Convergence of integral of $J_{\nu}(x)$ It is well known that, if $J_{\nu}$ denotes the Bessel function of the first kind, then we have
$$\displaystyle \int_{0}^{\infty} J_{0}(x) \ \mathrm{d}x = 1.$$
Moreover, since $\int J_{1}(x) = -J_{0}(x)$, then we also have
$$\displaystyle \int_{0}^{\infty} J_{1}(x) \ \mathrm{d}x = 1,$$
since $J_{0}(0) = 1$ and $J_{0}$ tends to zero at infinity. Can we say anything similar about $\displaystyle \int_{0}^{\infty}J_{\nu}(x)  \ \mathrm{d}x,$ where $\nu > 0$ is either an integer or a half-integer? Do all of these integrals converge? Are they listed somewhere in a table of integrals? Gradshteyn-Ryzhik lists some examples, but they only seem applicable to $\nu = 0$ or $1$.
 A: Assuming $\nu\in\mathbb{N}$, we have:
$$ \mathcal{L}\left(J_\nu(x)\right) = \frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^\nu},\qquad \mathcal{L}^{-1}(1)=\delta(s) \tag{1}$$
hence:
$$ \int_{0}^{+\infty}J_\nu(x)\,dx = \lim_{s\to 0^+}\frac{1}{\sqrt{1+s^2}\left(s+\sqrt{1+s^2}\right)^\nu} = \color{red}{1}.\tag{2}$$
I exploited a useful property of the Laplace transform. The first identity can be easily derived from the generating function for Bessel functions of the first kind:
$$ \exp\left(\frac{z}{2}\left(t-\frac{1}{t}\right)\right)=\sum_{n\in\mathbb{Z}}t^n J_n(z).\tag{3}$$
$(1)$ (then $(2)$) holds also if $\nu\in\mathbb{R}^+$ since $(1)$ follows from Bessel's differential equation, too.

Clarification: we have
$$ \int_{0}^{+\infty}J_\nu(x)\,dx = \lim_{s\to 0^+}\int_{0}^{+\infty}J_\nu(x)e^{-sx}\,dx $$
since $J_\nu(x)$ is an improperly Riemann-integrable function, because $J_\nu(x)$ is continuous (much more: entire) in a right neighbourhood of the origin and $$J_\nu(x)\approx\sqrt{\frac{2}{\pi x}}\cos(x-\theta_\nu)$$
for large $x$ (equation $(58)$ here), so $\int_{0}^{+\infty}J_\nu(x)\,dx$ is convergent by Dirichlet's test, since $\cos(x-\theta_\nu)$ is a function with mean zero.
