OFFSET
0,2
COMMENTS
Replacing 1/2 with any other number 0 < t < 1, the value of the integral is t - 1 - log(t).
REFERENCES
J.-M. Monier, Cours, Analyse, Tome 4, 2ème année, MP.PSI.PC.PT, Dunod, 1997, Exercice 4.3.14 pages 53 and 367.
FORMULA
Equals log(2) - 1/2 = A002162 - 1/2.
Equals Sum_{k>=1} 1/((2k-1)*(2k)*(2k+1)). - Bruno Berselli, Mar 16 2014
From Amiram Eldar, Jul 28 2020: (Start)
Equals Sum_{k>=0} (-1)^k/(k+3).
Equals Sum_{k>=2} 1/(k * 2^k).
Equals Sum_{k>=2} 1/(4*k^2 - 2*k).
Equals Sum_{k>=2} (zeta(k) - 1)/2^k.
Equals Sum_{k>=1} zeta(2*k + 1)/2^(2*k + 1). (End)
From Bernard Schott, Nov 22 2021: (Start)
Equals Sum_{k>=1} (S(k) - log(2)) when S(k) = Sum_{m=1..k} (-1)^(m+1) / m.
Equals Integral_{x=0..1} x/(1+x)^2 dx. (End)
Equals Sum_{k,m>=1} (-1)^(k+m)/(k+m). - Amiram Eldar, Jun 09 2022
Equals Integral_{x = 0..1} Integral_{y = 0..1} x*y/(x + y)^2 dy dx. - Peter Bala, Dec 12 2022
EXAMPLE
0.193147180559945309417232121458176568075500134360255254120680009493393621969...
MAPLE
(evalf(log(2) - 1/2), 111); # Bernard Schott, Nov 25 2021
MATHEMATICA
RealDigits[Log[2] - 1/2, 10, 111][[1]]
PROG
(PARI) log(2)-1/2 \\ Charles R Greathouse IV, Dec 27 2012
CROSSREFS
KEYWORD
AUTHOR
Robert G. Wilson v, Dec 27 2012
STATUS
approved