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Solve an integral equation

Short answer

By Fourier sine inversion, f(x) = (2/π) ∫_0^∞ F_s(ω) sin ωx dω = (2/π)[∫_0^1 sin ωx dω + 2∫_1^2 sin ωx dω] = (2/(πx)) (1 + cos x − 2 cos 2x) for x > 0.

Solve the integral equation ∫_0^∞ f(x) sin ωx dx = 1 for 0 ≤ ω < 1; 2 for 1 ≤ ω < 2; 0 for ω ≥ 2, using Fourier transform.

Write the equation as a Fourier sine transform: F_s(ω) = ∫₀^∞ f(x) sin ωx dx, where F_s(ω) = 1 for 0 ≤ ω < 1, 2 for 1 ≤ ω < 2, and 0 for ω ≥ 2. The inverse sine transform is f(x) = (2/π)∫₀^∞ F_s(ω) sin ωx dω.

f(x) = (2/π) [ ∫₀¹ 1·sin ωx dω + ∫₁² 2·sin ωx dω ]
     = (2/π) [ (1 − cos x)/x + 2(cos x − cos 2x)/x ]
     = (2/πx) [ 1 − cos x + 2cos x − 2cos 2x ]

Answer: f(x) = (2 / πx) (1 + cos x − 2 cos 2x), x > 0

Check: as x → 0, 1 + cos x − 2cos 2x → 0, so f stays finite (f(0) = 0 in the limit).

✓ Answer checked by Claude Opus 5.5 (high effort)

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