For real variable $x$, the function
\begin{equation}
f(x):=\sum\_{n=1}^\infty \frac{1}{n! n^n}(-x^2)^n
\end{equation}
clearly has infinite radius of convergence and defines a $C^\infty$ function on $\mathbb{R}$.


However, I wonder if this $f(x)$ is a bounded function on $\mathbb{R}$ as well. Also, is it possible to estimate the bound if it is indeed bounded?


I tried to apply summation by parts or the error bounds for an alternating series. But all I have found out is that the partial sums $\sum\_{n=1}^N \frac{1}{n!}(-x^2)^n$ are extremely slowly convergent to the Gaussian function..


Could anyone please help me with the above function $f(x)$?



Evaluating the sum $f(x):=\sum_{n=1}^\infty \frac{1}{n! n^n}(-x^2)^n$ and estimating bounds
As already stated by @Noam and @Alexandre in the comments above,
$$
f(x)= - x \int\_0^1 \mathrm d\tau \sqrt{-8\log \tau} \, J\_1\big(x \tau \sqrt{-8\log \tau}\big),\tag{1}
$$
with Bessel function $J\_1$. Here, I did one further substitution in order to simplify the expression. The function $f(x)$ oscillates at large $x$, and asymptotically seems to become
$$
f\_\infty(x)= \sqrt 2 \cos(2 e^{-1/2} x) + c + O(x^{-1/2}),\tag{2}
$$
with $c\approx -1.055$. The oscillation frequency stems from the oscillations of $J\_1(x \tau \sqrt{-8\log \tau})$ in the integrand at stationary maximal argument (see comment by @Noam), which is at $\tau=e^{-1/2}=0.60653\ldots$. The prefactor of $x^{-1/2}$ is around $-0.285$.


