 Solution

ID: past-exam-of-the-fourier-analysis-course-of-the-university-of-oxford/2023/3/b/i/solution

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}

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

, so

g_{t} = 2 t h_{t}

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}

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

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