pi}}{n(1-e^{10n\pi})}\cdots(13)$$

$$ln\left(\frac{2^{\frac{1}{2}}}{e^{\frac{\pi}{12}}}\right)= \sum_{m=1}^{\infty}\frac{1-tanh\left(\frac{m\pi}{2}\right)}{m}\cdots(14)$$

$$ln\left[\left(\frac{e^{\pi}\cdot{\pi}}{2^5}\right)^{\frac{1}{2}}\cdot{\frac{1}{\Gamma^2\left(\frac{3}{4}\right)}}\right]=\sum_{m=1}^{\infty}\frac{1-tanh(m\pi)}{m}\cdots(15)$$

x > 1

$$ln2=\sum_{n=1}^{\infty}\frac{1}{(2^x-x-1)n}\sum_{r=1}^{2^x-2}\frac{2^x-r-1} {(r+1)^{2n}}\cdots(1)$$

$$ln3=\sum_{k=0}^{\infty}\frac{1}{2^{2k}\cdot(2k+1)}\cdots(2)$$

$$ln2=\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{2^{2n}}+\frac{2}{3^{2n}} +\frac{2}{4^{2n}}+\frac{1}{5^{2n}}\right)\cdots(3)$$

$$ln\left(\frac{2\sqrt2}{e}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\left[\frac{1}{4}\left(\frac{1}{3^{2n}}\right)+ \frac{1}{8}\left(\frac{1}{5^{2n}}+\frac{1}{7^{2n}}\right)+\frac{1}{16}\left(\frac{1}{9^{2n}}+\frac{1}{11^{2n}}+ \frac{1}{13^{2n}}+\frac{1}{15^{2n}}\right)+\cdots\right]\cdots(4)$$

$$\sum_{k=1}^{\infty}\frac{\zeta(2k)\left[(2^{2k}-1)\zeta(2k)-2^{2k}\right]}{k}+ln\left(\frac{\pi}{2}\right)= \sum_{n=1}^{\infty}\frac{1}{n}\sum_{r=1}^{\infty}\sum_{a=3}^{\infty}\left[\frac{1}{(2ar-a-1)^{2n}} +\frac{2}{(2ar-a)^{2n}}+\frac{1}{(2ar-a+1)^{2n}}\right]\cdots(5)$$

'''Where K = 8.700... is the constant from polygon circumscribing.'''

$$ln\left(\frac{2K}{\pi}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{r=1}^{\infty} \sum_{k=1}^{\infty}\left[\frac{1}{(2rk+r-1)^{2n}}+\frac{2}{(2rk+r)^{2n}} +\frac{1}{(2rk+r+1)^{2n}}\right]\cdots(6)$$

$$ln\left[\sum_{n=0}^{\infty}\frac{(n!)^2}{\phi^{2n}(2n+1)!}\right]= \sum_{n=1}^{\infty}\frac{\zeta(2n)}{5^{2n}\cdot{n}}\cdots(7)$$

$$\sum_{n=1}^{\infty}\frac{1-\beta(n)}{n}=\sum_{n=1}^ {\infty}\frac{(-1)^{n+1}\eta(n)}{2^n\cdot{n}}\cdots(8)$$

$$ln\left(\frac{\pi}{\sqrt6}\right)=\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n}\left[\frac{\left(\frac{3^n+1}{2}\right)^2+\left(\frac{3^n-1}{2}\right)^2}{(6^n)^2}\right]\cdots(9)$$

$$ln\left[\frac{\Gamma\left(\frac{1}{4}\right)}{\pi}\right]+\frac{\gamma}{4}=\sum_{n=2}^{\infty}\frac{(-1)^n\lambda(n)}{n}\left(1-\frac{1}{2^n}\right)\cdots(10)$$

$$ln\left[\frac{4}{\Gamma\left(\frac{1}{4}\right)}\right]+\frac{3\gamma}{4}=\sum_{n=2}^{\infty}\frac{(-1)^n\lambda(n)}{n}\left(1+\frac{1}{2^n}\right)\cdots(11)$$

$$ln\left(\frac{\pi^2}{8}\cdot\sqrt{\frac{\pi}{e}}\right)=\sum_ {n=1}^{\infty}\frac{(3n+1)}{n(2n+1)}\cdot\left[1-\eta(2n)\right]\cdots(12)$$

$$ln\left(\frac{8}{\pi^2}\cdot\sqrt{\frac{\pi}{e}}\right)=\sum_{n=1}^{\infty}\frac{(n+1)}{n(2n+1)}\cdot\left[\eta(2n)-1\right]\cdots(13)$$

$$ln\left(\frac{4}{e}\right)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}\cdots(14)$$

Other

$$\pi=2arctan\left(\frac{1}{cos^2\phi}\right)+ arctan\left(\frac{2sin^6\phi+2sin^4\phi-4}{sin^8\phi-2sin^4\phi-4sin^2\phi}\right)\cdots(1)$$

$$\frac{tan^6\phi-tan^4\phi+2}{tan^6\phi-2tan^2\phi+4}\cdot{cos^2\phi}=\frac{sin^6\phi+sin^4\phi-2}{sin^6\phi-2sin^2\phi-4}\cdots(2)$$

$$\frac{cot^6\phi-cot^4\phi+2}{cot^6\phi-2cot^2\phi+4}\cdot{sin^2\phi}=\frac{cos^6\phi+cos^4\phi-2}{cos^6\phi-2cos^2\phi-4}\cdots(3)$$