Fourier sine integral of e^{-kx}

Fourier sine integral. For f(x) defined for x > 0 (extended as an odd function):

f(x) = (2/π) ∫₀^∞ B(λ) sin λx dλ , where B(λ) = ∫₀^∞ f(v) sin λv dv.

Step 1: find B(λ) for f(v) = e^{−Kv}, K > 0.

B(λ) = ∫₀^∞ e^{−Kv} sin λv dv = λ/(K² + λ²)

(Integration by parts twice gives I = λ/K² − (λ²/K²) I, hence I = λ/(K² + λ²).)

Step 2: representation.

e^{−Kx} = (2/π) ∫₀^∞ λ sin λx /(λ² + K²) dλ, x > 0

Step 3: deduction. Multiply by π/2:

∫₀^∞ λ sin λx /(λ² + K²) dλ = (π/2) e^{−Kx}, x > 0, K > 0 ∎

At x = 0 the integral is 0, while e⁰ = 1: the odd extension of f has a jump at x = 0 (value 0, the average of −1 and 1), so the formula holds for x > 0 only.