Closed form of $\int\limits_0^{2\pi} \prod\limits_{j=1}^n \cos(jx)dx$ and combinatorial link

Question

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, it is easy to see if for some $b$ it holds, it will also hold for $\overline{b}$ (each digit is inverted).

Answer 1

Hint. From $$\int\limits_{\gamma}f(z)dz=\int\limits_{a}^{b}f(\gamma(t))\gamma'(t)dt$$ with a transformation like $$\int\limits_0^{2\pi} \prod_{j=1}^n \cos(jx)dx= \int\limits_0^{2\pi} \prod_{j=1}^n \frac{e^{i j x}+e^{-i j x}}{2}dx=\\ \int\limits_0^{2\pi} \frac{1}{2^n} \cdot\prod_{j=1}^n \frac{1}{e^{i j x}} \cdot \prod_{j=1}^n \left(e^{2i j x}+1\right)dx=\\ \int\limits_0^{2\pi} \frac{1}{2^n} \cdot \frac{1}{e^{i \frac{n(n+1)}{2} x}} \cdot \prod_{j=1}^n \left(e^{2i j x}+1\right)dx=\\ \int\limits_0^{2\pi} \frac{1}{i2^n} \cdot \frac{1}{e^{i \frac{n(n+1)}{2} x+ix}} \cdot \left(\prod_{j=1}^n \left(e^{2i j x}+1\right)\right) \cdot ie^{i x}dx=\\ \int\limits_{|z|=1}\frac{1}{i2^n} \cdot \frac{1}{z^{\frac{n(n+1)}{2}+1}} \cdot \prod_{j=1}^n \left(z^{2 j }+1\right)dz=...$$ and noting $f(z)=\prod\limits_{j=1}^n \left(z^{2 j }+1\right)$, we have $$...=\frac{1}{i2^n} \int\limits_{|z|=1}\frac{f(z)}{z^{\frac{n(n+1)}{2}+1}} dz=...$$ recalling Cauchy's integral formula this is $$...=\frac{1}{i2^n}\cdot \frac{2 \pi i}{\left(\frac{n(n+1)}{2}\right)!}\cdot f^{\left(\frac{n(n+1)}{2}\right)}(0)= \frac{\pi}{2^{n-1}} \cdot \frac{1}{\left(\frac{n(n+1)}{2}\right)!} \cdot f^{\left(\frac{n(n+1)}{2}\right)}(0)$$

Answer 2

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 $\lim\limits_{n\to\infty} \int_{0}^{2\pi} \cos x \cos 2x\cdots \cos nx \space{dx}$

Answer 3

As pointed out in the comments, the result is

$$ I_n = a_n \frac{2\pi}{2^{n}} $$

where $a_n$ is the numbers of solutions of $\sum_{j=1}^n s_n \,j =0$ where $s_j \in \{1,-1\}$ (or number of ways of marking a subset of $\{ 1,2, \cdots n\}$ such that the sum of the marked subset equals the sum of the unmarked subset). This is given by OEIS A063865.

Asymptotics (for $n=0,3 \pmod 4$): $$I_n \approx \sqrt{24 \pi} \, n^{-3/2} $$

Ref