OFFSET
0,1
COMMENTS
The indefinite integral of x^(1/x) can be derived by expanding
x^(1/x) = exp(log(x)/x) = Sum_{n>=0} (log(x)/x)^n/n!,
then integrating term-by-term to get the double summation
F(x) = x + (1/2)*(log(x))^2
- (Sum_{n>=2} (n-1)^(-n-1) * x^(-n+1)/n!
* (Sum_{k=0..n} A350297(n,k)*(log(x))^k)).
We assume the constant of integration in F(x) is zero.
To compute definite integrals of x^(1/x) for ranges of x starting at x=0 or x=1, we would need the values
F(0) = lim {x->0} F(x) = -0.4203695887832022981324...
F(1) = 1 - Sum_{n>=2} (n-1)^(-n-1) = -0.06687278808178024266...
Note that the definite Integral_{t=0..1} x^(1/x) = F(1)-F(0) = A175999.
The calculation of F(0) requires some care. See the FORMULA below.
Since x*(1/x) is the inverse of the infinite exponentiation function E(y) = y^(y^(y^(...))) = x, the definite integrals of these two functions are related by
(Integral_{t=0..y} E(t)) + (Integral_{t=0..x) t*1/t)) = x*y.
Note that the repeated exponentiation in E(y) converges for 0 <= y < e^(1/e), but diverges for y > e^(1/e).
LINKS
Peter Luschny, Table of n, a(n) for n = 0..999
Jonathan Sondow and Diego Marques, Algebraic and transcendental solutions of some exponential equations, Annales Mathematicae et Informaticae 37 (2010) 151-164; see Figure 5.
Eric Weisstein's World of Mathematics, Steiner's Problem.
FORMULA
The calculation of F(0) requires some care, because terms in the formula for F(x) diverge for x=0, but converge for all x > 0, although convergence is progressively slower as x approaches zero. To calculate F(0), first choose a desired precision d (absolute error). Then choose any x such that 0 < x^(1/x) < d, and evaluate F(x) as defined above.
Since F(x)-F(0) < x^(1/x) < d, F(0)=F(x) to the desired precision.
For small x, the main summation terms initially increase in absolute value (but with alternating signs), reach a maximum of about 1/d at n = -log(x)/x, then decrease at an accelerating rate and reach a value of around d at n = -3.5911*log(x)/x, at which point the large terms have mostly cancelled out, and further terms are below precision level.
For example, for a final precision of d = 10^-50, the calculation must allow for intermediate terms and sums as large as 1/d = 10^50, so these terms must be evaluated to at least 100 digits. For 50 digits, F(.03) is a suitable choice, because .03^(1/.03) = 1.7273...*10^-51 < 10^-50. A few extra digits should also be allowed for round off error.
EXAMPLE
0.4203695887832022981324...
MAPLE
# The chosen parameters give about 100 exact decimal places.
Digits := 400: F := proc(b, x) local s, lnx; lnx := log(x);
s := add(add((n*lnx)^k / k!, k = 0..n+1) / (n*x)^(n+2), n = 1..b);
x - x^2 * s + lnx^2 / 2 end:
F(2000, 0.01); # Peter Luschny, Dec 27 2021
MATHEMATICA
RealDigits[N[Sum[1/n^(n+2), {n, 1, 100}] + Integrate[x^(1/x), {x, 0, 1}] - 1, 110]][[1]] (* Amiram Eldar, Dec 29 2021 *)
PROG
(Julia) # The chosen parameters give about 100 exact decimal places.
using Nemo
RR = RealField(1100)
function F(b::Int, x::arb)
lnx = log(x)
s = sum(gamma_regularized(RR(n+2), RR(n)*lnx) * RR(n)^(-n-2) for n in 1:b)
x + (lnx * lnx) / RR(2) - s
end
println( F(1400, RR(0.015)) ) # Peter Luschny, Dec 26 2021
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Robert B Fowler, Dec 26 2021
EXTENSIONS
More terms from Hugo Pfoertner, Dec 26 2021
STATUS
approved