I'm looking for a closed form for the following definite integral
$$ I(r,Q) := r \int_0^\infty dx J_1(r x) \left [ J_0(x) \right ]^Q $$
where $r$ is a positive real, $Q$ is a positive integer and $J_a$ is the Bessel function of the first kind.
One should have $I(r,Q)=1$ for $r\ge Q$ (the integral is a CDF).
A potentially useful formula is
$$ J_0(x) = \int_0^{2\pi} \frac{d\theta}{2 \pi} e^{ i x\sin\theta } \, . $$
Any idea?
The following answer expands on my comment.
I also included an evaluation for $I(r, 1) = 0$ when $ r < 1$ and for $I(r, 2)= \frac{2}{\pi} \, \arcsin \left(\frac{r}{2} \right) $ when $r < 2$. These two cases were mentioned by the user mattTheMathLearner in the comments.
The case $I(1,Q) = \frac{1}{Q+1}$ is straightforward since the derivative of $J_{0}(x)$ is $-J_{1}(x)$.
This approach is similar to the approach I used here.
Assume that $r>0$ and that $Q$ is a positive integer.
I'll show that $$r \int_{0}^{\infty}J_{1}(rx) J_{0}(x)^{Q} \, \mathrm dx =1, \quad r \ge Q \ge 2. $$
Let $$H_{\nu}^{(1)}(z)= J_{\nu}(z) + i Y_{\nu}(z) $$ be principal branch of the the Hankel function of the first kind of order $\nu$, which has a branch cut on the negative real axis.
On the upper side of the branch cut, where $z= xe^{\pi i}, \ x >0$,
$$\begin{align} J_{\nu}(x e^{ \pi i }) &= \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!\Gamma(m + \nu +1)} \left(\frac{xe^{ \pi i}}{2} \right)^{2m + \nu} \\ &= e^{\nu \pi i}\sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!\Gamma(m + \nu +1)} \left(\frac{x}{2} \right)^{2m + \nu} \\ &= e^{ \nu \pi i} J_{\nu}(x), \end{align}$$
and $$\begin{align} Y_{\nu}(xe^{\pi i}) &= \frac{J_{\nu}(xe^{\pi i}) \cos(\nu \pi)-J_{-\nu}(x e^{ \pi i})}{\sin(\nu \pi)} \\ &= \frac{e^{ \nu \pi i}J_{\nu}(x)\cos(\nu \pi) -e^{- \nu \pi i}J_{-\nu}(x)}{\sin(\nu \pi)} \\ &= \frac{e^{ \nu \pi i}J_{\nu}(x) \cos(\nu \pi)-e^{-\nu \pi i}J_{\nu}(x) \cos(\nu \pi) + e^{-\nu \pi i}J_{\nu}(x) \cos(\nu \pi)-e^{- \nu \pi i}J_{-\nu}(x)}{\sin(\nu \pi)} \\ &= 2i J_{\nu}(x) \cos(\nu \pi)+ e^{- \nu \pi i } \, \frac{J_{\nu}(x) \cos(\nu \pi) - J_{-\nu}(x)}{\sin(\nu \pi)} \\ &= 2i J_{\nu}(x) \cos(\nu \pi)+e^{- \nu \pi i} Y_{\nu}(x). \end{align} $$
Therefore, $$ \begin{align} H_{\nu}^{(1)}(xe^{\pi i}) &= J_{\nu}(xe^{\pi i})+ i Y_{\nu}(xe^{\pi i}) \\ &= (e^{\nu \pi i}-2\cos(\nu \pi)) J_{\nu}(x)+ ie^{- \nu \pi i}Y_{\nu}(x) \\ &= e^{-\nu \pi i} \left( -J_{\nu}(x) + i Y_{\nu}(x) \right). \end{align}$$
For $\nu =1$, we have $$H_{1}^{(1)}(xe^{ \pi i}) = J_{1}(x) - i Y_{1}(x), \quad x >0. $$
Since the function $J_{0}(x)^{Q}$ is even function of the real variable $x$, we may write $$ r \int_{0}^{\infty} J_{1}(rx) J_{0}(x)^{Q} \, \mathrm dx = \frac{r}{2} \, \Re \operatorname{PV} \int_{-\infty}^{\infty} H_{1}^{(1)}(rx) J_{0}(x)^{Q} \, \mathrm dx, $$ where the integration from $-\infty$ to $0$ is being performed on the upper side of the branch cut.
The principal value sign is needed because the imaginary part of the integrand is asymptotic to $-\frac{2}{r \pi x}$ near $x=0$, and thus the integral is not convergent in the traditional sense (See here.)
Let's integration the function $$f(z) = H_{1}^{(1)}(rz) J_{0}(x)^{Q}, \quad r \ge Q \ge 2, $$ around the contour consisting of the upper side of branch cut from $-R$ to $- \epsilon$, a small clockwise-oriented semicircle about the origin of radius $\epsilon$, the real axis from $\epsilon$ to $R$, and the the upper half of the semicircle $|z|=R$.
The integral on the large semicircle vanishes as $R \to \infty$. This has to do with the fact that the exponential growth in the magnitude of $J_{0}(z)^{Q}$ as $\Im(z) \to + \infty$ is countered by the exponential decay in the magnitude of $H_{1}^{(1)}(rz)$. (See here.)
(If $Q =1$, we need $r$ to be strictly greater than $Q$ for the integral to vanish. This allows us to use Jordan's lemma.)
Since $f(z)$ behaves like as constant times $\frac{1}{z}$ near the origin, the contribution from the small semicircle about the origin does NOT vanish as $\epsilon \to 0$. Similar to an indentation about a simple pole, it goes to $$- \pi i \lim_{z \to 0} z f(z) \overset{\spadesuit}{=} - \pi i \left(-\frac{2i}{r \pi } \right) = - \frac{2}{r}.$$
And since there are no singularities inside the contour, we have $$ \operatorname{PV} \int_{-\infty}^{\infty} H_{1}^{(1)}(rx) J_{0}(x)^{Q} \, \mathrm dx - \frac{2}{r} = 0, $$ from which is follows that $$r \int_{0}^{\infty}J_{1}(rx) J_{0}(x)^{Q} \, \mathrm dx = \frac{r}{2} \left(\frac{2}{r} \right)=1 $$ when $r \ge Q \ge 2$.
$\spadesuit$ https://dlmf.nist.gov/10.8#E1
To show that $$r\int_{0}^{\infty} J_{1}(rx) J_{0}(x) \, \mathrm dx = 0, \quad r <1,$$ integrate $J_{1}(rz)H_{0}^{(1)}(z)$ around the same contour and use Jordan's lemma. In this case the contribution from the indentation about the origin vanishes.
And to show that $$r \int_{0}^{\infty} J_{1}(rx) J_{0}(x)^{2} \, \mathrm dx = \frac{2}{\pi} \, \arcsin \left(\frac{r}{2} \right) , \quad r < 2,$$ we can used the integral representation $$J_{0}(x)^{2} = \frac{2}{\pi} \int_{0}^{\pi/2} J_{0}(2x \cos \theta) \, \mathrm d \theta $$ along with the previous results. (Also see here.)
Assume that $r <2$.
Then
$$ \begin{align} r \int_{0}^{\infty} J_{1}(rx) J_{0}(x)^{2} \, \mathrm dx &= \frac{2r}{\pi} \int_{0}^{\infty} J_{1}(rx) \int_{0}^{\pi/2} J_{0}(2x \cos \theta) \, \mathrm d \theta \, \mathrm d x \\ &= \frac{2r}{\pi} \int_{0}^{\pi/2} \int_{0}^{\infty} J_{1}(rx) J_{0}(2x \cos \theta) \, \mathrm dx \, \mathrm d \theta \\ &= \frac{2}{\pi} \int_{0}^{\pi/2} \frac{r}{2 \cos \theta} \int_{0}^{\infty} J_{1} \left(\frac{ru}{2 \cos \theta} \right) J_{0}(u) \, \mathrm d u \, \mathrm d \theta \\ &\overset{\clubsuit}{=} \frac{2}{\pi} \int_{\arccos (\tfrac{r}{2}) }^{\pi/2} \frac{r}{2 \cos \theta} \int_{0}^{\infty} J_{1} \left(\frac{ru}{2 \cos \theta} \right) J_{0}(u) \, \mathrm d u \, \mathrm d \theta \\ &= \frac{2}{\pi} \int_{\arccos (\tfrac{r}{2})}^{\pi/2} 1 \, \mathrm d \theta \\ &= \frac{2}{\pi} \left(\frac{\pi}{2} - \arccos \left(\frac{r}{2} \right) \right) \\ &= \frac{2}{\pi} \, \arcsin \left(\frac{r}{2} \right). \end{align}$$
The issue with this approach is the justification for switching the order of integration. It's not trivial.
$\clubsuit$ If $0 < \theta < \arccos \left(\frac{r}{2} \right)$, then $\frac{r}{2 \cos \theta} <1 $ and the value of the inner integral is zero.
