In both Mathematica 10.4.1.0 and 11.1.0.0
Integrate[r^2 Sin[r^-3], {r, 0, \[Infinity]}]
gives
Integrate::idiv: Integral of r^2 Sin[1/r^3] does not converge on {0,\[Infinity]}.
Now, using
Integrate[r^2 Sin[r^-3 c], {r, 0, \[Infinity]},
Assumptions -> {c \[Element] Reals}]
still gives
Integrate::idiv: Integral of r^2 Sin[c/r^3] does not converge on {0,\[Infinity]}.
in 11.1.0.0, but surprisingly gives
-(1/6) c (-2 + 2 EulerGamma + Log[c^2])
in 10.4.1.0. Which result is correct?
Using Rubi, you can investigate a bit further, what steps are required to find an antiderivative
<< Rubi`
int = Steps[Int[r^2 Sin[r^-3 * c], r]]
First, you can show that the solution is indeed one valid antiderivative of your expression:
In[31]:= D[int, r]
Out[31]= r^2 Sin[c/r^3]
Now, you can look at the limits of your integration bounds
In[32]:= Limit[int, r -> 0, Direction -> "FromAbove"]
Out[32]= ConditionalExpression[-(1/6) c (2 Log[c] - Log[c^2]), c \[Element] Reals]
This looks good, but this here not
In[33]:= Limit[int, r -> Infinity]
Out[33]= c \[Infinity]
Therefore, I suspect your version 10.4 result is not correct and the integral does not converge.
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.
Nope, at least right now, I wouldn’t know how to do this. The term (1 - 3 z^2) is a dealbreaker. If you use a*z^3 instead, an analytic solution is possible, but with your original expression, there are two terms including FresnelC that neither Rubi nor Mathematica is able to integrate.
The math is pretty clear. It diverges due to 1/r asymptotoc behavior at infinity. One can see this using Series for example.
Series[r^2*Sin[r^-3], {r, Infinity, 2}]
(* Out[2]= SeriesData[r,
DirectedInfinity[1], {1}, 1, 3, 1] *)
Thanks! Yes, the r integral diverges, but the 1-3z^2 term seems to rescue convergence in the 2D integral. This term alternates signs twice between z = -1 and +1, so it’s plausible that this is possible.
This is a funny situation to be in: to know the answer (helped by that faulty answer from Mathematica 10.4), but to have no clue how to get there, and no tool that might show the way.