Fourier integral representation

Fourier integral. f(x) = ∫₀^∞ [A(α) cos αx + B(α) sin αx] dα, where A(α) = (1/π)∫ f(v) cos αv dv and B(α) = (1/π)∫ f(v) sin αv dv.

For f(v) = e^(−v) (v > 0), 0 (v < 0):

A(α) = (1/π)∫₀^∞ e^(−v) cos αv dv = 1/[π(1 + α²)]
B(α) = (1/π)∫₀^∞ e^(−v) sin αv dv = α/[π(1 + α²)]

f(x) = (1/π) ∫₀^∞ (cos αx + α sin αx)/(1 + α²) dα

Deduction. Multiply both sides by π. The integral I(x) = ∫₀^∞ (cos ωx + ω sin ωx)/(1 + ω²) dω = π f(x), where at a jump f is replaced by the mean of its left and right limits. So:

• x > 0: f is continuous and equals e^(−x), so I = π e^(−x).
• x < 0: f = 0, so I = 0.
• x = 0: f jumps from 0 to 1, the integral converges to the mean (0 + 1)/2 = 1/2, so I = π/2.

Hence ∫₀^∞ (cos ωx + ω sin ωx)/(1 + ω²) dω = π e^(−x) for x > 0, π/2 for x = 0 and 0 for x < 0.