Fourier series: x + x²

For f(x) on (−π, π) of period 2π: f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx), a₀ = (1/π)∫f dx, aₙ = (1/π)∫f cos nx dx, bₙ = (1/π)∫f sin nx dx.

Here x is odd and x² is even, so the odd part contributes only to bₙ and the even part only to a₀ and aₙ.

a₀ = (1/π)∫₋π^π (x + x²) dx = (1/π)(2π³/3) = 2π²/3

aₙ = (1/π)∫₋π^π x² cos nx dx = (2/π)[x² sin nx/n + 2x cos nx/n² − 2 sin nx/n³]₀^π
   = (2/π)(2π cos nπ / n²) = 4(−1)ⁿ/n²

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

Fourier series

x + x² = π²/3 + Σ (n=1 to ∞) [ 4(−1)ⁿ cos nx / n² + 2(−1)ⁿ⁺¹ sin nx / n ]

= π²/3 − 4(cos x − cos 2x/2² + cos 3x/3² − …) + 2(sin x − sin 2x/2 + sin 3x/3 − …)

Check: at x = 0 the function is 0 and the series gives π²/3 − 4(π²/12) = 0.