Half range sine series: πx − x²

The half range sine series on (0, π) is f(x) = Σ bₙ sin nx with bₙ = (2/π)∫₀^π f(x) sin nx dx.

Coefficient for f(x) = πx − x². Integrate by parts with u = πx − x²:

bₙ = (2/π)∫₀^π (πx − x²) sin nx dx
   = (2/π){ [−(πx − x²)cos nx/n]₀^π + (1/n)∫₀^π (π − 2x) cos nx dx }

The first bracket is 0 (the factor πx − x² vanishes at 0 and π). For the remaining integral:

∫₀^π (π − 2x) cos nx dx = [(π − 2x) sin nx/n]₀^π + (2/n)∫₀^π sin nx dx
                        = 0 + (2/n)(1 − cos nπ)/n = 2(1 − (−1)ⁿ)/n²

So

bₙ = (2/π)(1/n)·2(1 − (−1)ⁿ)/n² = 4(1 − (−1)ⁿ)/(π n³)

• n even: bₙ = 0
• n odd: bₙ = 8/(π n³)

Series

πx − x² = (8/π) Σ (n odd) sin nx / n³

= (8/π)[ sin x + sin 3x/3³ + sin 5x/5³ + … ], 0 < x < π

Check: at x = π/2 the left side is π²/4; the right side is (8/π)(1 − 1/27 + 1/125 − …) = (8/π)(π³/32) = π²/4.