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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.

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Do we have some additional constraint like for instance $\alpha,\beta,\gamma \in \mathbb{Z}$? –  user17762 Jun 27 '12 at 1:06
For what it is worth. If we assume $\alpha \in \mathbb{Z}$, then the $1$D version of the problem, $$ \int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x\end{Vmatrix} \text{d}x = \pi $$ If we assume $\alpha, \beta \in \mathbb{Z}$, then the $2$D version of the problem, $$ \int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x&\sin \alpha y\\\sin \beta x&\sin \beta y\end{Vmatrix} \text{d}x\,\text{d}y = 2\pi^2 $$ Hence, the possible guess in $n$-D, assuming $\alpha_k \in \mathbb{Z}$ would be either $2^{n-1} \pi^n$ or $n! \pi^n$. –  user17762 Jun 27 '12 at 1:47
@Marvis: Unless any of the $\alpha_k$ are equal, in which case the determinant vanishes. –  user26872 Jun 27 '12 at 2:07
@oenamen Yes. True. I forgot to mention that. –  user17762 Jun 27 '12 at 2:23

2 Answers 2

up vote 4 down vote accepted

This is a bit too long to comment. A change in notation $(\alpha, \beta, \gamma) \to (a,b,c)$.

If $a=0$ or $b=0$ or $c=0$ or $a^2 = b^2$ or $b^2=c^2$ or $c^2 = a^2$, the integral is zero.

Let's assume $a,b,c \in \mathbb{Z} \backslash \{0\}$ and are distinct. $$ D(x,y,z,a,b,c) = {\det}^2\begin{Vmatrix}\sin a x&\sin a y&\sin a z\\\sin b x&\sin b y&\sin b z\\\sin c x&\sin c y&\sin c z\end{Vmatrix}\\ = (d_1(b,c,y,z) \sin(ax) + d_2(c,a,y,z) \sin(bx) + d_3(a,b,y,z) \sin(cx))^2 $$ Hence, $$\int_{-\pi}^{\pi}D(x,y,z,a,b,c) dx= \int_{-\pi}^{\pi} \left(d_1^2(b,c,y,z) \sin^2(ax) + d_2^2(c,a,y,z) \sin^2(bx) + d_3^2(a,b,y,z) \sin^2(cx) \right) dx$$ (The cross-terms integrals vanish by our initial assumption) Hence, $$\int_{-\pi}^{\pi}D(x,y,z,a,b,c) dx= \left(d_1^2(b,c,y,z) + d_2^2(c,a,y,z) + d_3^2(a,b,y,z) \right) \pi$$ Lets now consider $ \displaystyle \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz$. The other two can be computed using symmetry arguments. $$ \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz =\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left(\sin(by) \sin(cz) - \sin(cy) \sin(bz) \right)^2 dy dz $$ Again by our assumption, the integrals of the cross-terms vanish and hence we get $$ \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz =\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left(\sin^2(by) \sin^2(cz) + \sin^2(cy) \sin^2(bz) \right) dy dz = \pi^2 + \pi^2 = 2\pi^2$$ Hence, $$\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz = \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_2^2(c,a,y,z) dy dz = \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_3^2(a,b,y,z) dy dz = 2\pi^2$$ Hence, $$\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}D(x,y,z,a,b,c) dx dy dz= \left(2 \pi^2 + 2\pi^2 + 2\pi^2 \right) \pi = 6 \pi^3$$ To summarize, $$I(a,b,c) = \begin{cases} 0 & \text{ if $a=0$ or $b=0$ or $c=0$ or $a^2 = b^2$ or $b^2=c^2$ or $c^2 = a^2$}\\ 6 \pi^3 & \text{ if $a,b,c \in \mathbb{Z} \backslash \{0\}$ and are distinct}\\ ? & \text{else}\end{cases}$$

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Can this be generalized to situations when $a,b,c \in \mathbb{R}$ with certain constraints? For example, $\int_{-\pi}^{\pi}\sin^2 ax = \pi - \frac{\sin 2\pi a}{2a}$ so the integral of all terms with $\sin^2$ only will become $6(\pi - \frac{\sin 2\pi a}{2a})(\pi - \frac{\sin 2\pi b}{2b})(\pi - \frac{\sin 2\pi a}{2a})$ –  Andrew Salmon Jun 27 '12 at 2:28
@Marvis: I was starting to have a look at this but you beat me to it. Very nice! (+1) –  user26872 Jun 27 '12 at 2:29
@AndrewSalmon I don't think so. For one, if we set $a=b$, it should give us $0$ whereas the one you have doesn't. But I guess, yes you are right, I could just do that for all $a,b,c \in \mathbb{R}^3$! Thanks. I will update it accordingly! –  user17762 Jun 27 '12 at 2:29
@oenamen It is not complete since it is with the assumption $(a,b,c) \in \mathbb{Z}^3$. –  user17762 Jun 27 '12 at 2:30
@Marvis I meant the cases excluding those which you describe: $a^2=b^2$, $b^2=c^2$, $a^2=c^2$, or $abc=0$. Will the cases with terms including $\sin ax \sin bx \dots$ cancel out to $0$? –  Andrew Salmon Jun 27 '12 at 2:31

Exact solution and Mathematica code to produce it (< 10 seconds computing time):

M = {{Sin[a x], Sin[a y], Sin[a z]},
     {Sin[b x], Sin[b y], Sin[b z]},
     {Sin[c x], Sin[c y], Sin[c z]}};
Expand[Det[M]^2] /. Plus -> List;
Total[Integrate[%, {x, -Pi, Pi}, {y, -Pi, Pi}, {z, -Pi, Pi}]]

It looks terrible but for a computer no problem: $-\frac{8 \left(\pi -\frac{\sin (2 c \pi )}{2 c}\right) (b \cos (b \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (b \pi ))^2}{\left(a^2-b^2\right)^2}-\frac{8 (2 c \pi -\sin (2 c \pi )) (b \cos (b \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (b \pi ))^2}{\left(a^2-b^2\right)^2 c}+\frac{16 (a \cos (a \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (a \pi )) (2 c \cos (c \pi ) \sin (a \pi )-2 a \cos (a \pi ) \sin (c \pi )) (b \cos (b \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (b \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(b^2-c^2\right)}+\frac{16 (2 b \cos (b \pi ) \sin (a \pi )-2 a \cos (a \pi ) \sin (b \pi )) (a \cos (a \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (a \pi )) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(c^2-b^2\right)}+\frac{16 (a \cos (a \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (a \pi )) (a \cos (a \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (a \pi )) (2 c \cos (c \pi ) \sin (b \pi )-2 b \cos (b \pi ) \sin (c \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(b^2-c^2\right)}-\frac{(\sin (2 a \pi )-2 a \pi ) (2 b \pi -\sin (2 b \pi )) \left(\pi -\frac{\sin (2 c \pi )}{2 c}\right)}{2 a b}-\frac{(\sin (2 a \pi )-2 a \pi ) \left(\pi -\frac{\sin (2 b \pi )}{2 b}\right) (2 c \pi -\sin (2 c \pi ))}{2 a c}-\frac{\left(\pi -\frac{\sin (2 a \pi )}{2 a}\right) (\sin (2 b \pi )-2 b \pi ) (2 c \pi -\sin (2 c \pi ))}{2 b c}-\frac{8 \left(\pi -\frac{\sin (2 b \pi )}{2 b}\right) (c \cos (c \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (c \pi ))^2}{\left(a^2-c^2\right)^2}-\frac{8 (2 b \pi -\sin (2 b \pi )) (c \cos (c \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (c \pi ))^2}{b \left(a^2-c^2\right)^2}-\frac{8 \left(\pi -\frac{\sin (2 a \pi )}{2 a}\right) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))^2}{\left(b^2-c^2\right)^2}-\frac{8 (2 a \pi -\sin (2 a \pi )) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))^2}{a \left(b^2-c^2\right)^2}$

Perhaps someone can simplify it from here?

Edit: Confirmed

    a \[Element] Integers &&
    b \[Element] Integers &&
    c \[Element] Integers]

outputs $6\pi^3$.

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Thanks for your participation! –  user34574 Jun 27 '12 at 17:13
Maple does it too (but with an even more complicated-looking result). The result is also $6 \pi^3$ if $\alpha,\beta,\gamma$ are all of the form integer $+1/2$ (and again $\alpha \beta \gamma (\alpha^2-\beta^2)(\alpha^2-\gamma^2)(\beta^2-\gamma^2) \ne 0$). –  Robert Israel Jun 27 '12 at 18:18

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