Are there any non-constant "smooth bump functions" in "closed form" whose Fourier Transforms are also in closed form?

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Disclaimer: due mi ignorance about the topic I change the question to look for what I was really traying to ask, so there are a few answers that were right as it were described the original thing I write (I assumed that every bump function were smooth, which is false). Also I messed it up with requiring to be piecewise in just one compact-suported interval and $0$ outside, which I tried to fix it later (hope now it is understandable). Please be considered and don't downvote them, since was my fault. And also they were indeed useful since through them I realized I was asking something different to was I were intended to ask.

I am looking for the simplest cases possible of one-variable closed-form smooth bump functions $\in C_c^\infty$ [1] with known Fourier transforms in "closed form" (also the function itself). This means it can be described by commonly known functions (exponentials, polynomials, trigonometric, logarithmic, etc.), not defined by more than two piece-wise "steps" (within and outside the compact-supported domain), so things like: $$f(x) = \begin{cases} f_1(x),\,x_0 \leq x < x_1; \\ f_2(x),\,x_1 \leq x < x_2; \\ \,\,\,\,\,\vdots \\ f_n(x),\,x_{n-1} \leq x \leq x_n \end{cases}$$ are not allowed for $n \geq 3$.

As example, $f(x) = e^{\frac{1}{x^2-1}},\,|x|< 1 $ is a valid one, since is defined picewise at most in two pieces: its compact support on one piece defined with just one function, and $0$ outside.

If possible, an answer with domain in $[-1;\,1]$, and also if possible, the functions and its Fourier transforms made by functions that can be described "shortly" (since I want to plot them on Wolfam-Alpha and it didn't recognize functions that are "too long", like infinite sums of other simpler functions).

Beforehand, thanks you very much.


added later

I believe here are show a way to made bump functions $\in C_c^\infty$ that are not defined piecewise, but unfortunately, I don´t think its helps to find their Fourier Transforms.

Later I understood thanks to @CalvinKhor that its equivalent to define them piecewise, but since are continuous under the limits-based definition of continuity, I like them most since are simple to manipulate within differential equations.

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The triangular function $T(x)=1-|x|$ on the domain $[-1,1]$ has the simple Fourier transform $$ \hat T(\xi) = \frac{\sin^2(\pi\xi)}{\pi^2\xi^2}. $$ Indeed one can even write $$ T(x) = \tfrac{1}{2} \bigl(|x-1|+|x+1|\bigr)-2|x| $$ which explicitly evaluates to $0$ outside $[-1,1]$.

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If your smooth function is defined on the whole real line only using $+,-,\times,\div$, and finitely many of those functions (polynomials, exponentials, trigs, and their inverses) and without a piecewise definition, and without an infinite sum or integral or whatever, then your function is better than smooth: it’s analytic. But analytic and compact support implies identically zero by the identity theorem. So the only such function is the trivial function.