A209060 Decimal expansion of the triple integral Integral_{z = 0..1} Integral_{y = 0..1} Integral_{x = 0..1} 1/(x*y*z)^(x*y*z) dx dy dz.
1, 2, 1, 4, 8, 3, 7, 9, 9, 6, 0, 1, 7, 1, 6, 2, 7, 0, 0, 6, 9, 1, 1, 2, 0, 5, 2, 4, 8, 0, 2, 4, 2, 1, 2, 2, 2, 2, 3, 8, 2, 7, 3, 8, 8, 4, 9, 0, 5, 5, 6, 1, 1, 9, 9, 9, 4, 6, 1, 4, 2, 2, 9, 5, 2, 1, 1, 1, 4, 1, 3, 7, 5, 2, 4, 0, 0, 3, 7, 7, 1, 0, 5, 9, 1, 2, 1, 2, 4, 0, 0, 7, 7, 8, 8, 7, 4, 2, 1, 8, 3, 8, 1
Offset: 1
Examples
1.21483799601716270069...
Programs
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Mathematica
digits = 103; 1/2*NSum[ (1/n^n + 1/n^(n+1)), {n, 1, Infinity}, WorkingPrecision -> digits+10, NSumTerms -> 100] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 15 2013, from formula *) -
PARI
default( realprecision, 105); v = Vec( Str( suminf( n=1, n^-n + n^-(n+1)) / 20)); for( n=3, 105, print1( v[n],",")); /* Michael Somos, Mar 07 2012 */
Formula
The triple integral is most conveniently estimated from the identity Integral_{z = 0..1} Integral_{y = 0..1} Integral_{x = 0..1} 1/(x*y*z)^(x*y*z) dx dy dz = 1/2*Sum_{n = 1..oo} (1/n^n + 1/n^(n+1)).
Comments