Am I right in the Fourier series on $[-\pi; \pi]$? $$x^{2m}\sim\frac{\pi^{2m}}{2m+1}+2\sum\limits_{n=1}^{+\infty}\sum_{k=0}^{m-1}\frac{(-1)^k\pi^{2(m-k-1)}}{n^{2(k+1)}(2m-2k-1)!}\cos\pi n\cos nx$$
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To check this will be a lot of work. Why don't you show your basic calculations; with the integrals, the sines and cosines parts and etc.? – DonAntonio Oct 27 '18 at 11:25
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$b_n=0$, because $\int\limits_{-\pi}^{\pi}x^{2m}\sin nx=0$ – Кирилл Колокольцев Oct 27 '18 at 12:20