This series converges for all $x \in (-\infty, \infty)$, thus the function is analytic on the real line and defined by its Taylor series.

However, unlike the exponential function, this one is very elusive - I can't find any other representation for it - no integral, no differential equation, nothing.

Its derivatives exist (an infinite number of them), but it seems they can't be connected to the function itself in any other way.

I know that $s(1)-1$ is called 'Sophomore's Dream', because it has integral representation:

$$\sum_{k=1}^{\infty} \frac{1}{k^k}=\int_0^1 \frac{1}{t^t}dt$$

I don't know if the same method can be used to find a general integral representation for $s(x)$, but I have some hope.

The most interesting (in my opinion) property of this function - $s(x)$ and its derivatives all have exactly one zero (for $x_0<0$) and one minimum ($x_m<0$ and $s(x_m)<0$). They are connected in a sense, since obviously if $s(x)$ has a minimum, then $s'(x)$ has a zero.

Here is a plot of $s(x)$ for $x<0$:

enter image description here

I added $1$ to the sum in the definition of $s(x)$ for consistency - it can be seen from the form of its derivatives:

$$s(x)=1+x+\frac{x^2}{2^2}+\frac{x^3}{3^3}+\dots=1+\sum_{k=1}^{\infty} \frac{x^k}{k^k}$$

$$s'(x)=1+\frac{x}{2}+\frac{x^2}{9}+\frac{x^3}{64}+\dots=\sum_{k=1}^{\infty} \frac{k~x^{k-1}}{k^k}$$

$$s''(x)=\frac{1}{2}+\frac{2x}{9}+\frac{3x^2}{64}+\frac{4x^3}{625}+\dots=\sum_{k=1}^{\infty} \frac{k(k-1)~x^{k-2}}{k^k}$$

$$s'''(x)=\frac{2}{9}+\frac{3x}{32}+\frac{12x^2}{625}+\frac{5x^3}{1944}+\dots=\sum_{k=1}^{\infty} \frac{k(k-1)(k-2)~x^{k-3}}{k^k}$$

It is apparent, that $\lim_{n \to \infty} s^{(n)}(0) = 0$, however, the zero on the negative line actually moves to the left with each differentiation (see the position of the minimum). So I'm not sure, how the 'infinite derivative' of $s(x)$ would look.

I'd like to know if someone studied this function, and get a reference for it. But the main question is - what other definitions are possible for $s(x)$, except for the series?

Thanks to this great answer, I'm able to write the integral definition:

$$s(x)=1+\int_0^1 x ~u^{-u~x} du$$

So I consider my question answered.

By the way, if we define the function:

$$p(x)=\frac{s(x)-1}{x}=\int_0^1 u^{-u~x} du=\sum_{k=1}^{\infty} \frac{x^{k-1}}{k^k}$$

We get a very monotone (exponential-like) behavior, without any zeroes or minima. The function and several of its derivatives are plotted below.

enter image description here

  • 1
    $\begingroup$ I think this post and this post are somehow relevant to your function $s(x)$. $\endgroup$ – Sangchul Lee Mar 22 '16 at 21:19
  • $\begingroup$ Your answer to the first question also answers my question about integral definition. Maybe you can post the part about integral here, so I could accept it? $\endgroup$ – Yuriy S Mar 22 '16 at 21:36

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