A352769 Decimal expansion of Pi^2 * log(2).

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, 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, 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

Offset: 1

Author

Amiram Eldar, Apr 02 2022

Keywords

• nonn
• cons

Comments

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.

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

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

Examples

			6.84108846385711654484747915395409607129977904818791...
		

References

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

Links

• D. Bierens de Haan, Nouvelles tables d'intégrales définies, Leide, 1867, Table 206, eq. 10.
• M. L. Glasser, An Infinite Triple Summation, Problem 80-13, SIAM Review, Vol. 22, No. 3 (1980), pp. 363-364; Solutions by the proposer and by O. G. Ruehr, ibid., Vol. 23, No. 3 (1981), pp. 393-394.
• C. F. Lindman, Examen des nouvelles tables d'intégrales définies de m. Bierens de Haan, Amsterdam 1867, Stockholm, 1891, p. 99, Tab. 206, 10.
• D. Rainer and J. W. Serene, Free energy of superfluid He 3, Phys. Rev. B, Vol. 13, No. 11 (1976), pp. 4745-4765; Erratum, ibid., Vol. 18, No. 7 (1978), p. 3760.
• J. C. Rainwater, Evaluation of frequency sums for the free energy of superfluid He 3, Phys. Rev. B, Vol. 18, No. 7 (1978), pp. 3728-3729.

Crossrefs

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

Programs

Mathematica

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

PARI

Pi^2 * log(2) \\ Michel Marcus, Apr 02 2022

Formula

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

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

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