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Half range sine series: e^(2x)

Short answer

Compute bₙ = (2/π)∫₀^π e^(2x) sin nx dx = (2/π) n[1 − (−1)ⁿ e^(2π)]/(n²+4). The half-range sine series is e^(2x) = (2/π) Σ_{n=1}^∞ { n[1−(−1)ⁿ e^(2π)]/(n²+4) } sin nx.

Find the half range sine series for f(x) = e^(2x) in 0 < x < π.

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. Use ∫ eᵃˣ sin nx dx = eᵃˣ(a sin nx − n cos nx)/(a² + n²) with a = 2:

bₙ = (2/π)∫₀^π e^(2x) sin nx dx
   = (2/π)[ e^(2x)(2 sin nx − n cos nx)/(4 + n²) ]₀^π
   = (2/π)·[ −n(−1)ⁿ e^(2π) + n ]/(n² + 4)
   = (2/π)·n[1 − (−1)ⁿ e^(2π)]/(n² + 4)

Series

e^(2x) = (2/π) Σ (n=1 to ∞) { n[1 − (−1)ⁿ e^(2π)] / (n² + 4) } sin nx, 0 < x < π

= (2/π)[ (1 + e^(2π))/5 · sin x + 2(1 − e^(2π))/8 · sin 2x + 3(1 + e^(2π))/13 · sin 3x + … ]

For n = 1, 2, 3 the denominators n² + 4 are 5, 8 and 13.

✓ Answer checked by Claude Opus 5.5 (high effort)

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