Fourier transform of conjugate poisson kernel

Question

Let $$Q_t^{(j)}(x) = c_n \cdot \frac{x_j}{(t^2 + \vert x \vert^2)^{\frac{n+1}{2}}},$$ where $c_n = \Gamma(\frac{n+1}{2})/\pi^{\frac{n+1}{2}}$. What is $\widehat{Q_t^{(j)}}(\xi)$? I am having trouble computing this. Alternatively, if one knows $\widehat{Q_t^{(j)}}(\xi)$, is it easier to compute the inverse Fourier transform to get $Q_t^{(j)}(x)$?

Answer 1

Let me take the convention $$ \mathcal{F}(u)(y) = \widehat{u}(y) = \int_{\Bbb R^n} u(x)\,e^{-2i\pi\,x\cdot y}\,\mathrm d x. $$ I will work in vector notations and write $Q_t = (Q_t^{(j)})_{j\in\{1,\dots,n\}}$, and assume that $t>0$. First, notice that $Q_t(tx) = t^{-n} \,Q_1(x)$, so $Q_t(x) = t^{-n} \,Q_1(x/t)$. Hence by the scaling properties of the Fourier transform $$ \widehat{Q_t}(y) = t^{-n}\, \mathcal{F}(Q_1(x/t))(y) = \widehat{Q_1}(ty) $$ hence it suffices to compute the Fourier transform of $$ Q_1(x) = \frac{c_n\,x}{(1+|x|^2)^\frac{n+1}{2}} = \frac{c_n\,x}{\langle x\rangle^{n+1}} $$ where I used the notation $\langle x\rangle = (1+|x|^2)^\frac{1}{2}$. Now it is well known (see e.g. the proof I gave here) that $$ \mathcal{F}\!\left(\frac{c_n}{\langle x\rangle^{n+1}}\right)(y) = e^{-2\pi\,|y|}. $$ Hence, by the formula $\mathcal{F}(x\,u(x))(y) = \frac{i}{2\pi}\nabla\widehat{u}(y)$, we deduce $$ \widehat{Q_1}(y) = \frac{i}{2\pi}\nabla e^{-2\pi\,|y|} = -i \frac{y}{|y|}\,e^{-2\pi\,|y|}. $$ By the second equation, this gives $$ \widehat{Q_t}(y) = -i\, \frac{y}{|y|}\,e^{-2\pi\,t\,|y|}. $$