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).
Best Answer
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 $$...=\frac{1}{i2^n}\cdot \frac{2 \pi i}{\left(\frac{n(n+1)}{2}\right)!}\cdot f^{\left(\frac{n(n+1)}{2}\right)}(0)= \frac{\pi}{2^{n-1}} \cdot \frac{1}{\left(\frac{n(n+1)}{2}\right)!} \cdot f^{\left(\frac{n(n+1)}{2}\right)}(0)$$