Let $r : \mathbb { R } \rightarrow \mathbb { R }$, be a $2 \pi$-periodic, odd function, such that $\forall \theta \in ] 0 , \pi ] , r ( \theta ) = \frac { \pi - \theta } { 2 }$.
Deduce that $\sum _ { n = 0 } ^ { + \infty } \frac { 1 } { ( 2 n + 1 ) ^ { 2 } } = \frac { \pi ^ { 2 } } { 8 }$.