Find the value $$\sum_{n=1}^{\infty}\int_{0}^{\pi}\int_{0}^{\pi}(xy)^{k}[\cos n(x-y)-\cos n(x+y)] \, dx \, dy,\qquad k\in N^{+}$$
My idea: \begin{align} &\sum_{n=1}^{\infty}\int_{0}^{\pi}\int_{0}^{\pi}(xy)^{k}[\cos n(x-y) -\cos n(x+y)] \, dx \, dy\\ &=2\sum_{n=1}^{\infty}\left(\int_{0}^{\pi}x^k\sin(nx) \, dx\right)^2 \end{align}
Then we use $$ \sin{nx}=\dfrac{e^{inx}-e^{-inx}}{2i}$$ where $i=\sqrt{-1}$.
The series converges by Parseval's theorem, which also tells us its value. Begin with $$x^k = \sum_{n=1}^\infty b_n \sin nx\quad \text{where } \ b_n=\frac{2}{\pi}\int_0^\pi x^k \sin nx\,dx\tag1$$ Since $\sqrt{2/\pi}\sin nx$ are orthonormal functions on $[0,\pi]$, it follows that $$\sum \frac{\pi}2 b_n^2 = \int_0^\pi x^{2k}\,dx = \frac{\pi^{2k+1}}{2k+1} \tag2 $$ Hence, $$\sum_{n=1}^\infty \left(\int_0^\pi x^k \sin nx\,dx\right)^2 = \frac{\pi^{2k+2}}{4k+2}\tag3$$