An interesting power series

[Franklin T. Adams-Watters]

I'm not sure this really belongs on this mailing list, but I don't know where else to put it.

I've been looking at the function sum_{n=0}^{infinity} x^n/n^n.  Of particular interest is the behavior of the function for negative values of x.

Trivially, f(0) = 1.  f(1) =
2.291285997062663540407282590595600541498619368274522317310002445136944538765234455558817041129429709
= A073009 + 1.  f(-1) =
0.2165694892878655929407356134730245305923180098530690417445821772983998154108595543751357950277310610
= 1 - A083648.

For small negative values of x, f is decreasing, with a root at x =
-1.403761051217759055281232610485776603405490424751396415940373101155405927364061916874538968186718386.
It reaches a minimum at x =
-5.718368648420614152398427034613310546436549865574768585431493548397169770391911785604026826049891868,
where the value is
-0.6877336072666951621247175777384073185758142428273878687363150320321583420985527506265886721706864536,
after which it gradually heads back towards zero.  None of these three numbers are in the OEIS.

This function is closely related to the so-called "Sophomore's Dream" <http://mathworld.wolfram.com/SophomoresDream.html>.  Generalizing the formulas there, we have:

Integral_0^1 t^(x*t) dt = -1/x * Sum_{n=1}^{infinity} (-x)^n/n^n = (1 - f(-x))/x;

or, inverting,

f(x) = 1 + x * Integral_0^1 t^(-x*t) dt.

Evaluating numerically, it appears that as x -> -infinity, f(x) acts like -1/log(|x|); possibly -1/log(|x|) + O(log(log(|x|))/log(|x|)^2), although the latter term is far from certain.  This is equivalent to

limit x->infinity Integral_0^1 t^(x*t) dt = 1/x + 1/(x*log(x)) + O(1/(x*log(x)^2)) [if the error term is correct].

I have no idea how to evaluate this limit to see if this is correct.

It appears that as x -> infinity, f(x) ~ sqrt(x)*exp(x/e).  This is what one would expect from Stirling's approximation, but again I am unable to prove it.
-- 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645


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