Crazy Cosine Calculus!

This can be solved using the property

$f(2a - x) = f(x)$ $\Rightarrow \int_{0}^{2a}f(x)dx = 2\int_{0}^{a}f(x)dx$

Let $f(x) = \cos(x) \cos(2x) \dots\cos(mx)$ $\Rightarrow f(2\pi - x) = \cos(2\pi - x) \cos(4\pi - 2x) \dots\cos(2m\pi - mx)$ $\Rightarrow f(2\pi - x) = \cos(x) \cos(2x) \dots\cos(mx)$

(Since $\cos(2n\pi \pm \theta) = \cos(\theta)$ where $n$ is an integer )

$\Rightarrow f(2\pi - x) = f(x)$

Thus,

$I_m = 2\int_{0}^{\pi}\cos(x) \cos(2x) \dots\cos(mx) dx ...(i)$

Now, we know that $\int_{0}^{a}f(x)dx = \int_{0}^{a} f(a-x)dx$

Therefore

$I_m = 2\int_{0}^{\pi}\cos(\pi - x) \cos(2\pi - 2x) \dots\cos(m\pi - mx) dx$

Thus $I_m = +2\int_{0}^{\pi}\cos(x) \cos(2x) \dots\cos(mx) dx = I_m$

when $m = 4n+3$ or $m = 4n ; n \in Integers$

And

$I_m = -2\int_{0}^{\pi}\cos(x) \cos(2x) \dots\cos(mx) dx = -I_m$ $\Rightarrow 2 I_m = 0 \Rightarrow I_m = 0$

when $m = 4n+1$ or $m = 4n+2 ; n \in Integers$

Thus, $I_m \neq 0$ for

$m = 100 , 103 , 104 , 107 , 108$

Thus,

$Answer = 100+103+104+107+108 = \boxed{\color{#20A900}{522}}$

If we use the identity $\large{cos z = \frac{e^{iz} + e^{-iz}}{2}}$ , we can rewrite the integral as $\displaystyle \int_0^{2\pi} \displaystyle \prod_{k=1}^m \frac{e^{ikx} + e^{-ikx}}{2} dx = \frac{1}{2^m} \displaystyle \sum_{a_k = \pm 1} \displaystyle \int_0^{2\pi} e^{i(a_1 + 2a_2 + 3a_3 + \dots + ma_m)x} dx$ .

Furthermore, note that $\int_0^{2\pi} e^{inx} dx = 2\pi$ if $n = 0$ , otherwise it is equal to $0$ as $n$ ranges in the integers. Recall that $e^{ix} = \cos x + i \sin x$ , so the real and imaginary parts of the graph of $e^{inx}$ from $0$ to $2\pi$ is the cosine and sine graph, respectively, repeated $n$ times, so the integral is zero. Unless $n = 0$ , in which case it is the horizontal line $y = 1$ , which has an area of $2\pi$ .

Since our integral ranges over all $2^m$ m-tuples of $(a_1, a_2, \dots, a_m)$ with $a_k = \pm 1$ , it suffices to show that, for a given $m$ , there exists such a sequence $\{a_k\}$ such that $a_1 + 2a_2 + \dots + ma_m = 0$ .

So, $0 = a_1 + 2a_2 + \dots + ma_m \equiv 1 + 2 + \dots + m = \frac{m(m+1)}{2} \pmod 2$ , thus $4 | m(m+1)$ and $m \equiv 0, 3 \pmod 4$ .

Consider $m \equiv 0 \pmod 4$ . Then we find a solution in $(1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + \dots + ((m-3) - (m-2) - (m-1) + m)$

Consider $m \equiv 3 \pmod 4$ . Then we find a solution in $(1+2-3) + (4-5-6+7) + \dots + ((m-3)-(m-2)-(m-1)+m)$ .

Thus, $I_m \neq 0$ iff $m \equiv 0, 3 \pmod 4$ . This gives the solution set of $\{100, 103, 104, 107, 108\}$ , with sum $\boxed{522}$ . $\blacksquare$