Integrating Product of Sines

The finite product $\prod_{j=1}^{2N} \sin(jx)$ can be written as a sum of cosines, with $\prod_{j=1}^{2N} \sin(jx) \; \equiv \; \sum_{j\ge 0} a_{N,j}\cos(jx)$ in which case we have the integral $\int_0^{2\pi} \prod_{j=1}^{2N} \sin(jx)\,dx \; =\; 2\pi a_{N,0}$ We note that $\sin(2N-1)x\,\sin2Nx \; \equiv \; \tfrac12\big[\cos x - \cos(4N-1)x\big]$ and hence $\begin{aligned} \sum_{j \ge 0}a_{N,j}\cos(jx) & \equiv \; \frac12\big[\cos x - \cos(4N-1)x\big]\sum_{j \ge 0} a_{N-1,j} \cos(jx) \\ & \equiv \; \frac14\sum_{j \ge 0} a_{N-1,j}\big[\cos(j+1)x + \cos(j-1)x - \cos(j+4N-1)x - \cos(j - 4N+1)x\big] \end{aligned}$ and hence we can write $a_{N,j} \; = \; 4^{-N}b_{N,j}$ where $\sum_{j \ge 0}b_{N,j} \cos(jx) \; \equiv \; \sum_{j \ge 0}b_{N-1,j}\big[\cos(j+1)x + \cos|j-1|x - \cos(j+4N-1)x - \cos|j-4N+1|x\big]$ and we have $b_{0,0}=1$ with $b_{0,j} = 0$ for $j \ge 1$ .

It is a simple matter to run a computer programme to calculate $b_{1010,0}$ . The maximum required value of $j$ in $b_{N,j}$ is $\sum_{j=1}^N(4j-1) = N(2N+1)$ , so $j \le 2041210$ when $N = 1010$ , so we only need a finite-dimensional array $b_{i,j}$ to do the calculations. The end result is $b_{1010,0} = 1574408464$ , which makes the desired integral $2\pi a_{1010,0} \; = \; \frac{2\pi \times 2^4 \times 985900529}{2^{2020}} \; = \; \frac{985900529\pi}{2^{2015}}$

The answer is $985900529 + 2015 = \boxed{985902544}$ .

Finding the coefficient $b_{N,0}$ is a very hard problem. It is related to the problem of determining the number of subsets $A$ of $\{1,2,...,2N\}$ such that both $A$ and its complement have the same sum. The equal-sum subset problem (for a general set of integers) is NP-hard.