Past exam of the fourier analysis course of the University of Oxford / 2023 / 3 / b

Splitting the Fourier transform integral at zero gives two absolutely convergent integrals:Both $g_t$ and $\widehat{g_t}$ are integrable functions. Let $h_t(x)=(t^2+x^2{)}^{-1}$, so $\widehat{g_t}=2t{h}_t$. The duality property of the Fourier transform, or Fourier inversion in tempered distributions, givesThese Fourier transforms have continuous representatives by the Riemann-Lebesgue lemma, so the identity holds at every $\xi$. Since $g_t$ is an even function,

Put $h(x)=({\alpha }^2+x^2{)}^{-1}$ and form its period-one periodization of an integrable functionThe function $h$ is not a Schwartz function, so justify the needed Poisson summation formula directly. For every $j\ge 0$, its derivative satisfies $h^{(j)}(x)=O(|x{|}^{-2-j})$ as $|x|\to \infty$. Thus the Weierstrass M-test gives uniform convergence on $[0,1]$ of this series and each differentiated series. It follows that $H$ is a smooth period-one periodic function.

Absolute integrability and Fubini's theorem allow its Fourier coefficients to be computed by splitting the real line into unit intervals:These Fourier coefficients are absolutely summable. Their Fourier series defines a continuous periodic function with the same Fourier coefficients as $H$; uniqueness of Fourier coefficients in L1 identifies the two almost everywhere, and continuity then identifies them everywhere. Evaluating at zero gives the rigorously justified Poisson summation formulaRearranging provesThe last series is a geometric series with sum $(e^{2\pi \alpha }-1{)}^{-1}$. Equivalently, the left side is $\pi \coth (\pi \alpha )/(2\alpha )-1/(2{\alpha }^2)$, where $\coth$ is the hyperbolic cotangent.