Consider the improper integral
$$
f(r,t) =
\int_0^\infty \frac{\sin(\lambda t)}{\lambda^3} J_2(\lambda r) \mathrm{d}\lambda ,
$$
wherein $J_2$ is the 2nd order Bessel function of the first kind.
Clearly, the integral is convergent.
Using computer algebra systems, such as Maple, one gets
$$
f(r,t) = \frac{\pi r^2}{16} .
$$
However, using numerical integration, this value seems to be correct only if $r < t$.
Is there a way to find a closed analytical formula of the above integral when $r > t$.
Any help or advice is welcome.
Thanks
Best Answer
A possible solution is to use Poisson's integral $$\newcommand{\sgn}{\operatorname{sgn}}J_2(z)=\frac{z^2}{3\pi}\int_0^\pi\cos(z\cos\theta)\sin^4\theta\,d\theta$$ and switch the integrations (which is easy, though not entirely trivial, to justify): $$f(r,t)=\frac{r^2}{12}\int_0^\pi\big(\sgn(t+r\cos\theta)+\sgn(t-r\cos\theta)\big)\sin^4\theta\,d\theta.$$ Assume $r,t$ are positive. If $r<t$, the sum of $\sgn$'s is $2$ for all $\theta$, and we get your result.
Otherwise, introducing $\phi=\arcsin(t/r)$, we get (simplifiable if desired) $$f(r,t)=\frac{r^2}{6}\int_{\pi/2-\phi}^{\pi/2+\phi}\sin^4\theta\,d\theta=\frac{r^2}{96}(12\phi+8\sin2\phi+\sin4\phi).$$