Fourier integral of e^{-x}; ∫cos sx/(s²+1)

Fourier integral. For an even function f(x) the Fourier integral reduces to the cosine form:

f(x) = (2/π) ∫₀^∞ cos sx [ ∫₀^∞ f(v) cos sv dv ] ds

Here f(x) = e⁻ˣ for x ≥ 0 and f(−x) = f(x), so f(x) = e^{−|x|}, an even function.

Step 1: inner integral.

A(s) = ∫₀^∞ e^{−v} cos sv dv = 1/(1 + s²)

Step 2: representation.

e^{−|x|} = (2/π) ∫₀^∞ cos sx /(1 + s²) ds

Step 3: the integral. Multiplying by π/2 and taking x ≥ 0:

∫₀^∞ cos sx /(s² + 1) ds = (π/2) e^{−x}, x ≥ 0.

Since f is continuous everywhere (including x = 0), the representation holds for all x. At x = 0 it gives ∫₀^∞ ds/(1 + s²) = π/2 ✓.

(Numerical check at x = 1.5: the integral is 0.35049 = (π/2)e^{−1.5}.)