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?
 A: Expanding $\sin(xyz)$ into power series (converge for all values of $xyz$)
$$
I=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}(xyz)^{2n+1}dxdydz\\=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\bigg(\int_0^{\pi/2}x^{2n+1}dx\bigg)^3\\=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\frac{1}{(2n+2)^3}(\frac{\pi}{2})^{6n+6} 
\\=\frac{π^6}{512}   \ _3F_4(1, 1, 1;3/2, 2, 2, 2;-\frac{π^6 }{256})
$$
A: 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)
A: Compute the volume inside $[0,1]^3$ where $xyz\le p$.
$$
z=\left\{\begin{array}{cl}\frac p{xy}&\text{when }xy\ge p\\1&\text{when }xy\lt p\end{array}\right.\tag1
$$
The integral over the area where $z=1$ is
$$
\overbrace{\int_0^p1\,\mathrm{d}x}^{x\le p}+\overbrace{\int_p^1\frac px\,\mathrm{d}x}^{x\gt p,y\le\frac px}=p-p\log(p)\tag2
$$
The integral over the rest is
$$
\begin{align}
\overbrace{\int_p^1\int_{p/x}^1\frac p{xy}\,\mathrm{d}y\,\mathrm{d}x}^{x\gt p,y\gt\frac px\implies z\le\frac p{xy}}
&=p\int_p^1\frac{\log(x/p)}x\,\mathrm{d}x\\
&=\frac p2\log(p)^2\tag3
\end{align}
$$

Thus, the volume inside $[0,1]^3$ where $xyz\le p$ is
$$
v(p)=p-p\log(p)+\frac p2\log(p)^2\tag4
$$
The derivative of $(4)$ is
$$
v'(p)=\frac12\log(p)^2\tag5
$$
Therefore,
$$
\begin{align}
\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}\sin\left(xyz\right)\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x
&=\frac{\pi^3}8\int_0^1\int_0^1\int_0^1\sin\left(\frac{\pi^3}8xyz\right)\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x\tag6\\
&=\frac{\pi^3}8\int_0^1\sin\left(\frac{\pi^3}8p\right)\frac12\log(p)^2\,\mathrm{d}p\tag7\\
&=\frac{\pi^3}{16}\sum_{k=0}^\infty\int_0^1(-1)^k\frac{(\pi^3p/8)^{2k+1}}{(2k+1)!}\log(p)^2\,\mathrm{d}p\tag8\\
&=\frac{\pi^3}{16}\sum_{k=0}^\infty(-1)^k\frac{(\pi^3/8)^{2k+1}}{(2k+1)!}\frac2{(2k+2)^3}\tag9\\
&=\sum_{k=0}^\infty(-1)^k\frac{(\pi^3/8)^{2k+2}}{(2k+2)^2(2k+2)!}\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}(6)$: substitute $x,y,z\mapsto\frac\pi2x,\frac\pi2y,\frac\pi2z$
$\phantom{1}(7)$: apply $(5)$
$\phantom{1}(8)$: use the series expansion for $\sin(x)$
$\phantom{1}(9)$: $\int_0^1x^n\log(x)^2\,\mathrm{d}x=\frac2{(n+1)^3}$
$(10)$: simplify a bit
The last sum, $(10)$, can also be written as $\frac{\pi^6}{512}\,_3F_4\!\left(1,1,1;\frac32,2,2,2;-\frac{\pi^6}{256}\right)$, whose numerical value is approximately $1.4030717941246524706$.
