Gaussian function 的Fourier变换

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Wikipedia

The Fourier transform of a normal density $f$ with mean $\mu$ and standard deviation $\sigma$ is\[\hat f(t) = \int_{-\infty}^\infty f(x)e^{-itx} \, dx = e^{-i\mu t} e^{- \frac12 (\sigma t)^2}\]If the mean $\mu=0$, the first factor is 1, and the Fourier transform is, apart from a constant factor, a normal density on the frequency domain, with mean 0 and standard deviation $1/\sigma$. In particular, the standard normal distribution $\varphi$ is an eigenfunction of the Fourier transform.

同样的见The Fourier Transform & Its Applications, Bracewell第105页

Six transform pairs for reference are listed below. They are all well known, and the integrals are evaluated in Chapter 7; we content ourselves at this point with asserting that the following integrals may be verified.
$$
\begin{array}{lll}
\int_{-\infty}^{\infty} e^{-\pi x^2} e^{-i 2 \pi x s} d x=e^{-\pi s^2} & \text { and } & \int_{-\infty}^{\infty} e^{-\pi s^2} e^{+i 2 \pi s x} d s=e^{-\pi x^2} \\
\int_{-\infty}^{\infty} \operatorname{sinc} x e^{-i 2 \pi x s} d x=\Pi(s) & \text { and } & \int_{-\infty}^{\infty} \Pi(s) e^{+i 2 \pi s x} d s=\operatorname{sinc} x \\
\int_{-\infty}^{\infty} \operatorname{sinc}^2 x e^{-i 2 \pi x s} d x=\Lambda(s) & \text { and } & \int_{-\infty}^{\infty} \Lambda(s) e^{+i 2 \pi s x} d s=\operatorname{sinc}^2 x
\end{array}
$$
Thus the transform of the Gaussian function is the same Gaussian function, the transform of the sinc function is the unit rectangle function, and the transform of the sinc$^2$ function is the triangle function of unit height and area.

推导见 积分变换2020-2021 video slide 第80页
求 $f(x) = e^{−a^2x^2}$的Fourier变换.
\begin{aligned} \widehat{f}(s) & =\int_{-\infty}^{\infty} e^{-\mathrm i s x-a^{2} x^{2}} \mathrm{~d} x \\ & =\int_{-\infty}^{\infty} e^{-s^{2} / 4 a^{2}} e^{-a^{2}\left(x+\mathrm i s / 2 a^{2}\right)^{2}} d x \quad \text {(配方)} \\ & =e^{-s^{2} / 4 a^{2}} \int_{-\infty}^{\infty} e^{-a^{2} u^{2}} \mathrm{~d} u \quad\left(\text {代换} x+i s / 2 a^{2}=u\right) \\ & =\frac{\sqrt{\pi}}{a} \mathrm{e}^{-s^{2} / 4 a^{2}}\end{aligned}为什么可以换元$x+\mathrm{i}s / 2 a^2=u$呢?
[在积分第24页说in Theorem 5.8 (Substitution), the function $f$ may be complex-valued, but the substitution $g(t)$ is assumed to be real-valued. 没有保证复数代换]

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考虑矩形围道$\Gamma_{R}$, 顶点为$\pm R,\pm R+\frac{is}{2a^2}$
由Cauchy定理,\begin{equation}\int_{\Gamma_{R}} \exp \left(-a^{2} z^{2}\right) \mathrm{d} z=0\end{equation}
在矩形的竖直边$z=\pm R+\mathrm{i} y,0\leqslant y\leqslant\frac{s}{2a^2}$
$$\left|\int_{0}^{s / 2 a^{2}} \exp \left(-a^{2}(\pm R+\mathrm{i} y)^{2}\right) \mathrm i\rmd y\right| \leqslant \mathrm{e}^{-a^{2} R^{2}} \int_{0}^{s / 2 a^{2}}\left|\exp \left(a^{2} y^{2}\right)\right| \rmd y \rightarrow 0 \quad \text { as } R \rightarrow \infty$$
所以, 当$R \rightarrow \infty$, (1)变为\begin{equation}\int_{-\infty}^{\infty} \exp \left(-a^{2} x^{2}\right) \mathrm{d} x-\int_{-\infty}^{\infty} \exp \left(-a^{2}\left(x+\frac{\mathrm{i} s}{2 a^{2}}\right)^{2}\right) \mathrm{d} x=0\end{equation}根据Gaussian integral结论$$\int_{-\infty}^{\infty} \exp \left(-a^{2} x^{2}\right) \mathrm{d} x=\frac{\sqrt{\pi}}{a}$$(2)变为$$\frac{\sqrt{\pi}}{a}=\int_{-\infty}^{\infty} \exp \left(-a^{2}\left(x+\frac{\mathrm{i} s}{2 a^{2}}\right)^{2}\right) \mathrm{d} x=\exp \left(\frac{s^{2}}{4 a^{2}}\right) \hat{f}(s)$$即$$\hat{f}(s)=\frac{\sqrt{\pi}}{a} \exp \left(\frac{-s^{2}}{4 a^{2}}\right)$$

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积分Example 6.11发现是相同的一道题

Example 6.11. For \(s∈ℝ\), \(\int_{-∞}^∞e^{-isx}e^{-x^2}\rmd x\)

Note the integrand is continuous so measurable. \(| e^{-isx}e^{-x^2}|=e^{-x^2}∈ℒ^1(ℝ)\) [See this by noting e\(^{-x^2}<e^{-| x |}\) for \(| x |>1\)]

Let \(g_n(x)=\frac{(-isx)^n}{n!}e^{-x^2}\)

\(\sum_{n=0}^∞g_n(x)=e^{-isx}e^{-x^2}\) [convergence by power series expansion]

\(\sum_{n=0}^∞| g_n(x) |=e^{| sx |-x^2}≤e^{s^2/2}e^{-x^2/2}\) [using \(| sx |≤\frac{x^2+s^2}{2}\)]

\(\therefore \sum | g_n |∈ℒ^1(ℝ)\)

By Theorem 6.8[term by term integration]

\[\int_{-∞}^∞e^{-isx}e^{-x^2}\rmd x=\sum_{n=0}^∞\int_0^∞\frac{(-isx)^n}{n!}e^{-x^2}\rmd x\]

As \(\int_{-∞}^∞e^{-x^2}\rmd x=\sqrtπ\) [See later]

and \(\int_{-∞}^∞\frac{(-isx)^n}{n!}e^{-x^2}=0\) for \(n\) odd, can carefully use parts & induction

\[\int_{-∞}^∞\frac{(-isx)^n}{n!}e^{-x^2}=\left\{\begin{array}{ll}\frac{(2 m) ! \sqrt{π}}{4^m m!}& n=2 m, \text{even}\\ 0 & n \text{ odd}\end{array}\right.\]Thus\[\int_{-∞}^∞e^{-isx}e^{-x^2}\rmd x=\sum_{m=0}^∞\frac{(-is)^{2 m}\sqrt{π}}{4^m m!}=\sqrt{π}e^{-s^2/4}\]