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Cosine transform of e^{-x}; Parseval identity

Short answer

The cosine transform is F_c(s) = √(2/π) · 1/(1+s²). Parseval's identity gives ∫₀^∞ e^{-2x} dx = ∫₀^∞ (2/π)/(1+s²)² ds, so 1/2 = (2/π)∫₀^∞ ds/(1+s²)², hence ∫₀^∞ dx/(1+x²)² = π/4.

Find the Fourier cosine transform of f(x) = e^{-x}, x > 0 and hence, by Parseval's identity, show that ∫₀^∞ 1/(1 + x²)² dx = π/4.

Definition used. The Fourier cosine transform is

F_c(s) = √(2/π) ∫₀^∞ f(x) cos sx dx , with inverse f(x) = √(2/π) ∫₀^∞ F_c(s) cos sx ds.

Cosine transform of e⁻ˣ

Using ∫₀^∞ e^{−ax} cos sx dx = a/(a² + s²) with a = 1:

F_c(s) = √(2/π) · 1/(1 + s²)

Parseval's identity for cosine transforms

∫₀^∞ [f(x)]² dx = ∫₀^∞ [F_c(s)]² ds

Left side: ∫₀^∞ e^{−2x} dx = ½.

Right side: ∫₀^∞ (2/π) · 1/(1 + s²)² ds = (2/π) ∫₀^∞ ds/(1 + s²)².

Equating:

(2/π) ∫₀^∞ ds/(1 + s²)² = 1/2
∫₀^∞ ds/(1 + s²)²       = π/4

Changing the dummy variable s to x:

∫₀^∞ dx/(1 + x²)² = π/4 ∎

(Numerical check: the integral evaluates to 0.785398 = π/4.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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