A245637 Decimal expansion of Integral_{x = 1..infinity} 1/x^x dx.
7, 0, 4, 1, 6, 9, 9, 6, 0, 4, 3, 7, 4, 7, 4, 4, 6, 0, 0, 1, 1, 4, 4, 2, 1, 0, 7, 8, 5, 7, 1, 2, 3, 8, 1, 0, 5, 8, 7, 5, 9, 7, 2, 6, 8, 6, 9, 3, 4, 5, 6, 5, 5, 5, 4, 7, 8, 2, 9, 7, 6, 1, 5, 8, 4, 6, 0, 8, 7, 0, 7, 8, 3, 8, 1, 3, 3, 1, 9, 0, 7, 5, 0, 8, 1, 3, 7, 8, 8, 6, 6, 6, 0, 0, 3, 4, 1, 6, 8, 0, 7, 3, 1, 7
Offset: 0
Examples
0.704169960437474460011442107857123810587597268693456555478297615846...
References
- Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 4.6 Fransén-Robinson Constant, p. 263.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Peter Bala, Borel summation of a family of divergent series
- M. L. Glasser, A note on Beukers's and related integrals, Amer. Math. Monthly 126(4) (2019), 361-363.
- Math Overflow, Value of divergent sum Sum_{n >= 0} (-1)^n*n^n.
- G. N. Watson, Theorems stated by Ramanujan (VIII): Theorems on Divergent Series, Journal of the London Mathematical Society, Volume s1-4, Issue 2, April 1929, Pages 82-86.
Programs
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Mathematica
NIntegrate[1/x^x, {x, 1, Infinity}, WorkingPrecision -> 104] // RealDigits // First
Formula
From Peter Bala, Nov 10 2019: (Start)
Equals Integral_{x = 1..oo} x*(1 + log(x))/x^x dx - 1.
Equals Integral_{x = 1..oo} x*(1 - log^2(x))/x^x dx.
Conjecturally, equals 1 - Integral_{x = 1..oo, y = 1..oo} 1/(x*y)^(x*y) dx dy. [added Dec 21 2022: follows from Glasser's Theorem 1.] (End)
From Peter Bala, Dec 21 2022: (Start)
Equals 1 - Integral_{x = 1..oo} log(x)/x^x dx (since d/d(1/x^x) = -(1 + log(x))/x^x).
Equals the Borel sum of the divergent series 1 - 1^1 + 2^2 - 3^3 + 4^4 - .... See Watson, Section 5. Compare with the convergent series 1/1^1 - 1/2^2 + 1/3^3 - 1/4^4 + ... = Integral_{x = 0..1} x^x dx. See A083648.
More generally, for nonnegative integers a and b, the divergent series Sum_{n >= 0} (-1)^n*(a*n + b)^n is Borel summable to Integral_{x = 1..oo} x^(a-b-1)/x^(x^a) dx. (End)