Confusion with the Fourier Transform and Complex Differentiability: example with compact-supported function I have a misconception when applying the Fourier Transform to a compacted-supported function and the characteristics of the function obtained.

Intro
I am going to list what I believe is true so you can identify were I am making my conceptual mistake:

*

*The Fourier Transform of a square-integrable function which is compact-supported in the real line must be an entire analytic function due the Paley–Wiener theorem.

*An entire function is a complex-valued function that is holomorphic on the whole complex plane, so it is complex differentiable everywhere, so it is satisfying being Analytic, infinitely differentiable or Smooth, and its real-imaginary decomposition constituents fulfill the Cauchy–Riemann equations.

*If a function $g(z)$ is complex differentiable with the complex variable being described as $z=\sigma+iw$ so the function could be expressed as: $g(\sigma+iw)=u(\sigma,w)+iv(\sigma,w)$ with $u,\,v$ real-valued functions, then each constituents fulfill the Cauchy–Riemann equations as mentioned, which imply that each constituent function is individually an Harmonic function so both functions $u,\,v$ fulfill $\nabla^2 u = \frac{\partial^2 u}{\partial \sigma^2}+\frac{\partial^2 u}{\partial w^2}=0$ and $\nabla^2 v = \frac{\partial^2 v}{\partial \sigma^2}+\frac{\partial^2 v}{\partial w^2}=0$.

*The Fourier Transform $\hat{f}(iw)$ of an "even function" $f(t)=f(-t)$ is a real-valued function $\hat{f}(iw)\in \mathbb{R}$.

*The Fourier Transform $\hat{f}(iw)$ of an real valued function $f(t)$ fulfill that $\hat{f}(iw)^* = \hat{f}(-iw)$ is an Hermitian function.

I am using the electrician notation for the Fourier transform using the angular frequency including the imaginary unit as part of the variable.

Main text
With this, I am going to use the following example to present the problem:
$$f(t) = \left(\frac{1-t^2+|1-t^2|}{2}\right)^4 \tag{Eq. 1}\label{Eq. 1}$$
which is ploted here fulfilling is a real-valued function supported on the real line, but different from zero only in $t \in [-1,\,1]$ so it has compact-support. Also it can be seen is an "even function" due $(-t)^2 \equiv t^2$ for real-valued $t$, and it is also square-integrable since $\int\limits_{-\infty}^{\infty}\left|\left(\frac{1-t^2+|1-t^2|}{2}\right)^4\right|^2 dt = \frac{65536}{109395}\approx 0.6 \ll \infty$. So the function should be fulfilling all the $5$ points of the introduction.
As it can be seen here, the Fourier Transform of $f(t)$ of \eqref{Eq. 1} could been described using the Bessel function of First kind $J_{\nu}(w)$ and the Gamma function $\Gamma(w)$ as:
$$ \begin{array}{r c l}
\hat{f}(iw) = \int\limits_{-\infty}^{\infty} f(t)\ e^{-iwt}dt & = & \text{sgn}(w)\sqrt{\pi}\left(\frac{2}{w}\right)^{4+\frac{1}{2}}\Gamma(4+1)J_{4+\frac{1}{2}}(w)\\
& = & 24\sqrt{\pi}\,\text{sgn}(w)\left(\frac{2}{w}\right)^{\frac{9}{2}}J_{\frac{9}{2}}(w) \tag{Eq. 2}\label{Eq. 2}
\end{array}$$
Since this Fourier Transform $\hat{f}(iw)$ should be holomorphic, but is a real-valued function, I am confused how it is verified is being complex differentiable:
How is analyzed a real-valued function for complex differentiability?
Since the Fourier Transform $\hat{f}(iw) = \hat{f}(\{\sigma\equiv 0\}+iw)$ it will imply that $u(\sigma,w) = 24\sqrt{\pi}\,\text{sgn}(w)\left(\frac{2}{w}\right)^{\frac{9}{2}}J_{\frac{9}{2}}(w)$ and $v(\sigma,w)=0$: naturally the trivial solution zero function will fulfill $\nabla^2 v = 0$, but for reviewing $u(\sigma,w)$ I don´t know how to take the derivative respect to $\sigma$ since it is zero, neither I have a function related to $\sigma$ to make some manipulation. Even so, if I take $\frac{\partial^2}{\partial w^2}u(\sigma,w) \neq 0$ is definitely not zero as it can be seen here, so if the Fourier Transform is complex differentiable as is stated on the initial points, surely must be $\sigma$-related component that is missing.
Surely I am having a conceptual mistake, but I cannot figure it out so far. Hope you could explain what I am doing wrong, displaying which functions are going to be $u(\sigma,w)$ and $v(\sigma,w)$ for the Fourier Transform of \eqref{Eq. 2} showing in detail how is proved is complex differentiable. Beforehand, thanks you very much.
 A: Computing directly, we can find the Fourier transform of your function to be
$$\int_{-1}^1e^{-i\omega t}\:dt = 2\operatorname{sinc}(\omega)$$
$$ \implies \int_{-1}^1(1-4t^2+6t^4-4t^6+t^8)e^{-i\omega t}\:dt = 2\left(1+\left(\frac{d}{d\omega}\right)^2\right)^4\operatorname{sinc}(\omega)$$
Since this Fourier transform is a linear combination of derivatives of an analytic function, it too must be analytic. Notably this formula does not contain any Bessel functions so there must be an error in your calculation or source.
The conditions of the Paley-Wiener theorem treat $\omega\in\Bbb{C}$, not $\omega\in\Bbb{R}$ so this does indeed describe an entire function in $\omega = \mu + i\nu$, not $s=\sigma+i\omega$.
A: There are two parts to your difficulty.
1. How did you end up with a Bessel function?
Mathematica uses specific, very general families to represent common special functions internally, because those families can be integrated using a simple and consistent algorithm.  (I.e., rather than separate algorithms to integrate trig functions and exponentials, it has one algorithm to "rule them all").  Importantly, a given function in that family may not be defined everywhere: the results typically have poles or branch cuts.  When the results are then "massaged" back into human-readable form, the poles and branch cuts often cancel out.
Your result is a typical example: it originally appears expressed as a hypergeometric function.  For the given integer value of the parameter, that hypergeometric function is expressible as a Bessel function, which in turn can be written as a rational function in $w$ and the trig functions: $$J_{9/2}(w)=\sqrt{\frac{2}{\pi w}}\left(\frac{105\sin{w}}{w^4}-\frac{105\cos{w}}{w^3}-\frac{45\sin{w}}{w^2}+\sin{\!(w)}+\frac{10\cos{w}}{w}\right)\tag{a}$$
Notice the initial factor of $w^{-1/2}$.  When multiplied by $\left(\frac{2}{w}\right)^{9/2}$ in (Eq. 2), the two factors combine to produce the (relatively) nice term $\frac{16\sqrt{2}}{w^5}$…as long as $w>0$.  For negative $w$, the branch cuts that Mathematica chooses for $J_{9/2}$ and $\left(\frac{2}{\cdot}\right)^{9/2}$ combine to give give an extra factor of $-1$, the same way that $\sqrt{x^2}\neq x$ for negative $x$.  That's why the superficially discontinuous $\mathrm{sgn}$ appears in (Eq. 2).
Using expression (a) to eliminate the $\mathrm{sgn}$ is definitely the way to proceed here.  But you asked a more general question, so I'll pretend we don't know (a).
2. How to analyze a complex function for differentiability?
Theoretically, look for a Taylor series. $f$ is differentiable at $u_0+v_0i$ iff there exists a number $L\in\mathbb{C}$ such that, for any $(u,v)$ near $(u_0,v_0)$, $$f'(u+vi)=f(u_0+v_0i)+L((u+vi)-(u_0+v_0i))+o(|(u+vi)-(u_0+v_0i)|)$$  Notice that this is a two-dimensional limit (in $(u,v)$).  It is insufficient to test this on any single one-dimensional curve, such as the $u=u_0=0$ line.  That's a problem, because your computation only applies on the line of the positive imaginary axis.
Practically, the same as any function of one real variable: apply the rules of calculus to reduce to "atomic functions," and then use known properties of the atomic functions.
In your case, (Eq. 2) is a product of several known atomic functions: $$\hat{f}(iw)=24\sqrt{\pi}\cdot\mathrm{sgn}(w)\cdot\left(\frac{2}{w}\right)^{9/2}\cdot J_{9/2}(w)$$  Now apply the product rule: \begin{align*}
\hat{f}'(iw)=&\frac{d}{dw}(24\sqrt{\pi})\cdot\mathrm{sgn}(w)\cdot\left(\frac{2}{w}\right)^{9/2}\cdot J_{9/2}(w)+{} \\
&24\sqrt{\pi}\cdot(\mathrm{sgn})'(w)\cdot\left(\frac{2}{w}\right)^{9/2}\cdot J_{9/2}(w)+{} \\
&24\sqrt{\pi}\cdot\mathrm{sgn}(w)\cdot\frac{d}{dw}\left(\frac{2}{w}\right)^{9/2}\cdot J_{9/2}(w)+{} \\
&24\sqrt{\pi}\cdot\mathrm{sgn}(w)\cdot\left(\frac{2}{w}\right)^{9/2}\cdot J_{9/2}'(w)
\end{align*}  Some of these atomic functions have known derivatives: \begin{align*}
&\frac{d}{dw}(24\sqrt{\pi})=0 \\
&\frac{d}{dw}\left(\frac{2}{w}\right)^{9/2}=\frac{9}{2}\left(\frac{2}{w}\right)^{7/2}\left(-\frac{2}{w^2}\right)\quad\quad\quad(*) \\
&J_{9/2}'(w)=\frac{J_{7/2}(w)-J_{11/2}(w)}{2}
\end{align*} where the starred equation holds true away from whichever branch cut we end up choosing.
The function $\mathrm{sgn}$ does not have a known derivative, because it doesn't even make sense for most complex arguments.  Sure, $\mathrm{sgn}(1)=1$ and $\mathrm{sgn}(-1)=-1$, but what is $\mathrm{sgn}(i)$?  Any option that matches $\mathrm{sgn}$ on the real line will do; my first choice is $$\mathrm{sgn}(z)=\frac{\sqrt{z^2}}{z}$$ (with the branch cut along the negative real axis) which happens to work here.  Away from the branch cut, $$\mathrm{sgn}'(w)=\frac{\frac{2z}{2\sqrt{z^2}}\cdot z-\sqrt{z^2}}{z^2}=\frac{1}{\sqrt{z^2}}-\frac{\sqrt{z^2}}{z^2}\quad\quad\quad(\dagger)$$
Substituting the known derivatives, we now find that (Eq. 2) is differentiable… except possibly at $0$, where we might have a pole, and along the branch cut.  Those need to be checked individually for removable singularities.  You still don't need to fall back on Taylor series, though: by Cauchy's integral formula, if a function is holomorphic on a dense set and continuous everywhere, then it is entire.  So just check continuity by hand; it turns out this holds if we choose the branch cuts in ($*$) and ($\dagger$) to coincide.
