Thank you! Plotting the integrand also suggests a function that increases monotonically as r goes to infinity.
This brings me to my original problem, a 2D integral of the form
Integrate[
r^2 Sin[a r^-3 (1 - 3 z^2)], {r, 0, \[Infinity]}, {z, -1, 1},
Assumptions -> {a \[Element] Reals}]
Mathematica fails to find an answer. But numerically, it appears that this double integral converges nicely
In[56]:= R = 100; NIntegrate[
r^2 Sin[(1 - 3 z^2) r^-3], {r, 0, R}, {z, -1, 1}]
Out[56]= 0.106513
with a value
In[59]:= 4/27 (3 - Sqrt[12] ArcCoth[Sqrt[3]]) // N
Out[59]= 0.106513
Is there a way to convince Mathematica to solve the 2D integral? I suppose the integral over z needs to be done first, since the r integral (for any z except [PlusMinus]1/Sqrt[3]) diverges.