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Half range sine series: π − x

Short answer

For the half-range sine series on (0,π), bn = (2/π)∫_0^π (π-x) sin nx dx = 2/n. Thus π-x = 2 Σ_{n=1}^∞ (sin nx)/n, 0<x<π.

Obtain the half-range Fourier sine series for f(x) = π − x in the range 0 < x < π.

For a half-range sine series on (0, π):

f(x) = Σ bₙ sin nx,    bₙ = (2/π) ∫₀^π f(x) sin nx dx

For f(x) = π − x:

bₙ = (2/π) ∫₀^π (π − x) sin nx dx

Integrate by parts: ∫₀^π (π − x) sin nx dx = [−(π − x) cos nx / n]₀^π − ∫₀^π cos nx/n dx = (π/n) − 0 = π/n (the last integral is [sin nx/n²]₀^π = 0).

bₙ = (2/π)(π/n) = 2/n

Answer:

π − x = 2 Σ_{n=1}^∞ (sin nx)/n = 2 [ sin x + (sin 2x)/2 + (sin 3x)/3 + ... ],  0 < x < π

(Check: at x = π/2, π/2 = 2[1 − 1/3 + 1/5 − ...], i.e. π/4 = 1 − 1/3 + 1/5 − ... ✓.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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