Fourier transform of 1-x²; evaluate integral

Definition used. F(s) = ∫{−∞}^{∞} g(x) e^{−isx} dx, with inverse g(x) = (1/2π) ∫{−∞}^{∞} F(s) e^{isx} ds.

Fourier transform of g(x) = 1 − x² on |x| < 1

g is even, so the sine part vanishes:

F(s) = ∫₋₁¹ (1 − x²) cos sx dx = 2 ∫₀¹ (1 − x²) cos sx dx

Integrate by parts:

∫₀¹ (1 − x²) cos sx dx = [ (1 − x²) sin sx / s ]₀¹ + (2/s) ∫₀¹ x sin sx dx

The first term is 0. For the second:

∫₀¹ x sin sx dx = [ −x cos sx / s ]₀¹ + (1/s) ∫₀¹ cos sx dx = −cos s / s + sin s / s²

So ∫₀¹ (1 − x²) cos sx dx = (2/s)(sin s − s cos s)/s² = 2(sin s − s cos s)/s³, and

F(s) = 4 (sin s − s cos s) / s³

Evaluating the integral

Use inversion with g even:

g(x) = (1/2π) ∫₋∞^∞ F(s) cos sx ds = (1/π) ∫₀^∞ F(s) cos sx ds
     = (4/π) ∫₀^∞ (sin s − s cos s)/s³ · cos sx ds

Put x = ½ (inside |x| < 1), where g(½) = 1 − ¼ = ¾:

(4/π) ∫₀^∞ (sin s − s cos s)/s³ · cos(s/2) ds = 3/4
∫₀^∞ (sin s − s cos s)/s³ · cos(s/2) ds = 3π/16

The required integrand is (x cos x − sin x)/x³ cos(x/2), which has the opposite sign:

∫₀^∞ [(x cos x − sin x)/x³] cos(x/2) dx = −3π/16 ≈ −0.589

(Numerical check: the integral of (sin s − s cos s) cos(s/2)/s³ computes to 0.58905 = 3π/16 ✓.)