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A350358 Value of -F(0), where F(x) is the indefinite integral of x^(1/x).

%I A350358 #44 Mar 22 2025 08:41:19
%S A350358 4,2,0,3,6,9,5,8,8,7,8,3,2,0,2,2,9,8,1,3,2,4,3,9,3,8,1,8,1,8,0,8,8,1,
%T A350358 8,9,9,8,1,5,4,9,5,5,3,2,8,7,1,2,2,3,9,5,1,4,5,3,5,4,0,7,3,5,4,3,6,4,
%U A350358 0,1,2,0,8,1,2,1,8,7,0,6,2,2,7,3,1,5,1,4
%N A350358 Value of -F(0), where F(x) is the indefinite integral of x^(1/x).
%C A350358 The indefinite integral of x^(1/x) can be derived by expanding
%C A350358     x^(1/x) = exp(log(x)/x) = Sum_{n>=0} (log(x)/x)^n/n!,
%C A350358 then integrating term-by-term to get the double summation
%C A350358     F(x) = x + (1/2)*(log(x))^2
%C A350358              - (Sum_{n>=2} (n-1)^(-n-1) * x^(-n+1)/n!
%C A350358              * (Sum_{k=0..n} A350297(n,k)*(log(x))^k)).
%C A350358 We assume the constant of integration in F(x) is zero.
%C A350358 To compute definite integrals of x^(1/x) for ranges of x starting at x=0 or x=1, we would need the values
%C A350358     F(0) = lim {x->0} F(x) = -0.4203695887832022981324...
%C A350358     F(1) = 1 - Sum_{n>=2} (n-1)^(-n-1) = -0.06687278808178024266...
%C A350358 Note that the definite Integral_{t=0..1} x^(1/x) = F(1)-F(0) = A175999.
%C A350358 The calculation of F(0) requires some care. See the FORMULA below.
%C A350358 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
%C A350358     (Integral_{t=0..y} E(t) dt) + (Integral_{t=0..x} t^(1/t) dt) = x*y.
%C A350358 Note that the repeated exponentiation in E(y) converges for 0 <= y < e^(1/e), but diverges for y > e^(1/e).
%H A350358 Peter Luschny, <a href="/A350358/b350358.txt">Table of n, a(n) for n = 0..999</a>
%H A350358 Jonathan Sondow and Diego Marques, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_37_from151to164.pdf">Algebraic and transcendental solutions of some exponential equations</a>, Annales Mathematicae et Informaticae 37 (2010) 151-164; see Figure 5.
%H A350358 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SteinersProblem.html">Steiner's Problem</a>.
%F A350358 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.
%F A350358 Since F(x)-F(0) < x^(1/x) < d, F(0)=F(x) to the desired precision.
%F A350358 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.
%F A350358 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.
%e A350358 0.4203695887832022981324...
%p A350358 # The chosen parameters give about 100 exact decimal places.
%p A350358 Digits := 400: F := proc(b, x) local s, lnx; lnx := log(x);
%p A350358    s := add(add((n*lnx)^k / k!, k = 0..n+1) / (n*x)^(n+2), n = 1..b);
%p A350358    x - x^2 * s + lnx^2 / 2 end:
%p A350358 F(2000, 0.01); # _Peter Luschny_, Dec 27 2021
%t A350358 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 *)
%o A350358 (Julia) # The chosen parameters give about 100 exact decimal places.
%o A350358 using Nemo
%o A350358 RR = RealField(1100)
%o A350358 function F(b::Int, x::arb)
%o A350358     lnx = log(x)
%o A350358     s = sum(gamma_regularized(RR(n+2), RR(n)*lnx) * RR(n)^(-n-2) for n in 1:b)
%o A350358     x + (lnx * lnx) / RR(2) - s
%o A350358 end
%o A350358 println( F(1400, RR(0.015)) ) # _Peter Luschny_, Dec 26 2021
%Y A350358 Cf. A350297, A175999, A176000, A073229.
%K A350358 nonn,cons
%O A350358 0,1
%A A350358 _Robert B Fowler_, Dec 26 2021
%E A350358 More terms from _Hugo Pfoertner_, Dec 26 2021