FANDOM


$ \zeta(2)=\sum_{n=0}^{\infty} \frac{1}{n^2}= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}\! $(ゼータ関数)

$ \zeta(4)=\sum_{n=0}^{\infty} \frac{1}{n^4}= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90}\! $

$ \zeta(6)=\sum_{n=0}^{\infty} \frac{1}{n^6}= \frac{1}{1^6} + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + \cdots = \frac{\pi^6}{945}\! $

$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots = \frac{\pi^2}{8}\! $

$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^4} = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \frac{1}{7^4} + \cdots = \frac{\pi^4}{96}\! $

$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^6} = \frac{1}{1^6} + \frac{1}{3^6} + \frac{1}{5^6} + \frac{1}{7^6} + \cdots = \frac{\pi^6}{960}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n^2}= \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots = \frac{\pi^2}{12}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n^4}= \frac{1}{1^4} - \frac{1}{2^4} + \frac{1}{3^4} - \frac{1}{4^4} + \cdots = \frac{7\pi^4}{720}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n^6}= \frac{1}{1^6} - \frac{1}{2^6} + \frac{1}{3^6} - \frac{1}{4^6} + \cdots = \frac{31\pi^6}{30240}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3} = \frac{1}{1^3} - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + \cdots = \frac{\pi^3}{32}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^5} = \frac{1}{1^5} - \frac{1}{3^5} + \frac{1}{5^5} - \frac{1}{7^5} + \cdots = \frac{5\pi^5}{1536}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^7} = \frac{1}{1^7} - \frac{1}{3^7} + \frac{1}{5^7} - \frac{1}{7^7} + \cdots = \frac{61\pi^5}{184320}\! $

$ \sum_{p:prime} \frac{1}{p^2}= \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots = \frac{9}{2\pi^2}\! $

$ \sum_{p:prime} \frac{1}{p^4}= \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{5^4} + \frac{1}{7^4} + \cdots = \frac{15}{2\pi^4}\! $

$ \sum_{p:prime} \frac{1}{p^6}= \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{5^6} + \frac{1}{7^6} + \cdots = \frac{11340}{691\pi^6}\! $

$ \zeta(2k)=\sum_{n=0}^{\infty} \frac{1}{n^2k}= \frac{1}{1^2k} + \frac{1}{2^2k} + \frac{1}{3^2k} + \frac{1}{4^2k} + \cdots = \frac{2^2k|B_{2k}|\pi^2k}{2(2k)!}\! $

$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2k}= \frac{1}{1^2k} + \frac{1}{3^2k} + \frac{1}{5^2k} + \frac{1}{7^2k} + \cdots = \frac{(2^2k-1)|B_{2k}|\pi^2k}{2(2k)!}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n^2k}= \frac{1}{1^2k} - \frac{1}{2^2k} + \frac{1}{3^2k} - \frac{1}{4^2k} + \cdots = \frac{(2^2k-2)|B_{2k}|\pi^2k}{2(2k)!}\! $

$ \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2k}= \frac{1}{1^2k} - \frac{1}{2^2k} + \frac{1}{3^2k} - \frac{1}{4^2k} + \cdots = \frac{|E_{2k}|\pi^2k}{2^2k(2k)!}\! $

$ \sum_{p:prime} \frac{1}{p^k}= \frac{1}{2^k} + \frac{1}{3^k} + \frac{1}{5^k} + \frac{1}{7^k} + \cdots = \frac{\zeta^2(2k)-\zeta(4k)}{2\zeta(2k)\zeta(4k)}\! $

$ \prod_{n=1}^\infty (1+\frac{(-1)^{n+1}}{2n+1})=\frac{\pi \sqrt{2}}{4} $

$ \prod_{n=1}^\infty (1-\frac{1}{(6n)^2})=\frac{3}{\pi} $

$ \prod_{n=1}^\infty (1-\frac{3^2}{(2n)^2})=-\frac{2}{3\pi} $

$ \prod_{n=1}^\infty (1-\frac{5^2}{(2n)^2})=\frac{2}{5\pi} $

$ \prod_{p:odd prime} (1-\frac{(-1)^{\frac{p-1}{2}}}{p}) = \frac{4}{\pi} $

$ \prod_{p:odd prime} (1-\frac{(-1)^{\frac{p+1}{2}}}{p}) = \frac{2}{\pi} $

$ \prod_{c:composite} (1-\frac{1}{c^2})= \frac{\pi^2}{12} $

$ \prod_{c:odd composite} (1-\frac{1}{c^2})= \frac{\pi^3}{32} $

$ \sum_{n=1}^\infty \frac{n^{-11}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (11)+\frac{1453\pi}{851350500} $

$ \sum_{n=1}^\infty \frac{n^{-7}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (7)+\frac{19\pi}{113400} $

$ \sum_{n=1}^\infty \frac{n^{-3}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (3)+\frac{7\pi}{360} $

$ \sum_{n=1}^\infty \frac{n^{-1}}{e^{2n\pi}-1}=-\frac{\pi}{12}-\frac{1}{2} \log \frac{\omega}{\pi \sqrt{2}} $

$ \sum_{n=1}^\infty \frac{n}{e^{2n\pi}-1}=\frac{1}{24}-\frac{1}{8\pi} $

$ \sum_{n=1}^\infty \frac{n^{3}}{e^{2n\pi}-1}=-\frac{1}{240}+\frac{1}{80}(\frac{\omega}{\pi})^4 $

$ \sum_{n=1}^\infty \frac{n^{5}}{e^{2n\pi}-1}=\frac{1}{504} $

$ \sum_{n=1}^\infty \frac{n^{7}}{e^{2n\pi}-1}=-\frac{1}{480}+\frac{3}{160}(\frac{\omega}{\pi})^8 $

$ \sum_{n=1}^\infty \frac{n^{9}}{e^{2n\pi}-1}=\frac{1}{264} $

$ \sum_{n=1}^\infty \frac{n^{11}}{e^{2n\pi}-1}=-\frac{691}{65520}+\frac{89}{1040}(\frac{\omega}{\pi})^{12} $

$ \sum_{n=1}^\infty \frac{n^{13}}{e^{2n\pi}-1}=\frac{1}{24} $

$ \sum_{n=1}^\infty \frac{n^{15}}{e^{2n\pi}-1}=-\frac{3617}{16320}+\frac{43659}{5440}(\frac{\omega}{\pi})^{16} $

$ \sum_{n=1}^\infty \frac{n^{17}}{e^{2n\pi}-1}=\frac{43867}{28728} $

$ \sum_{n=1}^\infty \frac{n^{4k+1}}{e^{2n\pi}-1}=\frac{B_{4k+2}}{8k+4} (k \geq 1) $

$ \sum_{n=1}^\infty \frac{n^{4k-1}}{e^{2n\pi}-1}=\frac{1}{8k}(B_{4k}-(\frac{\omega}{\pi})^{4k} H_{4k}) (k \geq 1) $

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

$ =-\frac{1}{2}\zeta (4k-1)+\frac{(2\pi)^{4k-1}}{4(4k+4)!} (\binom{4k}{2k}(-1)^k B_{2k}^2 -2\sum_{m=0}^{k-1}\binom{4k}{2m}(-1)^m B_{2m} B_{4k-2m}) (k \geq 1) $

$ \int_0^{\frac{\pi}{2}}\frac{x^2dx}{\sin^2 x}=\pi\text{log}2 $

$ \int_0^{\frac{\pi}{2}} \text{log}^2(\cos x)dx=\frac{\pi^3}{24}+\frac{\pi \text{log}^22}{2} $

$ \int_0^\infty xe^{-x}\sqrt{1-e^{-2x}}dx=\frac{\pi(1+2\text{log}2)}{8} $

$ \int_0^\infty \frac{x^2dx}{\sqrt{e^x-1}}=4\pi\text{log}^22+\frac{\pi^3}{3} $

$ \int_0^1\frac{x^2dx}{(1+x^4)\sqrt{1-x^4}}=\frac{\pi}{8} $

$ \int_0^1\frac{\text{log}^2x}{1+x+x^2}dx=\frac{8\pi^3}{81\sqrt{3}} $

$ \int_0^1\frac{\text{log}(1+x^2)}{x^2}dx=\frac{\pi}{2}-\text{log}2 $

$ \int_0^1\frac{\text{log}(1+x^3)}{1-x+x^2}dx=\frac{2\pi\text{log}3}{\sqrt{3}} $

$ \int_0^1\int_0^1\frac{dxdy}{1-xy}=\frac{\pi^2}{6} $

$ \int_0^1\int_0^1\frac{dxdy}{1-x^2y^2}=\frac{\pi^2}{8} $

$ \int_0^1\int_0^1\frac{dxdy}{\sqrt{1+x^2+y^2}}=-\frac{\pi}{6}+\text{log}(2+\sqrt{3}) $

$ \int_0^1\int_0^1(\frac{x-1}{x+1})^2(\frac{y-1}{y+1})^2(\frac{xy-1}{xy+1})^2dxdy=5-\pi^2-4\text{log}2+16\text{log}^22 $

$ \int_0^1\int_0^1\int_0^1(x^2+y^2+z^2)^{-\frac{1}{2}}dxdydz=-\frac{\pi}{4}+\frac{3\text{log}(2+\sqrt{3})}{2} $

$ \int_0^1\int_0^1\int_0^1(x^2+y^2+z^2)^{\frac{1}{2}}dxdydz=-\frac{\pi}{24}+\frac{\sqrt{3}}{4}+\frac{\text{log}(2+\sqrt{3})}{2} $

$ \int_0^1\int_0^1\int_0^1(x^2+y^2+z^2)^{\frac{3}{2}}dxdydz=-\frac{\pi}{60}+\frac{2\sqrt{3}}{5}+\frac{7\text{log}(2+\sqrt{3})}{20} $