What is the minimal order of a “sigmoid-like” entire function? The following question is motivated by Is there a complex analytic function that acts like sigmoid on the reals?:

What is the minimal order of an entire function $f$ which is real-valued, increasing, and bounded on the real axis?

In the above-referenced Q&A an additional condition is imposed on the second derivative, which I have omitted here for simplicity.
The order of an entire function $f$ is defined as
$$
\rho(f) = \limsup_{r \to \infty} \frac{\log \log M(r, f)}{\log r}
$$
where $ M(r, f) = \max \{ |f(z)| : |z| = r \}$ is the maximum of $|f|$ on the circle with radius $r$.
The order of an entire function is equal to the order of its derivative (see, e.g., here), therefore an equivalent question is

What is the minimal order of an entire function $h$ which is real-valued, positive, and integrable on the real axis?

The answer must be a value between $1$ and $2$:

*

*The function $h(z) = \exp(-z^2)$ satisfies these conditions and has order $2$, so that is an upper bound for the minimal order. The corresponding function $f(z) = \int_0^z f(t) \, dt$ is – apart from a constant factor – the Gauss error function.


*On the other hand, the Phragmén–Lindelöf principle shows that $\rho(f) \ge 1$ for any entire function which is bounded on the real axis.
Some thoughts:

*

*The easiest way to construct admissible function $h$ of finite order would be to set $h(z) = \exp(p(z))$ with a polynomial $p$ with real coefficients. But in order to make $\int_{-\infty}^\infty h(x) \, dx$ finite, the degree of $p$ must be at least $2$, so that does not lead to a better upper bound.


*$f(z) = \sin(z)$ has order one, is real-valued and bounded on the real axis, but is not increasing on the real axis. The question is whether the monotony condition enforces a larger order or not.
Any examples or results which improve the lower or upper bound are welcome.
 A: The answer is $1$ as follows; first it is enough to find $g$ (entire non zero, order $1$) nonnegative and integrable on the real line, so $M=\int_{-\infty}^{\infty}g(x)dx < \infty, g(x)\ge 0, x \in \mathbb R$ as then $f(z)=\int_{[0,z]}g(t)dt$ works, since for $x<y, f(y)-f(x)=\int_x^yg(t)dt$ and $g(t)>0$ except possible at countably many discrete zeroes so in particular on some small interval in $(x,y)$ while clearly $-M < f(x) <M$, $f$ is entire and of order $1$ by general theory
But now pick any nonzero $h \in L^2(-a,a)$ with some $a>0$ and by Paley Wiener (very easy part - converse is the difficult part) $G(z)=\int_{-a}^ah(t)e^{izt}dt$ is an entire (nonzero) exponential function (order $1$ functions are also called exponential for obvious reasons) of type $a$ which is square integrable on any horizontal line, $\int_{-\infty}^{\infty}|G(x+iy)|^2dx < \infty$ for any fixed $y \in \mathbb R$.
If we take $h$ odd, then $G$ is conjugate invariant since $$G(x) =\int_{-a}^ah(t)(\cos xt+i\sin xt)=\int_{-a}^ah(t)\cos xtdt, x \in \mathbb R$$
Then if we take $g(z)=G(z)\overline {G(\bar z)}$ (or $g(z)=G^2(z)$ if $G$ is conjugate invariant), one clearly has $g$ entire non zero, $g(x) \ge 0$ for real $x$ and $\int_{-\infty}^{\infty}g(x)dx=\int_{-\infty}^{\infty}|G(x)|^2dx< \infty$, while $g$ is of order $1$ and type $2a$ by general theory
(note that if one wants, one can obtain a strictly positive $g$ by an elaboration of the above, picking a $y$ where $G$ has no zeroes on the horizontal line $x+iy$ and defining $G_1(z)=G(z+iy)$ and using the square integrability of $G$ on the horizontal line passing through $iy$) etc)
Edit later; choose $a=1, h(t)=t/|t|$ the signum of $t \ne 0$ so $G(z)=2\int_0^1\cos ztdt=\frac{2}{z}\sin z$ so $g(z)=4\frac{\sin^2(z)}{z^2}$ is the example of @reuns in the comment above
