= Fourier transform of the Bessel function J0
{c}
With the <Fourier transform> convention $\widehat f(\omega)=\int_{\mathbb R}f(t)e^{-i\omega t}\,dt$, the regular order-zero <Bessel function of the first kind> has the <tempered distribution> transform
$$
\boxed{\widehat{J_0}(\omega)=\frac{2\,\mathbf1_{(-1,1)}(\omega)}{\sqrt{1-\omega^2}}.}
$$
The density is <integrable> despite its endpoint singularities. <Fourier inversion> gives the equivalent <integral representation>
$$
J_0(t)=\frac1\pi\int_{-1}^{1}\frac{e^{it\omega}}{\sqrt{1-\omega^2}}\,d\omega=\frac1\pi\int_0^\pi\cos(t\cos\theta)\,d\theta.
$$
The second formula follows by $\omega=\cos\theta$ and cancellation of the odd imaginary part. It is the standard real integral representation of $J_0$.https://dlmf.nist.gov/10.9.E1[NIST Digital Library of Mathematical Functions, equation 10.9.1]{ref}
Indeed, transforming the order-zero <Bessel differential equation> $t y''+y'+ty=0$ gives $(1-\omega^2)Y'-\omega Y=0$. Inside $(-1,1)$, its solutions are $C(1-\omega^2)^{-1/2}$. The compactly supported choice also solves the equation as a <distribution>: rewrite it as $[(1-\omega^2)Y]'+\omega Y=0$, and observe that $(1-\omega^2)Y=C\sqrt{1-\omega^2}\,\mathbf1_{(-1,1)}$ is continuous and zero at the two endpoints. Its <distributional derivative> has no endpoint <Dirac delta> terms. The inverse is regular at $t=0$, and $y(0)=1$ fixes $C=2$.
The transformed equation alone also allows separate constants on $\omega<-1$ and $\omega>1$. Selecting the compactly supported density amounts to selecting the solution regular at the origin, rather than every possible singular solution of the <Bessel differential equation>.
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