What is the fourier expand for f(x)=x² on interval -π to π?

Note that L = π and f(x) = x².

So, we have

a₀ = (1/L) ∫(x = -L to L) f(x) dx

....= (1/π) ∫(x = -π to π) x² dx

....= (2/π) ∫(x = 0 to π) x² dx, since the integrand is even

....= (2/π) * (x³/3) {for x = 0 to π}

....= 2π²/3.

For n > 0:

an = (1/L) ∫(x = -L to L) f(x) cos(nπx/L) dx

....= (1/π) ∫(x = -π to π) x² cos(nπx/π) dx

....= (2/π) ∫(x = 0 to π) x² cos(nx) dx, since the integrand is even

....= (2/π) * [x² sin(nx)/n - 2x * -cos(nx)/n² + -sin(nx)/n³] {for x = 0 to π}, via int. by parts

....= (2/π) * [0 + 2π(-1)^n/n² + 0] - 0

....= 4(-1)^n/n².

bn = (1/L) ∫(x = -L to L) f(x) sin(nπx/L) dx

....= (1/π) ∫(x = -π to π) x² sin(nπx/π) dx

....= 0, since the integrand is odd.

So, f(x) ~ a₀/2 + Σ(n = 1 to ∞) [an cos(nπx/L) + bn sin(nπx/L)]

............= (1/2)(2π²/3) + Σ(n = 1 to ∞) [(4(-1)^n/n²) cos(nπx/π) + 0 sin(nπx/π)]

............= π²/3 + Σ(n = 1 to ∞) (4(-1)^n/n²) cos(nx).

I hope this helps!