Fourier cosine integral of e^{-kx}

Fourier cosine integral. For an even extension of f defined on x > 0:

f(x) = (2/π) ∫₀^∞ A(ω) cos ωx dω , where A(ω) = ∫₀^∞ f(v) cos ωv dv.

Step 1: find A(ω) for f(v) = e^{−kv}, k > 0.

A(ω) = ∫₀^∞ e^{−kv} cos ωv dv

Integrating by parts twice (or using the standard result):

A(ω) = k/(k² + ω²)

Step 2: representation.

e^{−kx} = (2k/π) ∫₀^∞ cos ωx /(k² + ω²) dω, x > 0.

Step 3: deduction. Multiply both sides by π/(2k):

∫₀^∞ cos ωx dω/(k² + ω²) = (π/(2k)) e^{−kx}, x > 0, k > 0 ∎

At x = 0 this gives ∫₀^∞ dω/(k² + ω²) = π/(2k), which is correct.