For each $n \geq 1$, denote $C_n= \int_0^\pi x^3cos(nx)dx$. Calculate $\sum_{n=1}^{\infty} {C_n}^2$
My suggestion: The given sequence is proportional to Fourier coefficients of $x^3$ expansion to an even function in $[-\pi,\pi]$. The sum that I get is: $\frac{9\pi^8}{224}$.
Am I in the right direction, how do I validate my answer?
Thanks
This follows easily from Parseval formula: assume $f(x)=a_0+\sum_{n\ge 1}a_n\cos nx, x \in [-\pi,\pi]$ is the Fourier series of the even extension of $x^3$ to $[-\pi, \pi]$.
Since for $n \ge1, a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nx dx$ it follows that $a_n=\frac{2C_n}{\pi}$ while $a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{\pi^3}{4}$
But Parseval says (easily seen by squaring, integrating and using orthogonality) that $$\int_{-\pi}^{\pi}f^2(x)dx=2\pi a_0^2+\pi\sum_{n \ge 1}a_n^2$$
Hence $$\frac{2\pi^7}{7}=\frac{\pi^7}{8}+\frac{4}{\pi}\sum_{n \ge 1}C_n^2$$ from which the claimed result $\sum_{n=1}^{\infty} {C_n}^2=\frac{9\pi^8}{224}$ follows.