Fourier series: x sin x

f(x) = x sin x is even on (−π, π) (x odd × sin x odd), so bₙ = 0.

Coefficients

a₀ = (2/π) ∫₀^π x sin x dx = (2/π)[−x cos x + sin x]₀^π = (2/π)(π) = 2

For n = 1:

a₁ = (2/π) ∫₀^π x sin x cos x dx = (1/π) ∫₀^π x sin 2x dx
   = (1/π)[−x cos 2x/2 + sin 2x/4]₀^π = (1/π)(−π/2) = −1/2

For n ≥ 2, use 2 sin x cos nx = sin(n+1)x − sin(n−1)x and ∫₀^π x sin mx dx = π(−1)^(m+1)/m:

aₙ = (1/π) ∫₀^π x [ sin(n+1)x − sin(n−1)x ] dx
   = (−1)ⁿ⁺² /(n+1) − (−1)ⁿ/(n−1)
   = (−1)ⁿ [ 1/(n+1) − 1/(n−1) ]
   = 2(−1)ⁿ⁺¹ / (n² − 1)

Series

x sin x = 1 − (1/2) cos x + 2 Σ_{n=2}^∞ [ (−1)ⁿ⁺¹ / (n² − 1) ] cos nx
        = 1 − (1/2) cos x − (2/3) cos 2x + (2/8) cos 3x − (2/15) cos 4x + ...

Check at x = π: π sin π = 0. The series gives 1 + 1/2 − 2 Σ_{n≥2} 1/(n² − 1) = 3/2 − 2(3/4) = 0 ✓.