Example of an integrable function with an entire extension and whose derivative only vanishes at infinity I am looking for a function $f : \mathbb{C} \to \mathbb{C}$ with the following properties:


*

*$f$ is entire.

*$\int_{-\infty}^\infty |f(t)| \ dt < \infty$ i.e. the restriction of $f$ to the real line is in $L^1(\mathbb{R})$. 

*$\lim_{|t| \to \infty} f'(t) = 0$ i.e. the restriction of $f'$ to the real line is in $C_0(\mathbb{R})$.

*$\int_{-\infty}^\infty |f'(t)| \ dt = \infty$  i.e. the restriction of $f'$ the real line is not in $L^1(\mathbb{R})$. 


So far I've at least had some success coming up with an example of an entire function $g$ whose restriction to the real line is $C_0(\mathbb{R})$, but not $L^1(\mathbb{R})$. For instance, $g(z) = \frac{e^{-z^2} - 1}{z}$ seems to do the trick. I'm not sure if there's an antiderivative $f$ of this $g$ with the desired property though.

Let
$$
f(z)=\int_0^z\frac{\sin(w^2)}{w}\,dw-\frac{\pi}{4}.
$$
$f$ is an even entire function. As the real variable $t \to \infty$, $f(t)$ behaves like $O(t^{-2})$, so, in particular,  that $\lim_{t\to\pm\infty}f(t)=0$ and $\int_{-\infty}^\infty|f(t)|\,dt<\infty$. On the other hand
$$
f'(z)=\frac{\sin(z^2)}{z}
$$
and
$$
\int_{-\infty}^\infty|f'(t)|\,dt=2\int_0^\infty\frac{|\sin(t^2)|}{t}\,dt=\int_0^\infty\frac{|\sin(u)|}{u}\,du=\infty.
$$
