TITLE: Integral of determinant
QUESTION [15 upvotes]: Good evening. I need help with this task
$$
\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x&\sin \alpha y&\sin \alpha z\\\sin \beta x&\sin \beta y&\sin \beta z\\\sin \gamma x&\sin \gamma y&\sin \gamma z\end{Vmatrix} \text{d}x\,\text{d}y\,\text{d}z
$$
where $\alpha,\beta,\gamma$ are integers.
Computations are horrible, I gave up.

REPLY [3 votes]: Consider the vector $x = [x_1,...,x_N].$
Let $A(x)$ and $B(x)$ be two $N\times N$ matrices such that $(A)_{ij} = f_i(x_j)$
and $(B)_{ij} = g_i(x_j)$ where $f_i(z)$ and $g_i(z)$ are any functions, $i=1,\cdots ,N;$
Then
$\int_{[a,b]^N}  det(A)det(B) dx = N! det(C)$
where $(C)_{ij} = \int_{[a,b]} f_i(z)g_j(z) dz$
In your case $A=B, a=-\pi, b=\pi, f_i(z) = g_i(z) = \sin(\alpha_i z)$
Since $\alpha_i$ are integers, the matrix $C$ is diagonal with diagonal elements
equal to 
$\int_{-\pi}^{\pi} \sin^2(\alpha_i z) dz = \pi$
Therefore the integral you are looking for gives $N! \pi^N$
For $N=3$ you have $6\pi^3$
If for some $i$ and $j$, $|\alpha_i|=|\alpha_j|$ the matrices $A$ and $B$ are rank deficient and therefore their determinant is $0$. It follows that the integral is $0$
This can be generalized for any $N$ and for real coefficients $\alpha_i$