A306617 - OEIS

%I #37 Mar 04 2019 02:03:38

%S 8,2,8,8,5,9,5,7,9,6,6,9,2,7,9,2,8,6,6,9,7,2,2,9,0,2,0,7,7,5,1,0,3,0,

%T 2,6,7,6,9,1,0,5,7,5,5,9,7,7,1,2,1,1,4,5,2,4,4,0,4,0,3,3,1,7,9,5,7,1,

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

%N Decimal expansion of a constant related to the asymptotics of A324425.

%C Ulrich Neumann found a closed form, see the "Mathematica Stack Exchange" link.

%H Vaclav Kotesovec, <a href="/A306617/b306617.txt">Table of n, a(n) for n = 0..1000</a> [Terms beyond 79 were computed using Ulrich Neumann's program]

%H Mathematica Stack Exchange, <a href="https://mathematica.stackexchange.com/questions/192410/how-to-compute-this-integral-with-a-better-precision">How to compute this integral with a better precision ?</a>

%F Equals limit_{n->infinity} (A324425(n)^(1/n^3))/n^2.

%e 0.828859579669279286697229020775103026769105755977121145244040331795...

%p evalf(exp(integrate(log(x^2 + y^2 + z^2), x = 0..1, y = 0..1, z = 0..1)), 20);

%p evalf(exp(integrate(-2 + 2*sqrt(y^2 + z^2) * arctan(1/sqrt(y^2 + z^2)) + log(1 + y^2 + z^2), y = 0..1, z = 0..1)), 20);

%t ixr = Exp[Integrate[1/3 (Log[1 + Sec[fi]^2] + (-7 + 3 Log[1 + Sec[fi]^2]) Sec[fi]^2 + 2 (Pi - 2 ArcTan[Sec[fi]]) Sec[fi]^3), {fi, 0, Pi/4}]]; Chop[N[ixr, 120]] (* A program by Ulrich Neumann added by _Vaclav Kotesovec_, Mar 03 2019. The calculation takes several minutes. *)

%o (PARI) exp(intnum(z=0, 1 ,intnum(y=0, 1, intnum(x=0, 1, log(x^2 + y^2 + z^2)))))

%o (PARI) exp(intnum(z=0, 1 ,intnum(y=0, 1, -2 + 2*sqrt(y^2 + z^2) * atan(1/sqrt(y^2 + z^2)) + log(1 + y^2 + z^2))))

%Y Cf. A324425.

%K nonn,cons

%O 0,1

%A _Vaclav Kotesovec_, Feb 28 2019

%E More terms computed by Ulrich Neumann added by _Vaclav Kotesovec_, Mar 03 2019