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

Theorem
Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:


 * $\map f x = x \paren {\pi - x}$

Then its half-range Fourier sine series can be expressed as:


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

Proof
By definition of half-range Fourier sine series:


 * $\displaystyle \map f x \sim \sum_{n \mathop = 1}^\infty b_n \sin n x$

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

Thus by definition of $f$:

Splitting it up into two:

and:

Thus:

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

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

Thus:

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