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Half range sine series: e^(ax)

Short answer

Using ∫ e^(ax) sin(nx) dx, the half-range sine series on (0,pi) is e^(ax) = (2/pi) sum_{n=1}^infinity [n(1-(-1)^n e^(a pi))/(a^2+n^2)] sin(nx), obtained by integration by parts.

Find the half-range Fourier sine series of f(x) = e^(ax) in the interval 0 ≤ x ≤ π.

Half-range sine series on (0, π): f(x) = Σ bₙ sin nx with

bₙ = (2/π) ∫₀^π e^(ax) sin nx dx

Use ∫ e^(ax) sin nx dx = e^(ax)(a sin nx − n cos nx)/(a² + n²):

∫₀^π e^(ax) sin nx dx = [ e^(ax)(a sin nx − n cos nx)/(a² + n²) ]₀^π
                      = [ −n(−1)ⁿ e^(aπ) + n ] / (a² + n²)
                      = n [ 1 − (−1)ⁿ e^(aπ) ] / (a² + n²)

bₙ = (2n/π) · [ 1 − (−1)ⁿ e^(aπ) ] / (a² + n²)

Answer:

e^(ax) = (2/π) Σ_{n=1}^∞ [ n (1 − (−1)ⁿ e^(aπ)) / (a² + n²) ] sin nx,   0 < x < π

✓ Answer checked by Claude Opus 5.5 (high effort)

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