A real smooth function $\varphi$ is said bell shaped iff as the Gaussian : $\varphi''$ is positive on $(-\infty,a) \cup (b,+\infty)$ and negative on $(a,b)$.
I'm interested in the bilateral Laplace transform of such a bell shaped function : $$\mathcal{L}[\varphi](s) = \int_{-\infty}^{+\infty} \varphi(x) e ^{-sx}dx$$ when $\varphi$ is $C^\infty_c$ or is a Schwartz function.
- has $\mathcal{L}[\varphi](s)$ some special properties due to the fact that $\varphi(x)$ is bell-shaped ?
- can you find an example where $\mathcal{L}[\varphi](s)$ has two zeros at the same $\text{Im}(s)$ in its domain of convergence ?
(note that $\phi(x) = \frac{\textstyle e^{\sigma x}}{\textstyle e^{\textstyle e^{x}}+1}$ is bell shaped, and $\mathcal{L}[\phi](\sigma-s) = \Gamma(s)(1-2^{1-s})\zeta(s)$)
I found an example of such a bell-shaped Schwartz function whose Laplace transform has two zeros at the same imaginary part in its domain of convergence :
$$\varphi(x) = e^{x/4}\left(\frac{1}{e^{e^x}-1}-\frac{c}{e^{c e^x}-1}\right)$$
$$\mathcal{L}[\varphi](1/4-s) = \Gamma(s) (1-c^{1-s}) \zeta(s)$$
choosing $c = e^{2 \pi /T} \approx 2.43$ where $T$ is the imaginary part of the first non-trivial zero of $\zeta(s)$, $\mathcal{L}[\varphi](s)$ converges for $Re(s) < 1/4$,
and we have $\mathcal{L}[\varphi](-1/4-iT) = 0$ (due to the $\zeta(1/2+iT)$ factor) and $\mathcal{L}[\varphi](-3/4-iT) = 0$ (due to the $1-c^{-i T}$ factor)
the conclusion is that $\varphi(x)$ is bell shaped :