A318741 - OEIS

A318741

Decimal expansion of Pi^3/48 + Pi*log(2)^2/4.

1, 0, 2, 3, 3, 1, 1, 0, 1, 2, 2, 3, 6, 3, 7, 0, 3, 2, 3, 0, 8, 4, 8, 2, 0, 5, 0, 4, 0, 8, 8, 4, 8, 6, 7, 3, 8, 3, 1, 8, 7, 2, 0, 9, 7, 6, 7, 4, 7, 3, 2, 8, 1, 3, 0, 3, 0, 5, 1, 3, 4, 2, 7, 6, 3, 6, 2, 9, 5, 3, 3, 4, 3, 9, 7, 5, 6, 0, 8, 6, 6, 8, 2, 9, 2, 3, 4

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OFFSET

1,3

COMMENTS

The first part of Ramanujan's question 308 in the Journal of the Indian Mathematical Society (III, 168) asked "Show that Integral_{t=0..Pi/2} t * cotan(t) * log(sin(t)) dt = -Pi^3/48 - Pi*log(2)^2/4".

LINKS

Table of n, a(n) for n=1..87.

V. S. Adamchik On Stirling numbers and Euler sums, J. Comput. Appl. Math. 79 (1997) 119-130. Equals [1/2,3] apart from factor 8/sqrt Pi.

B. C. Berndt, Y. S. Choi and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q308, JIMS III).

R. J. Mathar, Chebyshev expansion of x^m*(-log x)^l in the interval 0<=x<=1, arXiv:2408.15212 (2024)

Pedro Ribeiro, Problem 12051, The American Mathematical Monthly, Vol. 125, No. 6 (2018), p. 563; A Series Involving Central Binominal [sic] Coefficients, Solution to Problem 12051 by Hongwei Chen, ibid., Vol. 127, No. 2 (2020), pp. 182-183.

FORMULA

Equals Sum_{k>=0} binomial(2*k,k)/(4^k*(2*k+1)^3) (Ribeiro, 2018). - Amiram Eldar, Oct 04 2021

Equals 4F3(1/2,1/2,1/2,1/2 ; 3/2,3/2,3/2 ; 1) [Adamchik]. - R. J. Mathar, Aug 19 2024

Equals A196877/2. - R. J. Mathar, Aug 23 2024

EXAMPLE

1.0233110122363703230848205040884867383187209767473281303051342763...

MATHEMATICA

RealDigits[Pi^3/48 + Pi*Log[2]^2/4, 10, 100][[1]] (* Amiram Eldar, Oct 04 2021 *)

PROG

(PARI) Pi^3/48+Pi*log(2)^2/4

(PARI) -intnum(x=0, Pi/2, x*cotan(x)*log(sin(x)))

CROSSREFS