Integration of $\int_{0}^{R}\int_{-1}^{1} \mathrm{d}r\mathrm{d}zr^2\sin\bigg({\frac{1-3z^2}{r^3}}\bigg)$ I need to evaluate the following 2D integral:
$$\int_{0}^{R}\int_{-1}^{1} 
\mathrm{d}r\; \mathrm{d}z\;
r^2\sin\left(\frac{1-3z^2}{r^3}\right)$$
where $R$ tends to infinity.
We have made several different attempts and the knowledge we've gained so far is that:


*

*The $z$ integral must be completed first. The asymptotic behavior of the function $r^2\sin{\frac{(1-3z^2)}{r^3}}$ at large $r$ is $\frac{(1-3z^2)}{r}$ meaning the integral at large $r$ diverges for fixed $z$. The next term in the asymptotic series is on the order of $r^{-7}$. When integrating over z first, then the problematic $\frac{1}{r}$ terms cancel, because the $z$-integral over $1-3z^2$ is zero. What’s left is the $r^{-7}$ term, which means that the integrand decays quickly, and the integral is finite.

*The symmetry of z is positive for $-\frac{1}{\sqrt{3}} < z < \frac{1}{\sqrt{3}}$ and negative otherwise. This may cause convergence. 

*Using Mathematica version 10.4.1.0, doing the $r$ integral first and extending r-integration to infinity gives: 
$\frac{8\pi}{27}(3-\sqrt{12}\mathrm{acoth}(\sqrt{3}) = .106513$.
Though other, newer versions of Mathematica say that the integral diverges. When numerical integration is done, the given result is $.106513$, agreeing with the above result. However, we cannot find an analytical path to this solution. 
 A: From the definition of Cosine Integral: 
$$ 
\begin{align} 
\text{Ci}(x)&=-\int_{x}^{\infty}\frac{\cos{t}}{t}\,dt=\gamma+\log{x}+\int_{0}^{x}\frac{\cos{t}-1}{t}\,dt \\[8mm] 
I_r&=\int r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr \\[2mm] 
&\color{blue}{\small\left\{u=\sin\left(\frac{1-3z^2}{r^3}\right)\Rightarrow du=\frac{-3(1-3z^2)}{r^4}\cos\left(\frac{1-3z^2}{r^3}\right) dr,\quad dv=r^2 dr\Rightarrow v=\frac{r^3}{3}\right\}} \\[2mm] 
I_r&=\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)+(1-3z^2)\int\frac{1}{r}\,\cos\left(\frac{1-3z^2}{r^3}\right)\,dr \\[2mm] 
&\color{blue}{\small\left\{\frac{d}{dx}\left[\text{Ci}(x)\right]=\frac{\cos{x}}{x}\Rightarrow\frac{d}{dx}\left[\text{Ci}\left(\frac{1-3z^2}{r^3}\right)\right]=\frac{\cos\left(\frac{1-3z^2}{r^3}\right)}{\frac{1-3z^2}{r^3}}\frac{-3(1-3z^2)}{r^4}=-\frac{3}{r}\cos\left(\frac{1-3z^2}{r^3}\right)\right\}} \\[2mm] 
I_r&=\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)-\frac{1-3z^2}{3}\,\text{Ci}\left(\frac{1-3z^2}{r^3}\right)+const \\[8mm] 
\color{red}{I}&=\int_{-1}^{+1}\int_{0}^{\infty}r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr\,dz=2\int_{0}^{1}\int_{0}^{\infty}r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr\,dz\quad\color{blue}{\left\{\rightarrow z^2\right\}} \\[2mm] 
&=2\int_{0}^{1}\left[\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)-\frac{1-3z^2}{3}\,\text{Ci}\left(\frac{1-3z^2}{r^3}\right)\right]_{r=0}^{r=\infty}\,dz \\[2mm] 
&=2\int_{0}^{1}\left[\frac{1-3z^2}{3}\left(1-\gamma-\log\left(1-3z^2\right)\right)\right]\,dz \quad\color{blue}{\left\{\rightarrow\text{Ci}(x)=\gamma+\log{x}-\text{Cin}(x)\right\}} \\[2mm] 
&=\frac{2}{3}(1-\gamma)\int_{0}^{1}\left(1-3z^2\right)\,dz\,-\,\frac{2}{3}\int_{0}^{1}\left(1-3z^2\right)\left(\log\left(1-3z^3\right)\right)\,dz\qquad\color{blue}{\left\{\rightarrow\text{IBP}\right\}} \\[2mm] 
&=\frac{2}{3}(1-\gamma)\left[z-z^3\right]_{0}^{1}\,+\,\frac{2}{3}\left[\frac{2}{3}z+\left(z-z^3\right)\left(\frac{2}{3}+\log\left(1-3z^2\right)\right)-\frac{4}{3\sqrt{3}}\text{arctanh}(\sqrt3z)\right]_{0}^{1} \\[2mm] 
&=\color{red}{\frac{4}{9}-\frac{8}{9\sqrt{3}}\,\text{arctanh}{\sqrt{3}}}\,\approx{0.106513}
\end{align} 
$$
