Fourier series: e^(3x)

On (0, 2π) of period 2π: f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx), a₀ = (1/π)∫₀^{2π} f dx, aₙ = (1/π)∫₀^{2π} f cos nx dx, bₙ = (1/π)∫₀^{2π} f sin nx dx.

Use ∫ e^(3x) cos nx dx = e^(3x)(3 cos nx + n sin nx)/(9 + n²) and ∫ e^(3x) sin nx dx = e^(3x)(3 sin nx − n cos nx)/(9 + n²). At x = 2π, sin 2nπ = 0 and cos 2nπ = 1.

a₀ = (1/π)[e^(3x)/3]₀^{2π} = (e^(6π) − 1)/(3π)

aₙ = (1/π)·[ 3e^(6π) − 3 ]/(9 + n²) = 3(e^(6π) − 1)/[π(n² + 9)]

bₙ = (1/π)·[ −n e^(6π) + n ]/(9 + n²) = −n(e^(6π) − 1)/[π(n² + 9)]

Fourier series

e^(3x) = (e^(6π) − 1)/π · { 1/6 + Σ (n=1 to ∞) [ 3 cos nx − n sin nx ] / (n² + 9) }, 0 < x < 2π

Writing out: e^(3x) = ((e^(6π) − 1)/π)[ 1/6 + (3 cos x − sin x)/10 + (3 cos 2x − 2 sin 2x)/13 + … ]

At the ends x = 0 and 2π, the series converges to the mean ½(1 + e^(6π)).