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Fourier transform of a function

Short answer

The Fourier transform is F(ω) = ∫_{-1}^1 e^(iωx) dx = 2 sin ω / ω for ω≠0, and F(0)=2. In the symmetric convention with 1/√(2π), it is √(2/π) sin ω/ω.

Find the Fourier transform of f(x) = 1 if |x| < 1; 0 if |x| > 1.

The Fourier transform of f(x) is F(ω) = ∫₋∞^∞ f(x) e^(iωx) dx. (Some books use e^(−iωx) or a factor 1/√(2π); the real result below is the same up to that factor.)

For f(x) = 1 for |x| < 1 and 0 for |x| > 1:

F(ω) = ∫₋₁¹ e^(iωx) dx = [ e^(iωx)/(iω) ]₋₁¹
     = (e^(iω) − e^(−iω))/(iω)
     = 2 sin ω / ω

Answer: F(ω) = 2 sin ω / ω (ω ≠ 0), and F(0) = 2.

In the symmetric convention with the factor 1/√(2π), F(ω) = √(2/π) · sin ω / ω.

✓ Answer checked by Claude Opus 5.5 (high effort)

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