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Fourier cosine integral

Short answer

Using Fourier cosine integral, e^(-kx) = (2/π) ∫_0^∞ [k/(k²+ω²)] cos ωx dω. Hence ∫_0^∞ cos ωx/(k²+ω²) dω = π/(2k) e^(-kx), for x ≥ 0.

Find the Fourier cosine integral of f(x) = e^(−kx), k > 0, x > 0 and use it to evaluate ∫_0^∞ cos ωx/(k² + ω²) dω.

The Fourier cosine integral of an even extension of f(x), x > 0, is f(x) = (2/π)∫₀^∞ [∫₀^∞ f(v) cos ωv dv] cos ωx dω.

Inner integral for f(v) = e^(−kv):

∫₀^∞ e^(−kv) cos ωv dv = k/(k² + ω²)

Fourier cosine integral

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

Evaluation. Rearranging the above,

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

In particular, for x = 0 the integral ∫₀^∞ dω/(k² + ω²) = π/(2k).

✓ Answer checked by Claude Opus 5.5 (high effort)

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