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:
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 functon 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 of the absolute value of f(z) 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 has a minimum along the real axis at about z = -5.71837 where f takes the value of -1.68773.
The limit of f(z) as z approaches negative infinity along the real axis is minus one. $$\lim_{x\to-\infty} f(x) = -1$$ 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 $u = x(1-t)$, ranging from $0$ to $x$). $$x\int_0^1 t^{xt}\,dt = \int_0^x \left(1 - \frac{u}{x}\right)^{x\left(1-\frac{u}{x}\right)} du$$ For each fixed $u \geq 0$, as $x\to\infty$: $$\left(1-\frac{u}{x}\right)^{x\left(1-\frac{u}{x}\right)} = \underbrace{\left[\left(1-\frac{u}{x}\right)^x\right]^{\!\left(1-u/x\right)}}_{\to\; (e^{-u})^1} \;\longrightarrow\; e^{-u}$$, using the standard limit $(1-u/x)^x \to e^{-u}$. Using $\ln(1-s) \leq -s$ for $s \in [0,1)$: $$x\left(1-\frac{u}{x}\right)\ln\!\left(1-\frac{u}{x}\right) \leq -u\!\left(1-\frac{u}{x}\right) \leq -u + \frac{u^2}{x}$$ For $x \geq 2u$: $\;u^2/x \leq u/2$, so the integrand is $\leq e^{-u/2}$. Since $\int_0^\infty e^{-u/2}\,du < \infty$, the Dominated Convergence Theorem applies: $$\int_0^x \left(1-\frac{u}{x}\right)^{x(1-u/x)} du \;\longrightarrow\; \int_0^\infty e^{-u}\,du = 1$$ So $$\lim_{x\to+\infty} f(-x) = \lim_{x\to+\infty}\left(-\int_0^x\left(1-\frac{u}{x}\right)^{x(1-u/x)}du\right) = -1 \qquad \blacksquare$$