Half range cosine series: x sin x

The half-range cosine series on (0, π) uses the even extension: f(x) = a₀/2 + Σ aₙ cos nx, aₙ = (2/π)∫₀^π f(x) cos nx dx.

For f(x) = x sin x:

a₀ = (2/π) ∫₀^π x sin x dx = (2/π)[−x cos x + sin x]₀^π = 2
a₁ = (2/π) ∫₀^π x sin x cos x dx = (1/π) ∫₀^π x sin 2x dx = (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)

Answer:

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 + (1/4) cos 3x − (2/15) cos 4x + ...,   0 ≤ x ≤ π

(x sin x is already even, so this is also its Fourier series on (−π, π).)