Integral $\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx $ Hi I am trying to evaluate the integral
$$
\mathcal{I}(\omega)=\int_{-\infty}^\infty J^3_0(x) e^{i\omega x}\mathrm dx
$$
analytically.  We can also write
$$
\mathcal{I}(\omega)=\mathcal{FT}\big(J^3_0(x)\big)
$$ which is the Fourier Transform of the cube of Bessel function.  The Bessel function $J_0$ is given by
$$
J_0(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-ix\sin t} \mathrm dt.
$$
If it helps, we can represent the cube of the Bessel function by
$$
J^3_0(x)=-3\int J^2_0(x) J_1(x) \mathrm dx, \ \ \ \ \ J_1(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(t-x\sin t)} \mathrm dt.
$$
In general
$$
J_n(x)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(nt-x\sin t)}\mathrm  dt.
$$
The Fourier Transforms of the Bessel function and its square is given by
$$
\mathcal{FT}\big(J_0(x)\big)=\sqrt{\frac{2}{\pi}}\frac{\theta(\omega+1)-\theta(\omega-1)}{\sqrt{1-\omega^2}}
$$ 
and 
$$
\mathcal{FT}\big(J^2_0(x)\big)=\frac{\sqrt{2}K\big(1-\frac{\omega^2}{4}\big)\big(\theta(-\omega-2)-1\big)\big(\theta(\omega-2)-1\big)}{\pi^{3/2}}         
$$
where K is the elliptic-K function and $\theta$ is the heaviside step function.  However I need the cube...
 A: Actually more is true: the Fourier transform of $J_0(ax)J_0(x)^2$ for any $a\in \mathbb{R}$ can be expressed in elliptic integral $K(m) = \int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$.
Here is the explicit form of the already nontrivial case $a=1$: let $$\tag{1}G(w) = \frac{2 \sqrt{4-3 t}}{\pi ^2 w(1-t)}K(m_+)K(m_-)$$
with $$2v = 5-w^2+\sqrt{(w^2-9)(w^2-1)} \qquad t=\frac{v}{v-1}$$
$$4m_{\pm} = 2\pm v\sqrt{4-v} -(2-v)\sqrt{1-v}$$
then for $s\in \mathbb{R}$, our Fourier transform is $$\color{royalblue}{\int_0^\infty  J_0(t)^3 \cos(ts) dt} = -\lim_{\varepsilon\to 0^+}\Im[G(s+i\varepsilon)] $$
One has to be careful about branch problem though, see the end. Special cases:
$$\begin{aligned}\int_0^\infty J_0(x)^3 \cos x dx &= \frac{(2-\sqrt{2}) (\sqrt{2}+1)}{16 \pi ^3}\Gamma \left(\frac{1}{8}\right)^2 \Gamma \left(\frac{3}{8}\right)^2 \\
\int_0^\infty J_0(x)^3 \cos(\sqrt{5}x)dx &= \frac{1}{4\pi^3} \Gamma \left(\frac{3}{20}\right) \Gamma \left(\frac{7}{20}\right) \Gamma \left(\frac{9}{20}\right) \Gamma \left(\frac{21}{20}\right) \end{aligned}$$

Forget our notation of $G$ above, let
$$G(w) = \frac{1}{\pi^3}\int_{[0,\pi]^3} \frac{dx_1dx_2dx_3}{w-\cos x_1-\cos x_2 - \cos x_3}$$
then $G(w)$ is analytic in $\mathbb{C}-[-3,3]$. Using $\int_0^\pi e^{t\cos x} dx = \pi I_0(t)$ we have
$$G(w) = \frac{1}{\pi^3}\int_{[0,\pi]^3} \int_0^\infty e^{-t(w-\cos x_1-\cos x_2 - \cos x_3)} dx_i dt = \int_0^\infty e^{-tw} I_0(t)^3 dt$$
here we assume $w > 3$ to ensure convergence. By deforming contour, we can switch the path  of integral to be $\int_0^{i\infty}$ (integral over big circular arc $\to 0$ when $w>3$), so $$G(w) = i\int_0^\infty e^{-itw} J_0(t)^3 dt = -i\int_0^\infty e^{itw} J_0(t)^3 dt$$
Now RHS is analytic on $\Im w > 0$, so analytic continuation gives
$$G(w) = -i\int_0^\infty e^{itw} J_0(t)^3 dt \qquad \Im w > 0$$
In particular, we see for $s\in \mathbb{R}$, $\lim_{\varepsilon\to 0^+}G(s+i\varepsilon)$ exists, denote it by $G_{\Re}(s) - iG_{\Im}(s)$. Then above equality gives
$$G_{\Re}(s) = \int_0^\infty \sin(ts)J_0(t)^3 dt \qquad G_{\Im}(s) = \color{royalblue}{\int_0^\infty \cos(ts)J_0(t)^3 dt}$$
thus it suffices to show the triple integral actually equals RHS of $(1)$. The integral of $G_{\Re}(s)$ is not our goal, but we see it can be expressed as elliptic integral too.

The $G(w)$ is called Watson triple integral of simple cubic lattice. It turns out that the more general
$$G(w,a) = \frac{1}{\pi^3}\int_{[0,\pi]^3} \frac{dx_1dx_2dx_3}{w-a\cos x_1-\cos x_2 - \cos x_3}$$
called Watson triple integral of singly anisotropic cubic lattice can also be expressed in terms of $K$. Its closed-form evaluation is a culmination of ingenious symbolic computation, which will be only briefly recapitulated below, the reader are strongly encouraged to read the reference for this spectacular achievement.
[to be added later....]

$(1)$ holds at least for $w>3$ when principal branch of square root and $K$ is taken. Since $G(w)$ has no branch point on $\Im w > 0$ from the triple integral definition, $(1)$ is true for all $w \in \mathbb{C}-[-3,3]$ after analytic continuation.
For our concerned range $0<\Re w < 3, \Im w >0$, I think the only term that can cause deviation from principal branch is the term $\sqrt{(w^2-9)(w^2-1)}$ in $v$. After this has been considered, the following Mathematica code can calculate $G(w)$ in our range ($\Im w = 0$ not included)
G[w_] := Module[{x, y, v, t, m1, m2}, x = Re[w]; y = Im[w]; 
   If[5 x - x^3 + x*y^2 <= 0, 
    v = 1/2 (5 - w^2 + Sqrt[(w^2 - 9) (w^2 - 1)]), 
    v = 1/2 (5 - w^2 - Sqrt[(w^2 - 9) (w^2 - 1)])]; t = v/(v - 1); 
   m1 = 1/2 (1 + v*Sqrt[4 - v]/2 - (2 - v) Sqrt[1 - v]/2); 
   m2 = 1/2 (1 - v*Sqrt[4 - v]/2 - (2 - v) Sqrt[1 - v]/2); 
   2/Pi^2/w*Sqrt[4 - 3 t]/(1 - t)*EllipticK[m1] EllipticK[m2]];

