Closed form $\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin(xyz)\,dx\,dy\,dz$ This is part of an assignment from a multivariable calculus course. We've only just defined triple integrals and we're not using special functions or anything like that, so I'm pretty sure this was a mistake. Still, I'm interested to see if there exists some nice closed form.
The only progress I've managed to get is
$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin(xyz)\,dx\,dy\,dz=\int_{0}^{\frac{\pi}{2}}\dfrac{1}{z}\left(\ln\left(\dfrac{\pi^2}{4}z\right)-\text{Ci}\left(\dfrac{\pi^2}{4}z\right)+\gamma\right)dz\,,$$
where $\text{Ci}(x)$ is the cosine integral, but nothing else. Any ideas?

Incomplete (?) solution.
\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\sin(xyz)dxdydz\\
&\overset{u=\frac{2x}{\pi},v=\frac{2y}{\pi},w=\frac{2z}{\pi}}=\frac{\pi^3}{8}\int_0^1\int_0^1\int_0^1\sin\left(\frac{\pi^3uvw}{8}\right)dudvdw\\
&=-\frac{\pi^3}{8}\int_0^1\int_0^1 \ln x\sin\left(\frac{\pi^3xy}{8}\right)dxdy\\
&=-\frac{\pi^3}{16}\int_0^1\int_0^1 \ln (xy)\sin\left(\frac{\pi^3xy}{8}\right)dxdy\\
&=\frac{\pi^3}{16}\int_0^1 \ln^2 (z)\sin\left(\frac{\pi^3z}{8}\right)dz\\
&=\frac{\pi^3}{16}\int_0^1 \ln^2 z\left(\sum_{n=0}^\infty \frac{(-1)^n\pi^{6n+3}z^{2n+1}}{2^{6n+3}(2n+1)!}\right)dz\\
&=\frac{\pi^3}{16}\sum_{n=0}^\infty\left(\int_0^1 \frac{(-1)^n\pi^{6n+3}z^{2n+1}\ln^2 z}{2^{6n+3}(2n+1)!}\right)\\
&=\frac{\pi^3}{8}\sum_{n=0}^\infty \frac{(-1)^n\pi^{6n+3}}{2^{6n+3}(2n+2)^3(2n+1)!}\\
&=\frac{\pi^3}{8}\sum_{n=0}^\infty \frac{(-1)^n\pi^{6n+3}}{2^{6(n+1)}(n+1)^3(2n+1)!}\\
\end{align}
NB: I use the following result: If $f$ is a continuous function on $[0;1]$ then $\displaystyle \int_0^1 \int_0^1 f(xy)dxdy=-\int_0^1 f(x)\ln xdx$
(proof: change of variable $u(y)=xy$ then IBP)
