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 Incoming links: Duality property of the Fourier transform

Past exam of the fourier analysis course of the University of Oxford / 2023 / 3 / b / i / Solution 2026-10-25

Splitting the Fourier transform integral at zero gives two absolutely convergent integrals:

g_{t} (ξ) = \int_{0}^{\infty} e^{- (t + i ξ) x} d x + \int_{0}^{\infty} e^{- (t - i ξ) x} d x = \frac{1}{t + i ξ} + \frac{1}{t - i ξ} = \frac{2 t}{t ^{2} + ξ ^{2}} .

Both

g_{t}

and

g_{t}

are integrable functions. Let

h_{t} (x) = (t^{2} + x^{2})^{- 1}

, so

g_{t} = 2 t h_{t}

. The duality property of the Fourier transform, or Fourier inversion in tempered distributions, gives

2 t h_{t} (ξ) = g_{t} (ξ) = 2 π g_{t} (- ξ) .

These Fourier transforms have continuous representatives by the Riemann-Lebesgue lemma, so the identity holds at every

ξ

. Since

g_{t}

is an even function,

F_{x \to ξ} (\frac{1}{t ^{2} + x ^{2}}) = \frac{π}{t} e^{- t ∣ ξ ∣} .

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 4 / a / Solution 2026-10-18

Using Fourier inversion with the stated convention,

f (- ω) = \frac{1}{2 π} \int_{- \infty}^{\infty} f (t) e^{- iω t} d t .

Hence applying the transform twice gives the duality property of the Fourier transform:

F [f] (ω) = 2 π f (- ω) .

The formula holds directly for rapidly decreasing smooth functions, and extends through the usual L2 space or tempered distribution formulations when absolute convergence is unavailable.

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 4 / e / Solution 2026-10-18

α = 1

in part (d), obtaining

F [e^{- ∣ t ∣}] (ω) = 2/ (1 + ω^{2})

. Apply the duality property of the Fourier transform from part (a):

2 F [\frac{1}{1 + t ^{2}}] (ω) = 2 π e^{- ∣ ω ∣} .

Therefore

s (ω) = π e^{- ∣ ω ∣} .

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