we know that $x(\pi-x)=\dfrac{\pi^2}{6}-\displaystyle \sum_{n=1}^\infty \frac{\cos(2nx)}{n^2}$ $\forall$ $0 \le x \le \pi$
Prove: $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}= \frac{\pi^3}{32}$
I guess we would have to choose $x$ so that for $n$ even $\cos (2nx)$ is equal to $0$. But I'm still stuck with $(2n - 1)^3$. I think the derivation can be useful but I still can not prove it. Tks everyone
The cube in the denominator suggests antidifferentiating both sides: \begin{align} & \int x(\pi-x)\,dx = \int \left( \frac{\pi^2} 6 - \sum_{n=1}^\infty \frac{\cos(2nx)}{n^2} \right) \, dx \\[10pt] & \pi \frac{x^2} 2 - \frac{x^3} 3 = \frac{\pi^2} 6 x - \sum_{n=1}^\infty \frac{\sin(2nx)}{2n^3} \\[10pt] \text{If } x = \frac \pi 4 \text{ then this says } & \frac{\pi^3}{32} = \frac{\pi^3}{24} - \sum_{n=1}^\infty \frac{\cdots}{2n^3} \end{align} The numerator $\sin\left( n \dfrac \pi 2\right)$ is equal to $0$ when $n$ is even. You're left with odd-degree terms, so put $2n-1$ where $n$ was and observe that $\sin\left((2n-1)\dfrac\pi 2 \right)$ alternates between $\pm1.$