Fourier sine integral of e^{-x}cos x

Fourier sine integral. For f defined on x > 0:

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

Step 1: find B(ω) for f(v) = e^{−v} cos v.

Use sin ωv cos v = ½[sin (ω + 1)v + sin (ω − 1)v] and ∫₀^∞ e^{−v} sin bv dv = b/(1 + b²):

B(ω) = ½ [ (ω + 1)/(1 + (ω + 1)²) + (ω − 1)/(1 + (ω − 1)²) ]
     = ½ [ (ω + 1)/(ω² + 2ω + 2) + (ω − 1)/(ω² − 2ω + 2) ]

Combine over the common denominator (ω² + 2ω + 2)(ω² − 2ω + 2) = ω⁴ + 4:

numerator = (ω + 1)(ω² − 2ω + 2) + (ω − 1)(ω² + 2ω + 2) = (ω³ − ω² + 2) + (ω³ + ω² − 2) = 2ω³

B(ω) = ω³/(ω⁴ + 4)

Step 2: sine integral representation.

e^{−x} cos x = (2/π) ∫₀^∞ ω³ sin ωx /(ω⁴ + 4) dω, x > 0

Step 3: deduction. Multiplying by π/2:

∫₀^∞ ω³ sin ωx /(ω⁴ + 4) dω = (π/2) e^{−x} cos x, x > 0 ∎

(Numerical check of B(ω) at ω = 2.1: the integral gives 0.39496 = ω³/(ω⁴ + 4).)