ζ ( 2 ) = ∑ n = 0 ∞ 1 n 2 = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 {\displaystyle \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)
ζ ( 4 ) = ∑ n = 0 ∞ 1 n 4 = 1 1 4 + 1 2 4 + 1 3 4 + 1 4 4 + ⋯ = π 4 90 {\displaystyle \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}\!}
ζ ( 6 ) = ∑ n = 0 ∞ 1 n 6 = 1 1 6 + 1 2 6 + 1 3 6 + 1 4 6 + ⋯ = π 6 945 {\displaystyle \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}\!}
∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = π 2 8 {\displaystyle \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}\!}
∑ n = 0 ∞ 1 ( 2 n + 1 ) 4 = 1 1 4 + 1 3 4 + 1 5 4 + 1 7 4 + ⋯ = π 4 96 {\displaystyle \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}\!}
∑ n = 0 ∞ 1 ( 2 n + 1 ) 6 = 1 1 6 + 1 3 6 + 1 5 6 + 1 7 6 + ⋯ = π 6 960 {\displaystyle \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}\!}
∑ n = 0 ∞ ( − 1 ) n + 1 n 2 = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + ⋯ = π 2 12 {\displaystyle \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}\!}
∑ n = 0 ∞ ( − 1 ) n + 1 n 4 = 1 1 4 − 1 2 4 + 1 3 4 − 1 4 4 + ⋯ = 7 π 4 720 {\displaystyle \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}\!}
∑ n = 0 ∞ ( − 1 ) n + 1 n 6 = 1 1 6 − 1 2 6 + 1 3 6 − 1 4 6 + ⋯ = 31 π 6 30240 {\displaystyle \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}\!}
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) 3 = 1 1 3 − 1 3 3 + 1 5 3 − 1 7 3 + ⋯ = π 3 32 {\displaystyle \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}\!}
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) 5 = 1 1 5 − 1 3 5 + 1 5 5 − 1 7 5 + ⋯ = 5 π 5 1536 {\displaystyle \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}\!}
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) 7 = 1 1 7 − 1 3 7 + 1 5 7 − 1 7 7 + ⋯ = 61 π 5 184320 {\displaystyle \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}\!}
∑ p : p r i m e 1 p 2 = 1 2 2 + 1 3 2 + 1 5 2 + 1 7 2 + ⋯ = 9 2 π 2 {\displaystyle \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}\!}
∑ p : p r i m e 1 p 4 = 1 2 4 + 1 3 4 + 1 5 4 + 1 7 4 + ⋯ = 15 2 π 4 {\displaystyle \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}\!}
∑ p : p r i m e 1 p 6 = 1 2 6 + 1 3 6 + 1 5 6 + 1 7 6 + ⋯ = 11340 691 π 6 {\displaystyle \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}\!}
ζ ( 2 k ) = ∑ n = 0 ∞ 1 n 2 k = 1 1 2 k + 1 2 2 k + 1 3 2 k + 1 4 2 k + ⋯ = 2 2 k | B 2 k | π 2 k 2 ( 2 k ) ! {\displaystyle \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)!}\!}
∑ n = 0 ∞ 1 ( 2 n + 1 ) 2 k = 1 1 2 k + 1 3 2 k + 1 5 2 k + 1 7 2 k + ⋯ = ( 2 2 k − 1 ) | B 2 k | π 2 k 2 ( 2 k ) ! {\displaystyle \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)!}\!}
∑ n = 0 ∞ ( − 1 ) n + 1 n 2 k = 1 1 2 k − 1 2 2 k + 1 3 2 k − 1 4 2 k + ⋯ = ( 2 2 k − 2 ) | B 2 k | π 2 k 2 ( 2 k ) ! {\displaystyle \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)!}\!}
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) 2 k = 1 1 2 k − 1 2 2 k + 1 3 2 k − 1 4 2 k + ⋯ = | E 2 k | π 2 k 2 2 k ( 2 k ) ! {\displaystyle \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)!}\!}
∑ p : p r i m e 1 p k = 1 2 k + 1 3 k + 1 5 k + 1 7 k + ⋯ = ζ 2 ( 2 k ) − ζ ( 4 k ) 2 ζ ( 2 k ) ζ ( 4 k ) {\displaystyle \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)}\!}
∏ n = 1 ∞ ( 1 + ( − 1 ) n + 1 2 n + 1 ) = π 2 4 {\displaystyle \prod_{n=1}^\infty (1+\frac{(-1)^{n+1}}{2n+1})=\frac{\pi \sqrt{2}}{4}}
∏ n = 1 ∞ ( 1 − 1 ( 6 n ) 2 ) = 3 π {\displaystyle \prod_{n=1}^\infty (1-\frac{1}{(6n)^2})=\frac{3}{\pi}}
∏ n = 1 ∞ ( 1 − 3 2 ( 2 n ) 2 ) = − 2 3 π {\displaystyle \prod_{n=1}^\infty (1-\frac{3^2}{(2n)^2})=-\frac{2}{3\pi}}
∏ n = 1 ∞ ( 1 − 5 2 ( 2 n ) 2 ) = 2 5 π {\displaystyle \prod_{n=1}^\infty (1-\frac{5^2}{(2n)^2})=\frac{2}{5\pi}}
∏ p : o d d p r i m e ( 1 − ( − 1 ) p − 1 2 p ) = 4 π {\displaystyle \prod_{p:odd prime} (1-\frac{(-1)^{\frac{p-1}{2}}}{p}) = \frac{4}{\pi}}
∏ p : o d d p r i m e ( 1 − ( − 1 ) p + 1 2 p ) = 2 π {\displaystyle \prod_{p:odd prime} (1-\frac{(-1)^{\frac{p+1}{2}}}{p}) = \frac{2}{\pi}}
∏ c : c o m p o s i t e ( 1 − 1 c 2 ) = π 2 12 {\displaystyle \prod_{c:composite} (1-\frac{1}{c^2})= \frac{\pi^2}{12}}
∏ c : o d d c o m p o s i t e ( 1 − 1 c 2 ) = π 3 32 {\displaystyle \prod_{c:odd composite} (1-\frac{1}{c^2})= \frac{\pi^3}{32}}
∑ n = 1 ∞ n − 11 e 2 n π − 1 = − 1 2 ζ ( 11 ) + 1453 π 851350500 {\displaystyle \sum_{n=1}^\infty \frac{n^{-11}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (11)+\frac{1453\pi}{851350500}}
∑ n = 1 ∞ n − 7 e 2 n π − 1 = − 1 2 ζ ( 7 ) + 19 π 113400 {\displaystyle \sum_{n=1}^\infty \frac{n^{-7}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (7)+\frac{19\pi}{113400}}
∑ n = 1 ∞ n − 3 e 2 n π − 1 = − 1 2 ζ ( 3 ) + 7 π 360 {\displaystyle \sum_{n=1}^\infty \frac{n^{-3}}{e^{2n\pi}-1}=-\frac{1}{2}\zeta (3)+\frac{7\pi}{360}}
∑ n = 1 ∞ n − 1 e 2 n π − 1 = − π 12 − 1 2 log ω π 2 {\displaystyle \sum_{n=1}^\infty \frac{n^{-1}}{e^{2n\pi}-1}=-\frac{\pi}{12}-\frac{1}{2} \log \frac{\omega}{\pi \sqrt{2}}}
∑ n = 1 ∞ n e 2 n π − 1 = 1 24 − 1 8 π {\displaystyle \sum_{n=1}^\infty \frac{n}{e^{2n\pi}-1}=\frac{1}{24}-\frac{1}{8\pi}}
∑ n = 1 ∞ n 3 e 2 n π − 1 = − 1 240 + 1 80 ( ω π ) 4 {\displaystyle \sum_{n=1}^\infty \frac{n^{3}}{e^{2n\pi}-1}=-\frac{1}{240}+\frac{1}{80}(\frac{\omega}{\pi})^4}
∑ n = 1 ∞ n 5 e 2 n π − 1 = 1 504 {\displaystyle \sum_{n=1}^\infty \frac{n^{5}}{e^{2n\pi}-1}=\frac{1}{504}}
∑ n = 1 ∞ n 7 e 2 n π − 1 = − 1 480 + 3 160 ( ω π ) 8 {\displaystyle \sum_{n=1}^\infty \frac{n^{7}}{e^{2n\pi}-1}=-\frac{1}{480}+\frac{3}{160}(\frac{\omega}{\pi})^8}
∑ n = 1 ∞ n 9 e 2 n π − 1 = 1 264 {\displaystyle \sum_{n=1}^\infty \frac{n^{9}}{e^{2n\pi}-1}=\frac{1}{264}}
∑ n = 1 ∞ n 11 e 2 n π − 1 = − 691 65520 + 89 1040 ( ω π ) 12 {\displaystyle \sum_{n=1}^\infty \frac{n^{11}}{e^{2n\pi}-1}=-\frac{691}{65520}+\frac{89}{1040}(\frac{\omega}{\pi})^{12}}
∑ n = 1 ∞ n 13 e 2 n π − 1 = 1 24 {\displaystyle \sum_{n=1}^\infty \frac{n^{13}}{e^{2n\pi}-1}=\frac{1}{24}}
∑ n = 1 ∞ n 15 e 2 n π − 1 = − 3617 16320 + 43659 5440 ( ω π ) 16 {\displaystyle \sum_{n=1}^\infty \frac{n^{15}}{e^{2n\pi}-1}=-\frac{3617}{16320}+\frac{43659}{5440}(\frac{\omega}{\pi})^{16}}
∑ n = 1 ∞ n 17 e 2 n π − 1 = 43867 28728 {\displaystyle \sum_{n=1}^\infty \frac{n^{17}}{e^{2n\pi}-1}=\frac{43867}{28728}}
∑ n = 1 ∞ n 4 k + 1 e 2 n π − 1 = B 4 k + 2 8 k + 4 ( k ≥ 1 ) {\displaystyle \sum_{n=1}^\infty \frac{n^{4k+1}}{e^{2n\pi}-1}=\frac{B_{4k+2}}{8k+4} (k \geq 1)}
∑ n = 1 ∞ n 4 k − 1 e 2 n π − 1 = 1 8 k ( B 4 k − ( ω π ) 4 k H 4 k ) ( k ≥ 1 ) {\displaystyle \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)}
∑ n = 1 ∞ n − ( 4 k − 1 ) e 2 n π − 1 {\displaystyle \sum_{n=1}^\infty \frac{n^{-(4k-1)}}{e^{2n\pi}-1}}
= − 1 2 ζ ( 4 k − 1 ) + ( 2 π ) 4 k − 1 4 ( 4 k + 4 ) ! ( ( 4 k 2 k ) ( − 1 ) k B 2 k 2 − 2 ∑ m = 0 k − 1 ( 4 k 2 m ) ( − 1 ) m B 2 m B 4 k − 2 m ) ( k ≥ 1 ) {\displaystyle =-\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)}
∫ 0 π 2 x 2 d x sin 2 x = π log 2 {\displaystyle \int_0^{\frac{\pi}{2}}\frac{x^2dx}{\sin^2 x}=\pi\text{log}2}
∫ 0 π 2 log 2 ( cos x ) d x = π 3 24 + π log 2 2 2 {\displaystyle \int_0^{\frac{\pi}{2}} \text{log}^2(\cos x)dx=\frac{\pi^3}{24}+\frac{\pi \text{log}^22}{2}}
∫ 0 ∞ x e − x 1 − e − 2 x d x = π ( 1 + 2 log 2 ) 8 {\displaystyle \int_0^\infty xe^{-x}\sqrt{1-e^{-2x}}dx=\frac{\pi(1+2\text{log}2)}{8}}
∫ 0 ∞ x 2 d x e x − 1 = 4 π log 2 2 + π 3 3 {\displaystyle \int_0^\infty \frac{x^2dx}{\sqrt{e^x-1}}=4\pi\text{log}^22+\frac{\pi^3}{3}}
∫ 0 1 x 2 d x ( 1 + x 4 ) 1 − x 4 = π 8 {\displaystyle \int_0^1\frac{x^2dx}{(1+x^4)\sqrt{1-x^4}}=\frac{\pi}{8}}
∫ 0 1 log 2 x 1 + x + x 2 d x = 8 π 3 81 3 {\displaystyle \int_0^1\frac{\text{log}^2x}{1+x+x^2}dx=\frac{8\pi^3}{81\sqrt{3}}}
∫ 0 1 log ( 1 + x 2 ) x 2 d x = π 2 − log 2 {\displaystyle \int_0^1\frac{\text{log}(1+x^2)}{x^2}dx=\frac{\pi}{2}-\text{log}2}
∫ 0 1 log ( 1 + x 3 ) 1 − x + x 2 d x = 2 π log 3 3 {\displaystyle \int_0^1\frac{\text{log}(1+x^3)}{1-x+x^2}dx=\frac{2\pi\text{log}3}{\sqrt{3}}}
∫ 0 1 ∫ 0 1 d x d y 1 − x y = π 2 6 {\displaystyle \int_0^1\int_0^1\frac{dxdy}{1-xy}=\frac{\pi^2}{6}}
∫ 0 1 ∫ 0 1 d x d y 1 − x 2 y 2 = π 2 8 {\displaystyle \int_0^1\int_0^1\frac{dxdy}{1-x^2y^2}=\frac{\pi^2}{8}}
∫ 0 1 ∫ 0 1 d x d y 1 + x 2 + y 2 = − π 6 + log ( 2 + 3 ) {\displaystyle \int_0^1\int_0^1\frac{dxdy}{\sqrt{1+x^2+y^2}}=-\frac{\pi}{6}+\text{log}(2+\sqrt{3})}
∫ 0 1 ∫ 0 1 ( x − 1 x + 1 ) 2 ( y − 1 y + 1 ) 2 ( x y − 1 x y + 1 ) 2 d x d y = 5 − π 2 − 4 log 2 + 16 log 2 2 {\displaystyle \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}
∫ 0 1 ∫ 0 1 ∫ 0 1 ( x 2 + y 2 + z 2 ) − 1 2 d x d y d z = − π 4 + 3 log ( 2 + 3 ) 2 {\displaystyle \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}}
∫ 0 1 ∫ 0 1 ∫ 0 1 ( x 2 + y 2 + z 2 ) 1 2 d x d y d z = − π 24 + 3 4 + log ( 2 + 3 ) 2 {\displaystyle \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}}
∫ 0 1 ∫ 0 1 ∫ 0 1 ( x 2 + y 2 + z 2 ) 3 2 d x d y d z = − π 60 + 2 3 5 + 7 log ( 2 + 3 ) 20 {\displaystyle \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}}
(ゼータ関数)
