Yes. The infinite oscillatory integral under consideration is well-defined, by considering the asymptotic behavior of $J_n(x)$. To establish the evaluation
$\int_0^\infty J_0(t)\mathrm dt=1$
we treat this as the expression
$\lim_{c\to 0^+} \int_0^\infty \exp(-ct) J_0(t)\mathrm dt$
and then replace the Bessel function with the integral representation
$J_0(x)=\frac1{\pi}\int_0^\pi \exp(ix\cos\,u)\mathrm du$
to yield
$\lim_{c\to 0^+} \frac1{\pi}\int_0^\infty \exp(-ct) \int_0^\pi \exp(it\cos\,u)\mathrm du\mathrm dt$
after which,
$\lim_{c\to 0^+} \frac1{\pi} \int_0^\pi \frac1{c-i\cos\,u}\mathrm du=\lim_{c\to 0^+} \frac1{\sqrt{1+c^2}}=1$