Number of solutions of a Diophantine equation

Question

I'm currently studying about the solutions of the $x_1\pm x_2\pm\cdots\pm x_m=0$, where the $x_i$'s are positive integers. They started with a trigonometric identity \begin{equation} \cos(x_1) \cos(x_2)\cdots\cos(x_m)=\dfrac{1}{2^{m-1}}\sum \cos(x_1\pm x_2\pm\cdots\pm x_m) \end{equation} where the sum is taken over all possible choices of signs $+$ and $-$. Later, they arrived to the conclusion that the number of solutions of the equation $x_1\pm x_2\pm\cdots\pm x_m=0$ (all the possible choices of signs) is given by the integral \begin{equation} \dfrac{2^{m-1}}{2\pi}\int_{-\pi}^{\pi}\cos(x_1t) \cos(x_2t)\cdots\cos(x_mt)dt \end{equation} which of course can be simplified further (by using the fact that the integrant is an even function) as \begin{equation} \dfrac{2^{m-1}}{\pi}\int_{0}^{\pi}\cos(x_1t) \cos(x_2t)\cdots\cos(x_mt)dt \end{equation} How they did arrived to that integral representation? I can't figure out this. Any help given will be appreciated.

Answer 1

For every $t$, you have the given trigonometric identity $$\cos(x_1t) \cos(x_2t)\cdots\cos(x_mt)=\dfrac{1}{2^{m-1}}\sum \cos(x_1t\pm x_2t\pm\cdots\pm x_mt)$$

where, as you said, the sum is taken over all possible choices of signs $+$ and $−$.

Now integrate this between $-\pi$ and $\pi$ : you get $$\int_{-\pi}^{\pi}\cos(x_1t) \cos(x_2t)\cdots\cos(x_mt) dt=\dfrac{1}{2^{m-1}}\sum \int_{-\pi}^{\pi}\cos((x_1\pm x_2\pm\cdots\pm x_m)t) dt$$

But the integral $$\int_{-\pi}^{\pi}\cos((x_1\pm x_2\pm\cdots\pm x_m)t) dt$$ is equal to $2\pi$ if $x_1\pm x_2\pm\cdots\pm x_m = 0$, and is equal to $0$ otherwise (because the $x_i$'s are integers). So if $N$ denotes the number of solutions of the equation, you get directly that $$\int_{-\pi}^{\pi}\cos(x_1t) \cos(x_2t)\cdots\cos(x_mt) dt=\dfrac{1}{2^{m-1}} 2\pi N$$

so $$N = \frac{2^{m-1}}{2\pi}\int_{-\pi}^{\pi}\cos(x_1t) \cos(x_2t)\cdots\cos(x_mt) dt$$

which is the given integral representation.