Another one for http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm: 8*Catalan + Pi^2 == 12 + 256*Integrate[y/((-1 + E^(2*Pi*y))*(1 + 16*y^2)^2), {y, 0, Infinity}] More generally, a^s*HurwitzZeta[s, a] == (-1 + 2*a + s)/(2*(-1 + s)) + 2*a*Integrate[(Cos[t]^(-2 + s)*Sin[s*t])/(-1 + E^(2*a*Pi*Tan[t])), {t, 0, Pi/2}] so two more for the Catalan list: (1/16)*(8*Catalan + Pi^2) == 3/4 + (1/2)*Integrate[Sin[2*t]/(-1 + E^((1/2)*Pi*Tan[t])), {t, 0, Pi/2}] , (9/16)*(-8*Catalan + Pi^2) == 5/4 + (3/2)* Integrate[Sin[2*t]/(-1 + E^((3/2)*Pi*Tan[t])), {t, 0, Pi/2}] . Note the rapidity of convergence of the HurwitzZeta integral, especially if you boost a by an integer. Is this how Odlyzko et al do zeta numerics miles from the real axis? --rwg
On Sun, Jul 10, 2011 at 2:25 AM, Bill Gosper <billgosper@gmail.com> wrote:
Another one for http://www.cs.cmu.edu/~adamchik/articles/catalan/catalan.htm : 8*Catalan + Pi^2 == 12 + 256*Integrate[y/((-1 + E^(2*Pi*y))*(1 + 16*y^2)^2), {y, 0, Infinity}] More generally, a^s*HurwitzZeta[s, a] == (-1 + 2*a + s)/(2*(-1 + s)) + 2*a*Integrate[(Cos[t]^(-2 + s)*Sin[s*t])/(-1 + E^(2*a*Pi*Tan[t])), {t, 0, Pi/2}] so two more for the Catalan list: (1/16)*(8*Catalan + Pi^2) == 3/4 + (1/2)*Integrate[Sin[2*t]/(-1 + E^((1/2)*Pi*Tan[t])), {t, 0, Pi/2}] , (9/16)*(-8*Catalan + Pi^2) == 5/4 + (3/2)* Integrate[Sin[2*t]/(-1 + E^((3/2)*Pi*Tan[t])), {t, 0, Pi/2}] . Note the rapidity of convergence of the HurwitzZeta integral, especially if you boost a by an integer. Is this how Odlyzko et al do zeta numerics miles from the real axis?
Subtracting these last two, 10*Catalan - Pi^2 == -1 + Integrate[((2 + E^((1/2)*Pi*Tan[t]))*Sin[2*t])/ (1 + E^((1/2)*Pi*Tan[t]) + E^(Pi*Tan[t])), {t, 0, Pi/2}]
--rwg
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Bill Gosper