# Find integrals that are nonzero

We are given $I_m = \int_0^{2\pi} \cos(x) \cos(2x) \dots \cos(mx)dx$; i.e.,

$$I_m=\int_0^{2\pi}\prod_{k=1}^m\cos{k x}\,dx.$$

Task is to find such $m \in [1\dots 10]$, so $I_m \neq 0$.

I have computed all these integrals with the use of MATLAB and found out that $I_m \neq 0$ holds for $m \in \{3,4,7,8\}$. Those are cases, when expression after integration contains some other term, but $sin(x)$, something like: $\frac {x}{C}$, C - constant.

But this task is supposed to be solvable in a pure analytical approach. Maybe it is even possible to find the right $m$ values without computing all integrals, since it looks like it is nearly possible to do this just by hand.

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\begin{align} &2^{-m}\int_0^{2\pi}(e^{ix}+e^{-ix})(e^{2ix}+e^{-2ix})\cdots(e^{mix}+e^{-2mix})\,\mathrm{d}x\ne0\\ &=2^{-m}\int_0^{2\pi}e^{-i\frac{m(m+1)}{2}x}(e^{2ix}+1)(e^{4ix}+1)\cdots(e^{2mix}+1)\,\mathrm{d}x\ne0 \end{align} precisely when there is a term with exponent $0$ and that only happens when $\frac{m(m+1)}{2}$ can be written as the sum of even integers. That is when $m(m+1)$ is divisible by $4$. Therefore, the integral is non-zero precisely when $m\equiv0\pmod{4}$ or $m\equiv3\pmod{4}$.
Ahhh. I saw this formula $\cos(x)=\frac 1 2 (e^{ix}+e^{-ix})$ but I didn't tried it. – wf34 Apr 5 '14 at 6:48
Thank you for answer, I'm going to accept it. But I got some questions: how factoring out $e^{-i \frac{m(m+1)}{2}x}$ enables us to get 1 in all brackets? – wf34 Apr 5 '14 at 6:57
@wf34: factoring $e^{-ix}$ out of the first term, $e^{-2ix}$ out of the second term, and so on until factoring $e^{-mix}$ out of the $m^\text{th}$ term, accumulates $e^{-i\frac{m(m+1)}{2}x}$ outside and leaves $(e^{2ix}+1)$ in the first term, $(e^{4ix}+1)$ in the second term, and so on until $(e^{2mix}+1)$ in the $m^\text{th}$ term. This is because $1+2+\dots+m=\frac{m(m+1)}{2}$. – robjohn Apr 5 '14 at 7:03
Thanks, I see it now. I'm a bit ashamed to ask again, but why divisibility of $m(m+1)$ by 4 turn expression $-i\frac{m(m+1)}{2}x$ into 0? – wf34 Apr 5 '14 at 7:14
@wf34: in order to get a term with a $0$ in the exponent, $\frac{m(m+1)}{2}$ must be gotten as the sum of some subset of $\{0,2,4,\dots,2m\}$. The sums of all the subsets span all the even numbers from $0$ to $m(m+1)$. – robjohn Apr 5 '14 at 7:24