# Integral of products of cosines

Given $m+1$ integers $\alpha_0,\ldots,\alpha_m\geq 1$, I was trying to get a nice closed formula for the integral $$\int_0^\pi\cos(\alpha_1\theta)\cdots\cos(\alpha_m\theta)d\theta.$$ More precisely, I was trying to count the number of $m$-uplets $(\alpha_1\ldots,\alpha_m)$ with $1\leq \alpha_1\leq\ldots\leq\alpha_m\leq n$ such that the later integral is non-zero.

When $m=1$, this is easy because of the orthogonality relation $$\frac{2}{\pi}\int_0^\pi\cos(\alpha_0\theta)\cos(\alpha_1\theta)d\theta=\;\boldsymbol 1_{\alpha_0=\alpha_1},$$ and thus the number of such solutions is $\boldsymbol 1_{\alpha_0\leq n}$, that is the number of ways to partition $\alpha_0$ into one positive integer smaller than $n$.

If one looks instead at the integral $$\frac{1}{2\pi}\int_0^{2\pi}e^{i\alpha_0\theta}e^{-i\alpha_1\theta}\cdots e^{-i\alpha_m\theta} d\theta$$ then the number of solutions is the number of ways to partition $\alpha_0$ into $m$ integers smaller than $n$. I was somehow hopping that a similar type of solution would appear for the cosine integral, but maybe that is too optimistic ...

Any idea for the general case with the cosines ? Or a reference ? I have tried to use inductively the trigonometric formula $cos(a)cos(b)=\frac{1}{2}(cos(a+b)+\cos(a-b))$, but I don't see any structure appearing.

• I am dealing with the same question myself. Have you found something? – Giuseppe Negro Nov 18 '14 at 17:44
• @GiuseppeNegro I did not find what I wanted, but there are interesting things in the paper "PHASE TRANSITION AND FINITE-SIZE SCALING FOR THE INTEGER PARTITIONING PROBLEM" from Christian Borgs, Jennifer Chayes and Boris Pittel (see (3.5) and thereafter). Hope it helps. – Student Nov 18 '14 at 17:53
• Thank you, I'll have a look at it. – Giuseppe Negro Nov 19 '14 at 13:01

The question if this integral is zero, is one of the classical $NP$-complete Problems given in the book of Garey and Johnson, Computers and Intractability, 1979. p. 252. There called AN14 and with the upper limit of the integral $2\pi$ instead of $\pi$.