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Fourier coefficient bₙ of e^(−x)

Short answer

bₙ = n(1−e^(−2π)) / [π(1+n²)], obtained from bₙ = (1/π)∫₀^{2π} e^(−x) sin nx dx.

In the series f(x) = a₀/2 + Σ_{n=1}^∞ (aₙ cos nx + bₙ sin nx), find the coefficient bₙ for the function f(x) = e^(−x) where x ∈ (0, 2π).

For f(x) on (0, 2π), bₙ = (1/π)∫₀^{2π} f(x) sin nx dx.

With f(x) = e^(−x), use ∫ e^(−x) sin nx dx = e^(−x)(−sin nx − n cos nx)/(1 + n²):

bₙ = (1/π)[ e^(−x)(−sin nx − n cos nx)/(1 + n²) ]₀^{2π}
   = (1/π)·[ −n e^(−2π) + n ]/(1 + n²)          (sin 2nπ = 0, cos 2nπ = 1)

Answer: bₙ = n(1 − e^(−2π)) / [π(1 + n²)]

For example, b₁ = (1 − e^(−2π))/(2π) ≈ 0.159.

✓ Answer checked by Claude Opus 5.5 (high effort)

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