I find this function interesting. The power series coefficients, (1/n)n, decrease quite rapidly — faster than 1/n!. By the root test, for example, the infinite series converges everywhere (infinite radius of convergence), so the function is entire (holomorphic in the entire complex plane). To the best of my knowledge, it is unproven whether the value f(1) = 1.291285997… is transcendental or even irrational. See Weisstein, Eric W. “Sophomore’s Dream.” From MathWorld – A Wolfram Web Resource. Sophomore’s Dream.
The integral form is:
Laplace's method gives:
The plot above is a parametric plot of the real and imaginary parts of f(z), (Re(f(z)), Im(f(z))), for |z| between 0 and 5 and Arg(z) from 0 to 2π. As can be inferred from the above graph, where it pinches back to the real axis, the function is real at approximately f(-4.63878129507 + i*π) = -1.879188987.
It can be hard to see a pattern here, so another approach is to decompose it into separate plots.
The first two plots below show how f(z) maps the imaginary axis, i.e., points of the form 0+yi. It maps the imaginary axis into an expanding spiral shape. Increasing negative values of y map to a clockwise spiral (red), and positive values map to a counterclockwise spiral (blue). The third plot shows that f(z) maps circles of small radius into oval shapes, where the positive real side expands, and the negative real side is pinched in.
By Hadamard's factorization theorem (order 1, not of the form $e^{az+b}$), $f$ must have infinitely many complex zeros. It can be shown that the number of zeros inside a circle of radius r is $n(r) \sim r/(\pi e)$. The plots below suggest some of the values where f(z) = 0. For example, the absolute value of f(−4.30478021156984 ± 14.9450782537071i) is less than 10−12.
The function f $f$ has a unique real critical point at $z_c \approx -5.71837$ (minimum of $f$ on the negative real axis), with critical value: $f(z_c) \approx -1.68773$
Aside from the x-axis, the function f(z) is Real-valued along a series of rightward-opening curves. Far-right arms hug the discrete heights $\text{Im(z)}=k\pi e$; $f\to+\infty$ for even $k$, $f\to-\infty$ for odd $k$. The $k=0$ arm is the real axis. The $k=\pm1$ arms join across the axis into the symmetric $z_c$ horn — both go to $-\infty$, so it carries no zero. That “uses up” $\pm1$, forcing every higher pairing to be even-with-the-next-odd: $(2,3), (4,5), (6,7),\dots$ Each joins a $+\infty$ arm to a $-\infty$ arm, so $f$ sweeps all of $\mathbb{R}$ and the curve carries exactly one zero, $z_n$. Each zero sits just below the even line $2n\pi e$ (e.g. $z_1$ at $14.95 < 17.08$), and consecutive zeros are spaced by $\approx 2\pi e \approx 17.08$ — which reproduces $n(r)\sim r/(\pi e)$ exactly. The k=1 curve is shown below.
The limit of f(z) as z approaches negative infinity along the real axis is minus one.
A proof can be more easily crafted starting from the integral representation $f(-x) = -x\displaystyle\int_0^1 t^{xt}\,dt$ for $x > 0$. Then substitute $t = 1 - u/x$ so that $u = x(1-t)$, ranging from $0$ to $x$.
For each fixed $u \geq 0$, as $x\to\infty$:
This uses the standard limit $(1-u/x)^x \to e^{-u}$.
Using $\ln(1-s) \leq -s$ for $s \in [0,1)$:
For $x \geq 2u$, we have $\;u^2/x \leq u/2$, so the integrand is $\leq e^{-u/2}$. Since $\int_0^\infty e^{-u/2}\,du < \infty$, the Dominated Convergence Theorem applies:
So