∫ 0 1 ( 1 x + x n − 1 ln ( 1 − x n ) ) d x = γ n {\displaystyle \int _{0}^{1}\left({\frac {1}{x}}+{\frac {x^{n-1}}{\ln(1-x^{n})}}\right)\mathrm {d} x={\frac {\gamma }{n}}}
∫ 0 1 ( 1 x + 1 x + 1 + ⋯ 1 x + m − 1 + x n − 1 ln ( 1 − x n ) ) d x = γ n + ln ( m ) {\displaystyle \int _{0}^{1}\left({\frac {1}{x}}+{\frac {1}{x+1}}+\cdots {\frac {1}{x+m-1}}+{\frac {x^{n-1}}{\ln(1-x^{n})}}\right)\mathrm {d} x={\frac {\gamma }{n}}+\ln(m)}
∫ 0 1 ( 1 1 − x + x n ln ( x ) ) d x = γ + ln ( 1 + n ) {\displaystyle \int _{0}^{1}\left({\frac {1}{1-x}}+{\frac {x^{n}}{\ln(x)}}\right)\mathrm {d} x=\gamma +\ln(1+n)}
∫ 0 1 ( 1 1 − x + 1 2 − x + ⋯ + 1 m − x + x n − 1 ln ( x ) ) d x = γ + ln ( n m ) {\displaystyle \int _{0}^{1}\left({\frac {1}{1-x}}+{\frac {1}{2-x}}+\cdots +{\frac {1}{m-x}}+{\frac {x^{n-1}}{\ln(x)}}\right)\mathrm {d} x=\gamma +\ln(nm)}
∫ 0 1 ( 1 1 − x a + x b ln ( x a ) ) d x = F ( a , b ) {\displaystyle \int _{0}^{1}\left({\frac {1}{1-x^{a}}}+{\frac {x^{b}}{\ln(x^{a})}}\right)\mathrm {d} x=F(a,b)}
---
∫ 0 1 ∫ 0 1 ( 1 ( 1 − x ) ( 1 − y ) + x ln ( x y ) ) d x d y = − ln 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\left({\frac {1}{(1-x)(1-y)}}+{\frac {x}{\ln(xy)}}\right)\mathrm {d} x\mathrm {d} y=-\ln 2}
∫ 0 1 ∫ 0 1 ( x ( 1 + x ) ( 1 + y ) + y ln ( x y ) ) d x d y = − ln 2 ( 2 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\left({\frac {x}{(1+x)(1+y)}}+{\frac {y}{\ln(xy)}}\right)\mathrm {d} x\mathrm {d} y=-\ln ^{2}(2)}
∫ 0 1 ∫ 0 1 ( x ( 1 + x ) ( 1 + y ) ( 1 + x y ) + 1 ln ( x y ) ) d x d y = π 2 24 − 1 {\displaystyle \int _{0}^{1}\int _{0}^{1}\left({\frac {x}{(1+x)(1+y)(1+xy)}}+{\frac {1}{\ln(xy)}}\right)\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{24}}-1}
∫ 0 1 ∫ 0 1 ( 3 ( 1 + x ) ( 1 + y ) ( 1 − x y ) + 3 y 3 ln ( x y ) ) d x d y = ln 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\left({\frac {3}{(1+x)(1+y)(1-xy)}}+{\frac {3y^{3}}{\ln(xy)}}\right)\mathrm {d} x\mathrm {d} y=\ln 2}
∫ 0 1 ∫ 0 1 ( x n 1 − x y + x − y ln ( x y ) ) d x d y = H n n {\displaystyle \int _{0}^{1}\int _{0}^{1}\left({\frac {x^{n}}{1-xy}}+{\frac {x-y}{\ln(xy)}}\right)\mathrm {d} x\mathrm {d} y={\frac {H_{n}}{n}}}
∫ 0 1 ∫ 0 1 ( 1 1 − x y + x − y ln ( x y ) ) d x d y = ζ ( 2 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\left({\frac {1}{1-xy}}+{\frac {x-y}{\ln(xy)}}\right)\mathrm {d} x\mathrm {d} y=\zeta (2)}
∫ 0 ∞ arctan ( x ) ( 1 x − x 2 2 + x 2 ) d x = π ln ( ϕ 0 + ϕ 2 ϕ ) {\displaystyle \int _{0}^{\infty }\arctan \left({\sqrt {x}}\right)\left({\frac {1}{x}}-{\frac {x}{2^{2}+x^{2}}}\right)\mathrm {d} x=\pi \ln \left({\frac {\phi ^{0}+\phi ^{2}}{\phi }}\right)}
∫ 0 ∞ arctan ( 3 x ) ( 1 x − x 2 2 + x 2 ) d x = π ln 5 {\displaystyle \int _{0}^{\infty }\arctan(3{\sqrt {x}})\left({\frac {1}{x}}-{\frac {x}{2^{2}+x^{2}}}\right)\mathrm {d} x=\pi \ln 5}
∫ 0 ∞ x ( 2 n − 1 ) 2 + x 2 ⋅ tanh ( π 2 x ) ( 2 n ) 2 + x 2 d x = 2 H 2 n ′ + 1 n − ln ( 4 ) 4 n − 1 {\displaystyle \int _{0}^{\infty }{\frac {x}{(2n-1)^{2}+x^{2}}}\cdot {\frac {\tanh \left({\frac {\pi }{2}}x\right)}{(2n)^{2}+x^{2}}}\mathrm {d} x={\frac {2H_{2n}^{'}+{\frac {1}{n}}-\ln(4)}{4n-1}}}
∫ 0 ∞ ( 1 x − x ( 2 n − 1 ) 2 + x 2 ) tanh ( π 2 x ) d x = H n − 1 + ln ( 4 ) {\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}-{\frac {x}{(2n-1)^{2}+x^{2}}}\right)\tanh \left({\frac {\pi }{2}}x\right)\mathrm {d} x=H_{n-1}+\ln(4)}
∫ 0 ∞ ( 1 x − x ( 2 n ) 2 + x 2 ) tanh ( π 2 x ) d x = 2 H 2 n − H n {\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}-{\frac {x}{(2n)^{2}+x^{2}}}\right)\tanh \left({\frac {\pi }{2}}x\right)\mathrm {d} x=2H_{2n}-H_{n}}
∫ 0 1 ∫ 0 1 ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = π 2 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{4}}}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = π 2 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{8}}}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ln ( 1 y ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = π 2 − 4 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right)\ln \left({\frac {1}{y}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}-4}{8}}}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = π 2 + 4 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}+4}{8}}}
∫ 0 1 ∫ 0 1 ln ( 1 x y ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = π 2 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{xy}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{4}}}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x y ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = π 2 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{xy}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{2}}}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x ) ln 2 ( 1 y ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = 3 π 2 − 16 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{x}}\right)\ln ^{2}\left({\frac {1}{y}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {3\pi ^{2}-16}{8}}}
∫ 0 1 ∫ 0 1 ln 2 ( x y ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {x}{y}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=2}
∫ 0 1 ∫ 0 1 y ln ( y x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = ln 2 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}y\ln \left({\frac {y}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\ln 2}{2}}}
∫ 0 1 ∫ 0 1 y ln 2 ( y x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = 3 + ln 2 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}y\ln ^{2}\left({\frac {y}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {3+\ln 2}{4}}}
∫ 0 1 ∫ 0 1 arctan ( y x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln ( 1 x ) − ln ( 1 y ) d x d y = 2 β ( 3 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\arctan \left({\frac {y}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln \left({\frac {1}{x}}\right)-\ln \left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=2\beta (3)}
∫ 0 1 ∫ 0 1 ln ( 1 y ) ( ln ln ( 1 x ) − ln ln ( 1 y ) ) 2 ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = 2 β ( 3 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{y}}\right){\frac {\left(\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=2\beta (3)}
∫ 0 1 ∫ 0 1 ln 2 ( 1 y ) ( ln ln ( 1 x ) − ln ln ( 1 y ) ) 2 ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = ζ ( 2 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{y}}\right){\frac {\left(\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=\zeta (2)}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x y ) ( ln ln ( 1 x ) − ln ln ( 1 y ) ) 2 ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = 4 β ( 3 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{xy}}\right){\frac {\left(\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=4\beta (3)}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = π 2 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{8}}}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = π 2 16 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{16}}}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = π 2 16 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{16}}}
∫ 0 1 ∫ 0 1 ln ( y x ) ln ln ( 1 x ) − ln ln ( 1 y ) ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = π 2 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {y}{x}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{8}}}
∫ 0 1 ∫ 0 1 ln ( y x ) ln ( 1 x y ) ln ln ( 1 x ) − ln ln ( 1 y ) ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = π 2 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {y}{x}}\right)\ln \left({\frac {1}{xy}}\right){\frac {\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{8}}}
∫ 0 1 ∫ 0 1 ln ( 1 y ) ( ln 2 ln ( 1 x ) − ln 2 ln ( 1 y ) ) 2 ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = 8 β ( 3 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{y}}\right){\frac {\left(\ln ^{2}\ln \left({\frac {1}{x}}\right)-\ln ^{2}\ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=8\beta (3)}
∫ 0 1 ∫ 0 1 ln 2 ( 1 y ) ( ln 2 ln ( 1 x ) − ln 2 ln ( 1 y ) ) 2 ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = 4 ζ ( 2 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{y}}\right){\frac {\left(\ln ^{2}\ln \left({\frac {1}{x}}\right)-\ln ^{2}\ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=4\zeta (2)}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x y ) ( ln 2 ln ( 1 x ) − ln 2 ln ( 1 y ) ) 2 ln 2 ( 1 x ) + ln 2 ( 1 y ) d x d y = 16 β ( 3 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{xy}}\right){\frac {\left(\ln ^{2}\ln \left({\frac {1}{x}}\right)-\ln ^{2}\ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)+\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=16\beta (3)}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ( ln ln ( 1 x ) − ln ln ( 1 y ) ) 2 ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = ζ ( 2 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right){\frac {\left(\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=\zeta (2)}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x ) ( ln ln ( 1 x ) − ln ln ( 1 y ) ) 2 ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = ζ ( 2 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{x}}\right){\frac {\left(\ln \ln \left({\frac {1}{x}}\right)-\ln \ln \left({\frac {1}{y}}\right)\right)^{2}}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=\zeta (2)}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = π 2 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{xy}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{2}}}
∫ 0 1 ∫ 0 1 ln 2 ( y x ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {y}{x}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=2}
∫ 0 1 ∫ 0 1 ln 2 ( 1 x 3 y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = 2 ( π 2 + 1 ) {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln ^{2}\left({\frac {1}{x^{3}y}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=2(\pi ^{2}+1)}
∫ 0 1 ∫ 0 1 ln ( 1 x 3 y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = π 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x^{3}y}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y=\pi ^{2}}
∫ 0 1 ∫ 0 1 ln ( 1 x 2 y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = 3 π 2 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x^{2}y}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {3\pi ^{2}}{4}}}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ln ( 1 y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = π 2 − 4 8 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right)\ln \left({\frac {1}{y}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}-4}{8}}}
∫ 0 1 ∫ 0 1 ln ( 1 x ) ln ( 1 x y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = π 2 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{x}}\right)\ln \left({\frac {1}{xy}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{4}}}
∫ 0 1 ∫ 0 1 ln ( 1 x y ) ln ln 2 ( 1 x ) − ln ln 2 ( 1 y ) ln 2 ( 1 x ) − ln 2 ( 1 y ) d x d y = π 2 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}\ln \left({\frac {1}{xy}}\right){\frac {\ln \ln ^{2}\left({\frac {1}{x}}\right)-\ln \ln ^{2}\left({\frac {1}{y}}\right)}{\ln ^{2}\left({\frac {1}{x}}\right)-\ln ^{2}\left({\frac {1}{y}}\right)}}\mathrm {d} x\mathrm {d} y={\frac {\pi ^{2}}{2}}}
∫ 0 1 ∫ 0 1 2 x y ln ( x ) ln ( y ) ln ( x y ) ln ( x ( 2 n + 1 ) 2 y ) ln ( x y ) ln ln 2 ( x ) − ln ln 2 ( y ) ln 2 ( x ) − ln 2 ( y ) d x d y = ∑ k = 1 n k {\displaystyle \int _{0}^{1}\int _{0}^{1}2xy\ln(x)\ln(y)\ln(xy)\ln \left(x^{(2n+1)^{2}}y\right)\ln \left({\frac {x}{y}}\right){\frac {\ln \ln ^{2}(x)-\ln \ln ^{2}(y)}{\ln ^{2}(x)-\ln ^{2}(y)}}\mathrm {d} x\mathrm {d} y=\sum _{k=1}^{n}k}
n ≥ 1 {\displaystyle n\geq 1}
lim n → 0 1 n ( 1 + ζ ( 1 − n ) ζ ( 1 + n ) ) = 2 γ {\displaystyle \lim _{n\to 0}{\frac {1}{n}}\left(1+{\frac {\zeta (1-n)}{\zeta (1+n)}}\right)=2\gamma }
lim n → 0 1 n ( 1 + ζ ( 1 + n ) ζ ( 1 − n ) ) = − 2 γ {\displaystyle \lim _{n\to 0}{\frac {1}{n}}\left(1+{\frac {\zeta (1+n)}{\zeta (1-n)}}\right)=-2\gamma }
lim n → 0 1 n ( 1 − ( ζ ( 1 − n ) ζ ( 1 + n ) ) 2 k ) = 4 k γ {\displaystyle \lim _{n\to 0}{\frac {1}{n}}\left(1-\left({\frac {\zeta (1-n)}{\zeta (1+n)}}\right)^{2k}\right)=4k\gamma }
∫ 0 1 ∫ 0 1 ( x − 1 ) ( 2 x − y + 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = γ {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(2x-y+1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\gamma }
∫ 0 1 ∫ 0 1 ( x − 1 ) ( 2 y 2 − y + 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = γ {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(2y^{2}-y+1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\gamma }
∫ 0 1 ∫ 0 1 ( x − 1 ) ( 2 y 2 − y − 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln π 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(2y^{2}-y-1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln {\frac {\pi }{4}}}
∫ 0 1 ∫ 0 1 ( x − 1 ) ( 2 y 2 − x ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = 1 2 ln π 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(2y^{2}-x)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y={\frac {1}{2}}\ln {\frac {\pi }{2}}}
∫ 0 1 ∫ 0 1 ( x − 1 ) ( 2 x − x y + 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(2x-xy+1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln 2}
∫ 0 1 ∫ 0 1 ( x − 1 ) ( 2 y 2 − x y + 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(2y^{2}-xy+1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln 2}
∫ 0 1 ∫ 0 1 ( x − 1 ) ( x y − 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln π 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x-1)(xy-1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln {\frac {\pi }{4}}}
∫ 0 1 ∫ 0 1 ( x 2 − 1 ) ( x y − 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = − 1 2 ln 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(x^{2}-1)(xy-1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=-{\frac {1}{2}}\ln 2}
∫ 0 1 ∫ 0 1 ( x y − 1 ) ( 2 y 2 + x y − x − 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln e π {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(xy-1)(2y^{2}+xy-x-1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln {\frac {e}{\pi }}}
∫ 0 1 ∫ 0 1 ( x y − 1 ) ( 2 y 2 + x y + x ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln π 2 4 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(xy-1)(2y^{2}+xy+x)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln {\frac {\pi ^{2}}{4}}}
∫ 0 1 ∫ 0 1 ( x y − 1 ) ( 2 y 2 + x y + x − 1 ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(xy-1)(2y^{2}+xy+x-1)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln 2}
∫ 0 1 ∫ 0 1 ( x y − 1 ) ( 2 y 2 − x 2 y + x − y + a ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln 2 a π 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(xy-1)(2y^{2}-x^{2}y+x-y+a)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln {\frac {2^{a}\pi }{2}}}
∫ 0 1 ∫ 0 1 ( x y − 1 ) ( 2 y 2 − x 2 y ) ( 1 − x 2 y 2 ) ln ( x y ) ) d x d y = ln π 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(xy-1)(2y^{2}-x^{2}y)}{(1-x^{2}y^{2})\ln(xy))}}\mathrm {d} x\mathrm {d} y=\ln {\frac {\pi }{2}}}
∫ 0 1 ∫ 0 1 ( a y + 1 ) ( x − 1 ) ( 1 − x y ln ( x y ) ) d x d y = 1 + a − a γ {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(ay+1)(x-1)}{(1-xy\ln(xy))}}\mathrm {d} x\mathrm {d} y=1+a-a\gamma }
∫ 0 1 ∫ 0 1 ( a y − 1 ) ( x − 1 ) ( 1 − x y ln ( x y ) ) d x d y = a − ( a + 1 ) γ {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(ay-1)(x-1)}{(1-xy\ln(xy))}}\mathrm {d} x\mathrm {d} y=a-(a+1)\gamma }
∫ 0 1 ∫ 0 1 ( y − 1 ) ( a x 2 + 1 ) ( 1 − x y ln ( x y ) ) d x d y = a 2 ln 2 + γ {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(y-1)(ax^{2}+1)}{(1-xy\ln(xy))}}\mathrm {d} x\mathrm {d} y={\frac {a}{2}}\ln 2+\gamma }
∫ 0 1 ∫ 0 1 ( y 2 − 1 ) ( x 2 + 1 ) ( 1 − x y ln ( x y ) ) d x d y = 3 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(y^{2}-1)(x^{2}+1)}{(1-xy\ln(xy))}}\mathrm {d} x\mathrm {d} y={\frac {3}{2}}}
∫ 0 1 ∫ 0 1 ( y 2 + y − 2 ) ( x 2 + 1 ) ( 1 − x y ln ( x y ) ) d x d y = γ + 3 + ln 2 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(y^{2}+y-2)(x^{2}+1)}{(1-xy\ln(xy))}}\mathrm {d} x\mathrm {d} y=\gamma +{\frac {3+\ln 2}{2}}}
∫ 0 1 ∫ 0 1 ( y 2 + y + a ) ( x 2 − x ) ( 1 − x y ln ( x y ) ) d x d y = 1 + a ln 2 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(y^{2}+y+a)(x^{2}-x)}{(1-xy\ln(xy))}}\mathrm {d} x\mathrm {d} y={\frac {1+a\ln 2}{2}}}
∫ 0 1 ∫ 0 1 x 4 ( 1 2 − x + x 2 ) ( 1 4 − x + x 2 ) 4 + y 2 d x d y = π ln ( ϕ ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{4}}{\left({\frac {1}{2}}-x+x^{2}\right)\left({\frac {1}{4}}-x+x^{2}\right){\sqrt {4+y^{2}}}}}\mathrm {d} x\mathrm {d} y=\pi \ln(\phi )}
∫ 0 1 ∫ 0 1 x 2 ( 1 2 − x + x 2 ) ( 1 4 − x + x 2 ) 4 + y 2 d x d y = − 4 ln ( ϕ ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{2}}{\left({\frac {1}{2}}-x+x^{2}\right)\left({\frac {1}{4}}-x+x^{2}\right){\sqrt {4+y^{2}}}}}\mathrm {d} x\mathrm {d} y=-4\ln(\phi )}
∫ 0 1 ∫ 0 1 x 3 ( 1 2 − x + x 2 ) 2 ( 1 4 − x + x 2 ) 4 + y 2 d x d y = − 4 ln ( ϕ ) {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{3}}{\left({\frac {1}{2}}-x+x^{2}\right)^{2}\left({\frac {1}{4}}-x+x^{2}\right){\sqrt {4+y^{2}}}}}\mathrm {d} x\mathrm {d} y=-4\ln(\phi )}
∫ 0 1 ∫ 0 1 d x d y ( 1 2 − x + x 2 ) 4 − y 2 = π 2 6 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {\mathrm {d} x\mathrm {d} y}{\left({\frac {1}{2}}-x+x^{2}\right){\sqrt {4-y^{2}}}}}={\frac {\pi ^{2}}{6}}}
∫ 0 1 ∫ 0 1 x 2 ( 1 2 − x + x 2 ) 4 − y 2 d x d y = π 6 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{2}}{\left({\frac {1}{2}}-x+x^{2}\right){\sqrt {4-y^{2}}}}}\mathrm {d} x\mathrm {d} y={\frac {\pi }{6}}}
∫ 0 1 ∫ 0 1 x 3 ( 2 x 2 − 2 x + 1 ) 2 ( x − 1 2 ) 2 ( ϕ + y 2 ) 3 d x d y = − 1 ϕ 2 {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{3}}{(2x^{2}-2x+1)^{2}\left(x-{\frac {1}{2}}\right)^{2}{\sqrt {(\phi +y^{2})^{3}}}}}\mathrm {d} x\mathrm {d} y=-{\frac {1}{\phi ^{2}}}}
∫ 0 1 ∫ 0 1 x 3 ( 2 x 2 − 2 x + 1 ) 2 ( x − 1 2 ) 2 ( a + b y 2 ) 3 d x d y = − 1 a a + b {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{3}}{(2x^{2}-2x+1)^{2}\left(x-{\frac {1}{2}}\right)^{2}{\sqrt {(a+by^{2})^{3}}}}}\mathrm {d} x\mathrm {d} y=-{\frac {1}{a{\sqrt {a+b}}}}}
∫ 0 1 ∫ 0 1 x 3 ( 2 x 2 − 2 x + 1 ) 2 ( x − 1 2 ) 2 ( a − b y 2 ) 3 d x d y = − 1 a a − b {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{3}}{(2x^{2}-2x+1)^{2}\left(x-{\frac {1}{2}}\right)^{2}{\sqrt {(a-by^{2})^{3}}}}}\mathrm {d} x\mathrm {d} y=-{\frac {1}{a{\sqrt {a-b}}}}}
