This is not an answer, but rather a derivation of one of your results that you possibly don’t have a proof for (at least not in the forum anyway). Firstly consider the Hankel functions of the first and second kind; $H_{v}^{1}\left( z \right)={{J}_{v}}\left( z \right)+i{{Y}_{v}}\left( z \right)$ and $H_{v}^{2}\left( z \right)={{J}_{v}}\left( z \right)-i{{Y}_{v}}\left( z \right)$.
Note that $H_{v}^{1}\left( z \right)$ has zeros in the lower half plane, but not in the upper half plane or on the real axis (the zeros are distributed asymptotically to $-i\log \left( 2 \right)/2n$)
Note
$$\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)}\sim \frac{{{a}^{v}}}{{{b}^{v}}} \> as \> \left| u \right|\to 0$$
and
$$\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)}={{e}^{i\left( b-a \right)u}}\sqrt{\frac{a}{b}}\left( 1+O\left( {{\left| u \right|}^{-1}} \right) \right) \> as \> \left| u \right|\to \infty $$
Therefore consider a contour consisting of the real line, indented above z=0 with a small semi-circle of radius $\delta $, and a large semi-circle of radius R in the upper half plane. Therefore
$$\int\limits_{C}^{{}}{\frac{H_{v}^{1}\left( bz \right)}{H_{v}^{1}\left( az \right)z}dz}=\int\limits_{R}^{{}}{{}}+\int\limits_{-\infty }^{\infty }{{}}+\int\limits_{\delta }^{{}}{{}}=0$$
and hence
$$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=\frac{\pi i{{a}^{v}}}{{{b}^{v}}}$$
Here the integral is understood to be a principal value. Note the following properties
$${{Y}_{v}}\left( z{{e}^{m\pi i}} \right)={{e}^{-vm\pi i}}{{Y}_{v}}\left( z \right)+2i\sin \left( vm\pi \right)\cot \left( v\pi \right){{J}_{v}}\left( z \right)$$
$${{J}_{v}}\left( z{{e}^{m\pi i}} \right)={{e}^{-vm\pi i}}{{J}_{v}}\left( z \right)$$
From which it is noted
$$H_{v}^{1}\left( -z \right)={{J}_{v}}\left( -z \right)+i{{Y}_{v}}\left( -z \right)=-{{\left( -1 \right)}^{v}}{{J}_{v}}\left( z \right)+{{\left( -1 \right)}^{v}}\left( i{{Y}_{v}}\left( z \right) \right)={{\left( -1 \right)}^{v-1}}\left( {{J}_{v}}\left( z \right)-i{{Y}_{v}}\left( z \right) \right)={{\left( -1 \right)}^{v-1}}H_{v}^{2}\left( z \right)$$
From this we break the integral in two, i.e.
$$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=-\int\limits_{0}^{\infty }{\frac{H_{v}^{1}\left( -bu \right)}{H_{v}^{1}\left( -au \right)u}du}+\int\limits_{0}^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}$$
Or using the results above
$$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=\int\limits_{0}^{\infty }{\frac{H_{v}^{1}\left( bu \right)H_{v}^{2}\left( au \right)-H_{v}^{2}\left( bu \right)H_{v}^{1}\left( au \right)}{H_{v}^{1}\left( au \right)H_{v}^{2}\left( au \right)u}du}$$
Simplifying this one obtains
$$\int\limits_{-\infty }^{\infty }{\frac{H_{v}^{1}\left( bu \right)}{H_{v}^{1}\left( au \right)u}du}=2i\int\limits_{0}^{\infty }{\frac{{{J}_{v}}\left( au \right){{Y}_{v}}\left( bu \right)-{{J}_{v}}\left( bu \right){{Y}_{v}}\left( au \right)}{\left( J_{v}^{2}\left( au \right)+Y_{v}^{2}\left( au \right) \right)u}du}$$
Finally
$$\int\limits_{0}^{\infty }{\frac{{{J}_{v}}\left( au \right){{Y}_{v}}\left( bu \right)-{{J}_{v}}\left( bu \right){{Y}_{v}}\left( au \right)}{\left( J_{v}^{2}\left( au \right)+Y_{v}^{2}\left( au \right) \right)u}du}=\frac{\pi {{a}^{v}}}{2{{b}^{v}}}$$
This result is listed in Bateman. As for your integral, I have several results, but nothing I am happy with as yet.
It turns out that the Fourier transform of $J_0^3$ can still be expressed
in terms of complete elliptic integrals, but it's considerably
more complicated than the formula for ${\cal FT}(J_0^2)$:
for starters, it involves the periods of a curve $E$ defined over ${\bf C}$
but (except for a few special values of $\omega$) not over ${\bf R}$.
Assume $|\omega| < 3$, else $I(\omega) = 0$. Then the relevant curve is
$$
E : Y^2 = X^3
- \bigl(\frac{3}{4} f^2 + \frac{27}{2} f - \frac{81}{4}\bigr) X^2
+ 9 f^3 X
$$
where
$$
f = \frac12 \bigl( e + 1 + \sqrt{e^2-34e+1} \bigr)
$$
and
$$
e = \bigl( |\omega| + \sqrt{\omega^2-1} \, \bigr)^2.
$$
Let $\lambda_1, \lambda_2$ be generators of the period lattice of $E$
with respect to the differential $dx/y$ (note that these are twice
the periods that gp reports, because gp integrates $dx/2y$
for reasons coming from the arithmetic of elliptic curves). Then:
if $|\omega| \leq 1$ then
$$
I(\omega) =
\left|\,f\,\right|^{5/2}\, \left|\,f-1\right| \frac{\Delta}{(2\pi)^2},
$$
where $\Delta = \bigl|{\rm Im} (\lambda_1 \overline{\lambda_2}) \bigr|$
is the area of the period lattice of $E$. If $1 \leq |\omega| \leq 3$ then
$$
I(\omega) = \left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2}
\frac{\Delta'}{(2\pi)^2},
$$
where $\Delta' = \bigl| {\rm Re}(\lambda_1 \overline{\lambda_2}) \bigr|$
for an appropriate choice of generators $\lambda_1,\lambda_2$
(these "appropriate" generators satisfy $|\lambda_1|^2 = \frac32 |\lambda_2|^2$,
which determines them uniquely up to $\pm$ except for finitely many
choices of $\omega$).
The proof, alas, is too long to reproduce here, but here's the basic idea.
The Fourier transform of $J_0$ is $(1-\omega^2)^{-1/2}$ for $|\omega|<1$
and zero else. Hence the Fourier transforms of $J_0^2$ and $J_0^3$
are the convolution square and cube of $(1-\omega^2)^{-1/2}$.
For $J_0^2$, this convolution square is supported on $|\omega| \leq 2$,
and in this range equals
$$
\int_{t=|\omega|-1}^1 \left( (1-t^2) (1-(|\omega|-t)^2) \right)^{-1/2} \, dt,
$$
which is a period of an elliptic curve [namely the curve
$u^2 = (1-t^2) (1-(|\omega|-t)^2)$], a.k.a. a complete eliptic integral.
For $J_0^3$, we likewise get a two-dimensional integral, over a hexagon
for $|\omega|<1$ and a triangle for $1 \leq |\omega| < 3$, that is
a period of the K3 surface
$$
u^2 = (1-s^2) (1-t^2) (1-(|\omega|-s-t)^2).
$$
(The phase change at $|\omega|=1$ was already noted here in a
now-deleted partial answer.)
In general, periods of K3 surfaces are hard to compute, but this one
turns out to have enough structure that we can convert the period
into a period of the surface $E \times \overline E$ where $\overline E$
is the complex conjugate.
Now to be honest I have only the formulas for the "correspondence" between
our K3 surface and $E \times \overline E$, which was hard enough to do,
but didn't keep track of the elementary multiplying factor
that I claim to be $\left|\,f\,\right|^{5/2}\, \left|\,f-1\right|$
or $\left|\,f\,\right|^{-4}\, \left|\,f-1\right|^5 (3/2)^{13/2}$.
I obtained these factors by comparing numerical values for the few
choices of $\omega$ for which I was able to compute $I(\omega)$
to high precision (basically rational numbers with an even numerator
or denominator); for example $I(2/5)$ can be computed in gp
in under a minute as
intnum(x=0,5*Pi,2*cos(2*x/5) * sumalt(n=0,besselj(0,x+5*n*Pi)^3))
There were enough such $c$, and the formulas are sufficiently simple,
that they're virtually certain to be correct.
Here's gp code to get $e$, $f$, $E$, and generators $\lambda_1,\lambda_2$
of the period lattice:
e = (omega+sqrt(omega^2-1))^2
f = (sqrt(e^2-34*e+1)+(e+1)) / 2
E = ellinit( [0, -3/4*f^2-27/2*f+81/4, 0, 9*f^3, 0] )
L = 2*ellperiods(E)
lambda1 = L[1]
lambda2 = L[2]
NB the last line requires use of gp version 2.6.x; earlier versions
did not directly implement periods of curves over $\bf C$.
For $\omega=0$ we have $e=1$, $f=3$, and $E$ is the curve
$Y^2 = X^3 - 27 X^2 + 243 X = (X-9)^3 + 3^6$,
so the periods can be expressed in terms of beta functions and
we recover the case $\nu=0$ of Question 404222, How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$? .
Best Answer
Changing $x=a\sinh t$ in the integral, \begin{equation} I=\int_\beta^{\gamma}e^{ika\cosh t}\,dt \end{equation} where $\beta=\sinh^{-1}b/a$ and $\gamma=\sinh^{-1}(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<\sigma<\pi$ \begin{equation} H_0^{(1)}(ka,w)=\frac{2}{i\pi}\int_w^{\infty+i\sigma}e^{ika\cosh t}\,dt \end{equation} Then \begin{equation} I=\frac{2}{i\pi}\left[ H_0^{(1)}(ka,\beta)- H_0^{(1)}(ka,\gamma) \right] \end{equation} It can also be expressed as \begin{equation} I= K_0(-ika,\beta)- K_0(-ika,\gamma) \end{equation} where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).