共轭泊松核相关函数$Q_t^{(j)}(x)$的傅里叶变换计算及逆变换可行性咨询

Hey there, let's unpack this problem together—computing the Fourier transform of $Q_t^{(j)}(x)$ is totally manageable once we leverage some key Fourier transform properties and known results about the Poisson kernel!

First, let's recall the definition you gave:
$$Q_t^{(j)}(x) = c_n \cdot \frac{x_j}{(t^2 + \vert x \vert²⁾{\frac{n+1}{2}}}, \quad c_n = \frac{\Gamma\left(\frac{n+1}{2}\right)}{\pi^{\frac{n+1}{2}}}$$

Notice this is closely related to the Poisson kernel, defined as:
$$P_t(x) = c_n \cdot \frac{t}{(t^2 + \vert x \vert²⁾{\frac{n+1}{2}}}$$
A standard harmonic analysis result tells us the Fourier transform of the Poisson kernel is straightforward:
$$\widehat{P_t}(\xi) = e^{-2\pi t \vert \xi \vert}$$

We can rewrite $Q_t^{(j)}(x)$ in terms of $P_t(x)$ to simplify things:
$$Q_t^{(j)}(x) = \frac{x_j}{t} P_t(x)$$

Next, use the critical Fourier transform property for multiplication by $x_j$: for any function $h(x)$,
$$\widehat{x_j h(x)}(\xi) = i \frac{\partial}{\partial \xi_j} \widehat{h}(\xi)$$

Applying this to our case (where $h(x) = P_t(x)$), we get:
$$\widehat{Q_t^{(j)}}(\xi) = \frac{1}{t} \cdot i \frac{\partial}{\partial \xi_j} \widehat{P_t}(\xi)$$

Now compute the partial derivative of $\widehat{P_t}(\xi) = e^{-2\pi t \vert \xi \vert}$ with respect to $\xi_j$. Remember $\vert \xi \vert = \sqrt{\xi_1^2 + \dots + \xi_n^2}$, so $\frac{\partial \vert \xi \vert}{\partial \xi_j} = \frac{\xi_j}{\vert \xi \vert}$. Using the chain rule:
$$\frac{\partial}{\partial \xi_j} e^{-2\pi t \vert \xi \vert} = e^{-2\pi t \vert \xi \vert} \cdot (-2\pi t) \cdot \frac{\xi_j}{\vert \xi \vert}$$

Plugging this back in, the $t$ terms cancel out, leaving us with the final result:
$$\widehat{Q_t^{(j)}}(\xi) = -2\pi i \cdot \frac{\xi_j}{\vert \xi \vert} e^{-2\pi t \vert \xi \vert}$$

On the inverse Fourier transform question #

If you already knew $\widehat{Q_t^{(j)}}(\xi)$, would computing the inverse transform to get back $Q_t^{(j)}(x)$ be easier? Honestly, probably not. The inverse transform would require evaluating an $n$-dimensional integral:
$$Q_t^{(j)}(x) = (2\pi)^{-n} \int_{\mathbb{R}^n} -2\pi i \cdot \frac{\xi_j}{\vert \xi \vert} e^{-2\pi t \vert \xi \vert} e^{2\pi i x \cdot \xi} d\xi$$
While you could compute this using polar coordinates (taking advantage of radial symmetry), it's more involved than the forward transform approach we used above—since we could piggyback on the known Poisson kernel Fourier transform and simple differential properties. The forward direction is definitely the smoother path here.

备注：内容来源于stack exchange，提问作者user1147800