Fourier 变换例解

本文旨在摘录一些需要技巧的 Fourier 变换计算.

Bessel 位势

问题. 已知 $\hat{B}=\frac1{1+|y|^2}$，其中 $y\in \mathbb{R}^n$. 求 $B=\left(\frac1{1+|y|^2}\right)^\vee$.
解. 利用如下积分

$$\frac1{1+|y|^2} = \int_0^\infty e^{-t(1+|y|^2)} dt,$$

则

$$B=\frac1{(2\pi)^{n/2}} \int_0^\infty e^{-t} \left(\int_{\mathbb{R}^n}e^{ix\cdot y-t|y|^2}dy\right) dt.$$

接下来我们把这个积分按一个坐标一个坐标拆开来计算. 为此，我们令 $z=\sqrt{b} x-\frac{a}{2\sqrt{b}} i$，计算

$$\frac{e^{-a^2/4b}}{\sqrt{b}}\int_\Gamma e^{-z^2} dz =\int_{-\infty}^{+\infty} e^{iax-bx^2} dx,$$

这里 $\Gamma$ 是围道 ${\operatorname{Im}(z)=-\frac{a}{2b^{1/2}} }$，积分方向自左向右. $e^{-z^2}$ 是全纯函数，所以我们可以形变到横轴，得到

$$\int_{-\infty}^{+\infty} e^{iax-bx^2} dx= e^{-a^2/4b}\left(\frac{\pi}{b}\right)^{1/2}.$$

于是

$$\int_{\mathbb{R}^n}e^{ix\cdot y-t|y|^2} dy= \prod_{j=1}^n \int_{-\infty}^{+\infty} e^{i x_j y_j-t y_j^2} d y=\left(\frac{\pi}t\right)^{n/2}e^{-\frac{|x|^2}{4t}}.$$

故

$$B(x) = \frac1{2^{n/2}}\int_0^{+\infty} \frac{e^{-t-\frac{|x|^2}{4t}}{t^{n/2}}}dt.$$

这个 $B$ 称为 Bessel 位势. 摘自 [1].

至于这个积分为什么可以换序，我差点去 Stack Exchange 上问了，好在最后好像想出来了. 以下是我差点发出去的提问：
How can I justify the interchange of integrals in the derivation of the Bessel potential in Evans’ Partial Differential Equations?

In his Partial Differential Equations, p187, Evans adopts the following method to calculate the inverse Fourier transform of $\frac1{1+|y|^2}$, where $y\in \mathbb{R}^n$: he uses $\frac1{1+|y|^2}=\int_0^\infty e^{-t(1+|y|^2)}\,dt$, and claims that

$$\left(\frac1{1+|y|^2}\right)^\vee=\frac1{(2\pi)^{n/2}} \int_0^\infty e^{-t} \left(\int_{\mathbb{R}^n}e^{ix\cdot y-t|y|^2}\,dy\right) dt,\quad (*)$$

thus shifts attention to the evaluation of $\int_{\mathbb{R}^n}e^{ix\cdot y-t|y|^2}\,dy$.
However I can’t see why (*) is correct.

My attempt: I try to simplify the notation to $\int \left(\int_0^\infty f(t,y) \,dt\right) e^{ixy} dy$, and let $I_b^a(y) =\int_b^a f(t,y) \,dt$. Then we have to show

$$\int\lim_{a\to +\infty} I_0^a(y)e^{ixy} \,dy=\int_0^{+\infty}\int f(t,y)e^{ixy} \,dydt.$$

Maybe we should first admit that

$$\int I_0^b(y)e^{ixy} \,dy=\int_0^{b}\int f(t,y)e^{ixy} \,dydt$$

and then prove that

$$\int\lim_{a\to +\infty} I_0^a(y)e^{ixy}\,dy=\lim_{a\to +\infty} \int I_0^a(y)e^{ixy} \,dy$$

which is the same as

$$\left|\int I_b^\infty e^{ixy} dy\right| \text{ can be arbitrarily small as long as } b \text{ is sufficiently large,}$$

which in turn is the same as

$$\int\frac{e^{-b(1+|y|^2)+ixy}}{1+|y|^2}\,dy \to 0.$$

So I think it remains to show
$
\int\frac{e^{-b(1+|y|^2)}}{1+|y|^2}\,dy\to 0
$.
So I rewrite the integral as

$$\int \frac{e^{-b(1+|y|^2)}}{1+|y|^2} \,dy=\int_0^\infty \int_{|y|=R} \frac{e^{-b(1+R^2)}}{1+|R|^2} dydR=\int_0^\infty \frac{e^{-b(1+R^2)}\omega_n R^{n-1}}{1+R^2} dR.$$

Then we can use Gamma function to prove that it is so.

$e^{-|x|^2}$ 的 Fourier 变换

从上面的计算（配方法）中我们看出 $e^{-t|\lambda|^2}$ 的 Fourier 反变换为 $\frac1{(2\pi)^{n/2}}\left(\frac{\pi}t\right)^{n/2}e^{-\frac{|x|^2}{4t}}=\left(\frac1{2t}\right)^{n/2}e^{-\frac{|x|^2}{4t}}$. 令 $t=\frac1{4A}$，则看出 $e^{-A|x|^2}$ 的 Fourier 变换为 $(2A)^{-n/2} e^{-\frac{|\lambda|^2}{4A}}$. 这一结果需要牢牢记住.

另一种方法见 [2]，我们用分部积分法，设 $f(x)=e^{-x^2}$，

$$\hat{f}(\lambda)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2}e^{-ix\lambda}\,dx=\frac1{\sqrt{2\pi}}\left(\frac1{-i\lambda}\left.e^{-x^2}e^{-ix\lambda}\right|_{-\infty}^{+\infty}-\frac{-2}{-i\lambda}\int_{-\infty}^{+\infty} xe^{-x^2}e^{-ix\lambda}\,dx\right),$$

所以实际上 $\hat{f}(\lambda)=\frac{2i}{\lambda}(xf(x))^{\vee}=-\frac2\lambda\frac{d}{d\lambda}\hat{f}(\lambda)$.
又 $\hat{f}(0)=\frac1{\sqrt{2}}$. 我们解 ODE 可得

$$\hat{f}(\lambda)=\frac1{\sqrt{2}}e^{-\lambda^2/4}.$$

也许一种更对称的写法是 $e^{-x^2/2}\mapsto e^{-\lambda^2/2}$.

注意正负号！

参考文献

1. Evans. Partial Differential Equations.
2. 周蜀林. 偏微分方程.