Half-Range Fourier Sine Series/x by Pi minus x over 0 to Pi

Theorem
Let $f \left({x}\right)$ be the real function defined on $\left({0 \,.\,.\, \pi}\right)$ as:


 * $f \left({x}\right) = x \left({\pi - x}\right)$

Then its Fourier series can be expressed as:


 * $\displaystyle f \left({x}\right) \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \left({2 r + 1}\right) x} {\left({2 r + 1}\right)^3}$

Proof
By definition of half-range Fourier sine series:


 * $\displaystyle f \left({x}\right) \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$

where for all $n \in \Z_{> 0}$:
 * $b_n = \displaystyle \frac 2 \pi \int_0^\pi f \left({x}\right) \sin n x \rd x$

Thus by definition of $f$:

Splitting it up into two:

and:

Thus:

When $n$ is even, $\left({-1}\right)^n = 1$ and so $\dfrac {4 \left({1 - \left({-1}\right)^n}\right)} {\pi n^3} = 0$.

When $n$ is odd, $n$ can be expressed as $n = 2 r + 1$ for $r \ge 0$.

Thus:

Hence:
 * $\displaystyle f \left({x}\right) \sim \frac 8 \pi \sum_{r \mathop = 0}^\infty \frac {\sin \left({2 r + 1}\right) x} {\left({2 r + 1}\right)^3}$