Fourier Transform Intro - Fourier Integral Theorem_fourierformer: transformer meets generalized fouri-优快云博客

本文深入探讨了Fourier积分定理的核心概念，包括定理的前提条件、结论及Fourier变换对连续函数的应用。通过引入Fourier变换对连续函数的表述，展示了定理在数学分析中的重要性。证明过程利用了Riemann-Lebesgue引理，验证了定理的有效性。

https://people.math.osu.edu/gerlach.1/math/BVtypset/node31.html

The Fourier Integral Theorem The mathematically more precise statement of this theorem is as follows:

Theorem 23.1 (Fourier's Integral Theorem)

Given:(i)

$f$ is function piecewise continuous on every bounded closed interval of the $x$ -axis.

(ii)

At each point $x$ the function $f$ has both a left derivative $f'_L(x)$ and a right derivative $f'_R(x)$ ,

(iii)

$f\in L^1(-\infty,\infty)$ , i.e. $f$ is absolutely integrable along the $x$ -axis:

$\displaystyle \int_{-\infty}^{\infty}\vert f(x)\vert dx<\infty$

Conclusion:

$\displaystyle \underbrace{\lim_{K\to\infty}\int_{-K}^Kdk\frac{e^{ikx}}{\sqrt{2\... ...2\pi}}f(t)dt}_{ \displaystyle\hat f(k)} =\frac{1}{2} \left[f(x^-)+f(x^+)\right]$

(239)

Comments:

This result can be restated as a Fourier transform pair,

$\displaystyle \hat f(k)=\int^\infty_{-\infty}\frac{e^{-ikt}}{\sqrt{2\pi}}f(t)dt$ (240)

$\displaystyle f(x)=\int^\infty_{-\infty}\frac{e^{ikx}}{\sqrt{2\pi}}\hat f(k)dk$ (241)

whenever $f$ is continuous.
By interchanging integration order and letting $K=1/\alpha$ one has

$\displaystyle f(x)=\lim_{\alpha\to 0} \int^\infty_{-\infty} \underbrace{ \int_{... ...pha}dk \frac{e^{ik(x-t)}}{2\pi}}_{ \displaystyle\delta_\alpha(x-t)} f(t)\, dt~.$ (242)

This equation holds for all continuous functions $f\in L^1(-\infty,\infty)$ . Thus $\delta_\alpha(x-t)$ is another delta convergent sequence:

$\displaystyle \delta(x-t)=\lim_{\alpha\to 0}\delta_\alpha(x-t)=\int^\infty_{-\infty} \frac{e^{ik(x-t)}}{2\pi}dk~.$ (243)

It is of course understood that one first do the integration over $t$ before taking to the indicated limit.
Either one of the two equations, Eq.(2.42) or (2.43), is a generalized completeness relation for the set of ``wave train'' functions,

$\displaystyle \left\{ \frac{e^{ikx}}{\sqrt{2\pi}}\colon -\infty <k<\infty\right\}~.$

However, these functions are not normalizable, i.e., they $\not\in L^2(-\infty , \infty )$ . Instead, as Eq.(2.43) implies, they are said to be `` $\delta$ -function normalized''.

Proof of the Fourier integral theorem:

The proof of the Fourier integral theorem presupposes that the Fourier amplitude $\hat f(k)$ is well-defined for each $k$ . That this is indeed the case follows from the finiteness of $\vert\hat f(k)\vert$ : ( following loss "=" )

$\displaystyle \vert\hat f(k)\vert\le \int^\infty_{-\infty}\vert\frac{e^{-ikt}}{... ...\vert dt \frac{1}{\sqrt{2\pi}}\int^\infty_{-\infty}\vert f(t)\vert dt<\infty~.$

The last inequality is an expression of the fact that $f\in L^1(-\infty,\infty)$ . Thus $\hat f(k)$ is well-defined indeed.

The proof of the Fourier integral theorem runs parallel to the Fourier series theorem on page $[*]$ . We shall show that

$\displaystyle \lim_{K\to\infty}S_K(x)=\frac{1}{2}\left[f(x^-)+f(x^+)\right]~,$

where

$\displaystyle S_K(x)$	$\displaystyle = \int^\infty_{-\infty}dt\, f(t)\int_{-K}^K\frac{e^{ik(x-t)}}{2\pi}dk$
	$\displaystyle =\int^\infty_{-\infty}dt\, f(t) \left. \frac{e^{ik(x-t)}}{2\pi i(x-t)}\right\vert^K_{-K}$
	$\displaystyle =\int^\infty_{-\infty}dt\, f(t) \frac{\sin K(x-t)}{\pi (x-t)}$
	$\displaystyle =\underbrace{\int^x_{-\infty} f(t) \frac{\sin K(x-t)}{\pi (x-t)}d... ...x)} +\underbrace{\int^\infty_{x} f(t) \frac{\sin K(x-t)}{\pi (x-t)}dt}_{I_K(x)}$

The evaluation of the integrals is done by shifting the integration variable. For the second integral one obtains

$\displaystyle I_K(x)$	$\displaystyle =\int^\infty_0 f(x+u) \frac{\sin Ku}{\pi u}du$
	$\displaystyle =\int^\infty_0 \frac{f(x+u)-f(x^+)}{\pi u}\sin Ku\,du + \frac{f(x^+)}{\pi}\int^\infty_0 \frac{\sin Ku}{u}du ~.$

Using the fact that

$\displaystyle \int^\infty_0 \frac{\sin Ku}{u}du=\frac{\pi}{2}~,$

and the fact that

$\displaystyle \frac{f(x+u)-f(x^+)}{\pi u}\equiv G(u)$

is piecewise continuous everywhere, including at $u=0$ , where

$\displaystyle G(0)\equiv \lim_{u\to 0^+}\frac{f(x+u)-f(x^+)}{\pi u}=f'_R(x)$

is the right hand derivative of $f$ at $x$ , one finds that

$\displaystyle \lim_{K\to\infty}I_K(x)$	$\displaystyle =\lim_{K\to\infty} \left\{ \int^\infty_0 G(u)\sin Ku \,du +\frac{f(x^+)}{\pi}\frac{\pi}{2}\right\}$
	$\displaystyle =0+\frac{1}{2}f(x^+)$