Complex Fourier series; verify convolution

Fourier series in complex form

A periodic function f(x) of period 2L (satisfying the Dirichlet conditions) can be written as

f(x) = Σ_{n=−∞}^{∞} cₙ e^{inπx/L}

with complex coefficients

cₙ = (1/2L) ∫_{−L}^{L} f(x) e^{−inπx/L} dx, n = 0, ±1, ±2, …

It comes from the real form by writing cos θ = (e^{iθ} + e^{−iθ})/2 and sin θ = (e^{iθ} − e^{−iθ})/2i, giving cₙ = (aₙ − ibₙ)/2 and c₋ₙ = (aₙ + ibₙ)/2, c₀ = a₀/2. For L = π: f(x) = Σ cₙ e^{inx}, cₙ = (1/2π) ∫_{−π}^{π} f e^{−inx} dx.

Verifying the convolution theorem for f = g = e^{−x²}

For non-periodic functions the convolution theorem belongs to the Fourier transform, F(s) = ∫ f(x) e^{−isx} dx, with (f * g)(x) = ∫ f(u) g(x − u) du and F{f * g} = F(s)G(s).

Right side. F(s) = G(s) = √π e^{−s²/4}, so F(s)G(s) = π e^{−s²/2}.

Left side. Since u² + (x − u)² = 2(u − x/2)² + x²/2,

(f * g)(x) = e^{−x²/2} ∫ e^{−2(u − x/2)²} du = √(π/2) e^{−x²/2}

The transform of e^{−x²/2} is √(2π) e^{−s²/2}, hence

F{f * g} = √(π/2) · √(2π) e^{−s²/2} = π e^{−s²/2}

The two sides are equal, so the theorem is verified. ∎