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?
Best Answer
Expanding $\sin(xyz)$ into power series (converge for all values of $xyz$) $$ I=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}(xyz)^{2n+1}dxdydz\\=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\bigg(\int_0^{\pi/2}x^{2n+1}dx\bigg)^3\\=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\frac{1}{(2n+2)^3}(\frac{\pi}{2})^{6n+6} \\=\frac{π^6}{512} \ _3F_4(1, 1, 1;3/2, 2, 2, 2;-\frac{π^6 }{256}) $$