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Fourier integral representation

Short answer

Fourier integral representation: f(x) = (1/π) ∫_0^∞ (cos αx + α sin αx)/(1+α^2) dα. Setting x = 0 where f(0) averages 1/2 gives 1/2 = (1/π)∫_0^∞ dα/(1+α^2), so ∫_0^∞ dx/(1+x^2) = π/2.

Obtain the Fourier integral of the function f(x) = e^(−x) for x > 0; 0 for x < 0. Hence evaluate ∫_0^∞ 1/(1 + x²) dx.

The Fourier integral of f(x) is f(x) = ∫₀^∞ [A(α) cos αx + B(α) sin αx] dα, with A(α) = (1/π)∫ f(v) cos αv dv and B(α) = (1/π)∫ f(v) sin αv dv.

Coefficients. f(v) = e^(−v) for v > 0, and 0 for v < 0.

A(α) = (1/π)∫₀^∞ e^(−v) cos αv dv = 1 / [π(1 + α²)]
B(α) = (1/π)∫₀^∞ e^(−v) sin αv dv = α / [π(1 + α²)]

Fourier integral

f(x) = (1/π) ∫₀^∞ (cos αx + α sin αx) / (1 + α²) dα

This equals e^(−x) for x > 0, 0 for x < 0, and the mean value 1/2 at the jump x = 0.

Evaluation of ∫₀^∞ dx/(1 + x²). Put x = 0 in the integral. The value there is the average (0 + 1)/2 = 1/2:

1/2 = (1/π) ∫₀^∞ dα / (1 + α²)

Answer: ∫₀^∞ dx / (1 + x²) = π/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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