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%I #14 Mar 10 2016 17:43:31
%S 5,3,9,2,9,8,6,7,6,5,5,7,6,5,2,7,0,2,9,8,7,5,5,4,9,9,2,5,5,1,7,9,0,6,
%T 0,9,7,9,8,8,5,4,8,3,5,6,0,9,8,8,0,1,7,4,0,8,7,2,3,3,7,5,0,3,4,7,4,4,
%U 8,1,2,3,5,1,2,0,4,0,2,3,1,9,1,8,3,5,3,1,2,8,5,6,5,0,2,0,1,8,8,7
%N Decimal expansion of integral(t=0,1,zeta(t)+1/(1-t))=0.539298676...
%C The function zeta(t)+1/(1-t)) being almost linear between 0 and 1, the integral is near 1/4 + EulerGamma/2 = 0.5386... - _Jean-François Alcover_, Nov 08 2012
%p Digits := 100 ; H := proc(n) option remember ; if n < 1 then 0 ; elif n = 1 then 1; else procname(n-1)+1/n ; fi; end: bn := proc(n) n*(1-gamma-H(n-1))-1/2+add(binomial(n,k)*(-1)^k*Zeta(k),k=2..n) ; end: x := 0.0 ; for n from 0 do ints := expand(binomial(s,n)) ; ints := int(ints,s=0..1) ; x := x+evalf((-1)^n*bn(n)*ints) ; printf("%.40f\n",1+x) ; od: # J Comp Appl Math 220 (2008) 58. - _R. J. Mathar_, May 19 2009
%t RealDigits[ N[ Integrate[Zeta[t] + 1/(1-t), {t, 0, 1}] , 105]][[1]] (* _Jean-François Alcover_, Nov 08 2012 *)
%o (PARI) intnum(t=0,1,zeta(t)+1/(1-t)) \\ _Charles R Greathouse IV_, Mar 10 2016
%K cons,nonn
%O 0,1
%A _Benoit Cloitre_, Apr 07 2009
%E Leading zero removed by _R. J. Mathar_, May 19 2009
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