1

I've been given the half-range sine series

$$f(\pi-t)=\sum_{n=1}^\infty \frac{\sin(2n-1)t}{(2n-1)^3}, \ \ \ \ \ \ \ 0\le t\le \pi$$

The value of series $\sum_{n=1}^\infty \frac{1}{(2n-1)^6}$ needs to be determined.

Since $f(x)$ is not directly known, how are we supposed to apply Parseval's Theorem. Any help is appreciated.

Leucippus
  • 26,329
arcane_rt5
  • 49
  • The point is that $f(t)$ is known, since its formal second derivative constantly equals $\frac{\pi}{4}$ on $(0,\pi)$. Have a look at this very similar question: https://math.stackexchange.com/questions/2502046/compute-zeta6-using-fourier-series – Jack D'Aurizio Nov 12 '17 at 10:25
  • In particular $f(t)=\frac{\pi}{8}t(\pi-t)$ over $(0,\pi)$ and the problem boils down to computing the elementary integral $\int_{0}^{1}x^2(1-x)^2,dx$. – Jack D'Aurizio Nov 12 '17 at 10:29

1 Answers1

0

Using the definition of the zeta function, you have $$\sum_{i=1}^\infty\frac{1}{(2i-1)^n}=(1-2^{-n})\,\zeta (n)$$ and the nice values for even $n$ $$\left( \begin{array}{cc} n & \zeta (n) \\ 2 & \frac{\pi ^2}{6} \\ 4 & \frac{\pi ^4}{90} \\ 6 & \frac{\pi ^6}{945} \\ 8 & \frac{\pi ^8}{9450} \\ 10 & \frac{\pi ^{10}}{93555} \end{array} \right)$$

Claude Leibovici
  • 260,315