A352769 - OEIS

%I #17 Jan 08 2025 11:35:12

%S 6,8,4,1,0,8,8,4,6,3,8,5,7,1,1,6,5,4,4,8,4,7,4,7,9,1,5,3,9,5,4,0,9,6,

%T 0,7,1,2,9,9,7,7,9,0,4,8,1,8,7,9,1,3,5,1,5,3,2,4,1,3,1,8,4,8,5,1,7,1,

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

%N Decimal expansion of Pi^2 * log(2).

%C Rainer and Serene (1976) used the sum that is given in the first formula in the calculation of the free energy of superfluid Helium-3. They evaluated the sum by 6.8.

%C Rainwater (1978) found the integral representation of this sum, which is given in the second formula, and evaluated it by 6.84109 +- 0.00001.

%C Glasser and Ruehr (1981) proved the sum is equal to this constant.

%D Murray S. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, Philadelphia, 1987, pp. 187-188.

%H D. Bierens de Haan, <a href="https://archive.org/details/nouvetaintegral00haanrich/page/n329/mode/2up">Nouvelles tables d'intégrales définies</a>, Leide, 1867, Table 206, eq. 10.

%H M. L. Glasser, <a href="https://doi.org/10.1137/1022061">An Infinite Triple Summation</a>, Problem 80-13, SIAM Review, Vol. 22, No. 3 (1980), pp. 363-364; <a href="https://doi.org/10.1137/1023076">Solutions</a> by the proposer and by O. G. Ruehr, ibid., Vol. 23, No. 3 (1981), pp. 393-394.

%H C. F. Lindman, <a href="https://babel.hathitrust.org/cgi/pt?id=wu.89062908546&amp;view=1up&amp;seq=103&amp;size=150">Examen des nouvelles tables d'intégrales définies de m. Bierens de Haan, Amsterdam 1867</a>, Stockholm, 1891, p. 99, Tab. 206, 10.

%H D. Rainer and J. W. Serene, <a href="https://doi.org/10.1103/PhysRevB.13.4745">Free energy of superfluid He 3</a>, Phys. Rev. B, Vol. 13, No. 11 (1976), pp. 4745-4765; <a href="https://doi.org/10.1103/PhysRevB.18.3760">Erratum</a>, ibid., Vol. 18, No. 7 (1978), p. 3760.

%H J. C. Rainwater, <a href="https://doi.org/10.1103/PhysRevB.18.3728">Evaluation of frequency sums for the free energy of superfluid He 3</a>, Phys. Rev. B, Vol. 18, No. 7 (1978), pp. 3728-3729.

%F Equals Sum_{i,j,k, positive and negative odd integers} sign(i) * sign(j) * sign(k) * sign(i+j-k)/(i^2*j^2).

%F Equals -8 * Integral_{x=0..1} arctanh(x)*log(x)/(x*(1-x^2)) dx - 7*zeta(3)/2.

%F Equals Integral_{x=0..Pi/2} (4*x^2*cos(x) - x*(Pi-x))/sin(x) dx (Bierens de Haan, 1867; Lindman, 1891).

%e 6.84108846385711654484747915395409607129977904818791...

%t RealDigits[Pi^2*Log[2], 10, 100][[1]]

%o (PARI) Pi^2 * log(2) \\ _Michel Marcus_, Apr 02 2022

%Y Cf. A000796 (Pi), A002162 (log(2)), A002117 (zeta(3)), A002388 (Pi^2), A086054 (Pi*log(2)).

%K nonn,cons

%O 1,1

%A _Amiram Eldar_, Apr 02 2022