Half range sine series: lx − x²

To show: for f(x) = lx − x² on (0, l), the half-range sine series is (8l²/π³) Σ_{n=0}^∞ sin((2n+1)πx/l)/(2n+1)³.

Coefficients

Half-range sine series: f(x) = Σ bₙ sin(nπx/l), with bₙ = (2/l) ∫₀^l (lx − x²) sin(nπx/l) dx. Let k = nπ/l (so kl = nπ). Integrate by parts twice:

∫₀^l (lx − x²) sin kx dx = [ −(lx − x²) cos kx/k ]₀^l + (1/k) ∫₀^l (l − 2x) cos kx dx
 = 0 + (1/k) { [ (l − 2x) sin kx/k ]₀^l + (2/k) ∫₀^l sin kx dx }
 = (1/k)(2/k) [ (1 − cos kl)/k ]
 = 2 (1 − (−1)ⁿ)/k³
 = 2 l³ (1 − (−1)ⁿ)/(nπ)³

(The boundary terms vanish because sin kl = sin nπ = 0 and lx − x² = 0 at both ends.)

bₙ = (2/l) · 2l³ (1 − (−1)ⁿ)/(n³π³) = 4l² (1 − (−1)ⁿ)/(n³π³)

For even n, bₙ = 0. For odd n, put n = 2m + 1: bₙ = 8l²/(π³(2m + 1)³).

Result

lx − x² = (8l²/π³) Σ_{m=0}^∞ [1/(2m + 1)³] sin((2m + 1)πx/l),   0 < x < l

which is the required series. ∎ (Check at x = l/2: l²/4 = (8l²/π³)(1 − 1/27 + 1/125 − ...) → π³/32 = 1 − 1/3³ + 1/5³ − ..., a known result ✓.)