TITLE: Closed form of $\int\limits_0^{2\pi} \prod\limits_{j=1}^n \cos(jx)dx$ and combinatorial link
QUESTION [8 upvotes]: I have been trying to find a closed form for this integral:
$$I_n = \int\limits_0^{2\pi} \prod_{j=1}^n \cos(jx)dx$$
The first values are: $I_1=I_2=0,I_3=\frac{\pi}{2}, I_4=\frac{\pi}{4}, I_5=I_6=0, I_7=\frac{\pi}{8}, I_8=\frac{7\pi}{64}$
I am not able to see here a clean pattern except that for $n=4k+1,4k+2$ the integral should be zero. If someone could give me a hint I would appreciate it.
EDIT
As suggested by Winther in the comments, the problem can be viewed from a combinatorial standpoint. Looking at the complex exponential representation one gets $2^n$ integrals of the form $\int_0^{2\pi}e^{iNx}dx$, which is only nonzero, if $N=0$. The integral evaluates to $\frac{M\pi}{2^{n-1}}$, where $M$ is the number of nonzero integrals.
So one needs to find $M$, which is the number of binary numbers $b$ for which holds that
$$\sum_{k=1}^n (2b_k-1)k = 0$$
where $b_k$ is the k-th digit of $b$.
With this is is easy to see if for some $b$ it holds, it will also hold for $\overline{b}$ (each digit is inverted).

REPLY [3 votes]: The integral is equal to $\frac{A_n\pi}{2^{n-1}}$ where $A_n$ is the number of subsets of   $\{1,2,3,\dots,n\}$ whose sum is $\frac{n(n+1)}4.$
In particular, if $n\equiv 1,2\pmod 4,$ the $\frac{n(n+1)}4$ is not an integer, so there can be no such aubsets, so the integral is zero in that case.
You get this value by representing $\cos nx =\frac12 \left(e^{inx}+e^{-inx}\right)$ and realize the integral is zero for all terms expanding the produc, except for the constant term, which has coefficient $\frac{A_n}{2^n}.$
I don’t think there is an easy way to represent this term. There is an upper bound $\binom n{\lfloor n/2\rfloor}.$
See this answer for details. Compute $\lim\limits_{n\to\infty} \int_{0}^{2\pi} \cos x \cos 2x\cdots \cos nx \space{dx}$