From what I am aware of, let the function $\varphi$ be $$\varphi(x)=\frac{\cos(x)}{\pi^2-4x^2}.$$ One can easily show that $\varphi\in L^1$ and its tail has a polynomial decay $\varphi\sim x^{-2}$. Its Fourier transform is $$\hat\varphi(s)=c\cos(\pi s/2),\quad s\in(-1,1)$$ with a proper constant $c$ depending on the convention of Fourier transform. Also it is easily seen that $\hat\varphi$ has a compact support.
My question is, is there a generalisation of this test function such that $\varphi_k\in L^1$, $\hat\varphi_k$ has a compact support, and $\varphi_k$ decays like $x^{-k}$ for a fixed $k$? The previous one is like $\varphi_2$.
For even values of $k$ you may just consider
$$\varphi_{2m}(x) =\frac{\cos(x)}{\prod_{n=1}^{m}\left(1-\frac{4x^2}{(2n-1)^2\pi^2}\right)}=\prod_{n > m}\left(1-\frac{4x^2}{(2n-1)^2\pi^2}\right)$$ given by the Weierstrass product for the cosine function. For odd values of $k>1$ you can do essentially the same with the Weierstrass product for the sine function.