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Fourier transform of a function

Short answer

The Fourier transform of e^(-a|x|) is 2a/(a²+ω²). Applying Parseval’s identity to f and g(x)=e^(-b|x|) gives ∫_{-∞}^∞ 4ab/[(a²+ω²)(b²+ω²)] dω = 2π ∫ e^{-(a+b)|x|} dx = 4π/(a+b). Thus ∫_0^∞ dx/[(a²+x²)(b²+x²)] = π/[2ab(a+b)].

Find the Fourier transform of f(x) = e^(−ax) and use the result to show ∫_0^∞ dx/((a² + x²)(b² + x²)) = π/(2ab(a + b)).

Reading of the question. The function is taken as f(x) = e^(−a|x|) with a > 0 (so that its transform exists on the whole line and the Parseval identity applies). Use F(ω) = ∫₋∞^∞ f(x) e^(iωx) dx.

Fourier transform

F(ω) = ∫₀^∞ e^(−ax) e^(iωx) dx + ∫₋∞^0 e^(ax) e^(iωx) dx
     = 1/(a − iω) + 1/(a + iω)
     = 2a / (a² + ω²)

Likewise, for g(x) = e^(−b|x|), G(ω) = 2b/(b² + ω²).

Parseval's (Plancherel) identity. For real even functions: ∫₋∞^∞ F(ω)G(ω) dω = 2π ∫₋∞^∞ f(x)g(x) dx.

Right side : 2π ∫₋∞^∞ e^(−(a+b)|x|) dx = 2π · 2/(a + b) = 4π/(a + b)
Left side  : 4ab ∫₋∞^∞ dω /[(a² + ω²)(b² + ω²)] = 8ab ∫₀^∞ dω /[(a² + ω²)(b² + ω²)]

Equating: 8ab ∫₀^∞ dω /[(a² + ω²)(b² + ω²)] = 4π/(a + b).

Answer: ∫₀^∞ dx / [(a² + x²)(b² + x²)] = π / [2ab(a + b)]

✓ Answer checked by Claude Opus 5.5 (high effort)

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