The Fourier transform of sinx/x and (sinx/x)^2 and more

In this post

We are going to evaluate the Fourier transform of \(\frac{\sin{x}}{x}\) and \(\left(\frac{\sin{x}}{x}\right)^2\). And it turns out to be a comprehensive application of many elementary theorems in real and complex analysis. It is a good thing to make sure that you can compute and understand all the identities in this post by yourself in the end. Also, you are expected to be able to recall what all words in italics mean.

To be clear, by the Fourier transform of \(f\) we actually mean

\[ \hat{f}(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-itx}dx. \]

This is a matter of convenience. Indeed, the coefficient \(\frac{1}{\sqrt{2\pi}}\) is superfluous, but without this coefficient, when computing the Fourier inverse, one has to write \(\frac{1}{2\pi}\) on the other side. Instead of making the transform-inverse unbalanced, we write \(\frac{1}{\sqrt{2\pi}}\) all the time and pretend it is not here.

We say a function \(f\) is in \(L^1\) if \(\int_{-\infty}^{+\infty}|f(x)|dx<+\infty\). As a classic exercise in elementary calculus, one can show that \(\frac{\sin{x}}{x} \not\in L^1\) but \(\left(\frac{\sin{x}}{x}\right)^2 \in L^1\).

Problem 1

For real \(t\), find the following limit:

\[ \lim_{A \to \infty}\int_{-A}^{A}\frac{\sin{x}}{x}e^{itx}dx. \]

Since \(\frac{\sin{x}}{x}e^{itx}\not\in L^1\), we cannot evaluate the integral over \(\mathbb{R}\) in the ordinary sense (the reader can safely ignore this if he or she has no background in Lebesgue integration at this moment, but do keep in mind that being in \(L^1\) is a big matter). However, for given \(A>0\), the integral over \([-A,A]\) is defined, and we evaluate this limit as \(A \to \infty\) to get what we want (by abuse of language). The reader is highly encouraged to write down calculation and supply pictures that should've been here.

We will do this using contour integration. Since the complex function \(f(z)=\frac{\sin{z}}{z}e^{itz}\) is entire, by Cauchy's theorem, its integral over \([-A,A]\) is equal to the one over the path \(\Gamma_A\) by going from \(-A\) to \(-1\) along the real axis, from \(-1\) to \(1\) along the lower half of the unit circle, and from \(1\) to \(A\) along the real axis (why?). Since the path \(\Gamma_A\) avoids the origin, we are safe to use the identity

\[ 2i\sin{z}=e^{iz}-e^{-iz}. \]

Replacing \(\sin{z}\) with \(\frac{1}{2i}(e^{itz}-e^{-itz})\), we get

\[ I_A(t)=\int_{\Gamma_A}f(z)dz=\int_{\Gamma_A}\frac{1}{2iz}(e^{i(t+1)z}-e^{i(t-1)z})dz. \]

If we put \(\varphi_A(t)=\int_{\Gamma_A}\frac{1}{2iz}e^{itz}dz\), we see \(I_A(t)=\varphi_A(t+1)-\varphi_A(t-1)\). It is convenient to divide \(\varphi_A\) by \(\pi\) since we therefore get

\[ \frac{1}{\pi}\varphi_A(t)=\frac{1}{2\pi i}\int_{\Gamma_A}\frac{e^{itz}}{z}dz \]

and we are cool with the divisor \(2\pi i\).

Now, close the path \(\Gamma_A\) in two ways. First, by the semicircle from \(A\) to \(-Ai\) to \(-A\); second, by the semicircle from \(A\) to \(Ai\) to \(-A\), which finishes a circle with radius \(A\). For simplicity we denote the two paths by \(\Gamma_U\) and \(\Gamma_L\). Again by the Cauchy theorem, the first case gives us an integral with value \(0\), thus by Cauchy's theorem,

\[ \frac{1}{\pi}\varphi_A(t)=\frac{1}{2\pi i}\int_{-\pi}^{0}\frac{\exp{(itAe^{i\theta})}}{Ae^{i\theta}}dAe^{i\theta}=\frac{1}{2\pi}\int_{-\pi}^{0}\exp{(itAe^{i\theta})}d\theta. \]

Notice that

\[ \begin{aligned} |\exp(itAe^{i\theta})|&=|\exp(itA(\cos\theta+i\sin\theta))| \\ &=|\exp(itA\cos\theta)|\cdot|\exp(-At\sin\theta)| \\ &=\exp(-At\sin\theta) \end{aligned} \]

hence if \(t\sin\theta>0\), we have \(|\exp(iAte^{i\theta})| \to 0\) as \(A \to \infty\). When \(-\pi < \theta <0\) however, we have \(\sin\theta<0\). Therefore we get

\[ \frac{1}{\pi}\varphi_{A}(t)=\frac{1}{2\pi}\int_{-\pi}^{0}\exp(itAe^{i\theta})d\theta \to 0\quad (A \to \infty,t<0). \]

(You should be able to prove the convergence above.) Also trivially

\[ \varphi_A(0)=\frac{1}{2}\int_{-\pi}^{0}1d\theta=\frac{\pi}{2}. \]

But what if \(t>0\)? Indeed, it would be difficult to obtain the limit using the integral over \([-\pi,0]\). But we have another path, namely the upper one.

Note that \(\frac{e^{itz}}{z}\) is a meromorphic function in \(\mathbb{C}\) with a pole at \(0\). For such a function we have

\[ \frac{e^{itz}}{z}=\frac{1}{z}\left(1+itz+\frac{(itz)^2}{2!}+\cdots\right)=\frac{1}{z}+it+\frac{(it)^2z}{2!}+\cdots. \]

which implies that the residue at \(0\) is \(1\). By the residue theorem,

\[ \begin{aligned} \frac{1}{2\pi{i}}\int_{\Gamma_L}\frac{e^{itz}}{z}dz&=\frac{1}{2\pi{i}}\int_{\Gamma_A}\frac{e^{itz}}{z}dz+\frac{1}{2\pi}\int_{0}^{\pi}\exp(itAe^{i\theta})d\theta \\ &=1\cdot\operatorname{Ind}_{\Gamma_L}(0)=1. \end{aligned} \]

Note that we have used the change-of-variable formula as we did for the upper one. \(\operatorname{Ind}_{\Gamma_L}(0)\) denotes the winding number of \(\Gamma_L\) around \(0\), which is \(1\) of course. The identity above implies

\[ \frac{1}{\pi}\varphi_A(t)=1-\frac{1}{2\pi}\int_{0}^{\pi}\exp{(itAe^{i\theta})}d\theta. \]

Therefore, when \(t>0\), since \(\sin\theta>0\) when \(0<\theta<\pi\), we get

\[ \frac{1}{\pi}\varphi_A(t)\to 1 \quad(A \to \infty,t>0). \]

But as is already shown, \(I_A(t)=\varphi_A(t+1)-\varphi_A(t-1)\). To conclude,

\[ \lim_{A\to\infty}I_A(t)= \begin{cases} \pi\quad &|t|<1, \\ 0 \quad &|t|>1 ,\\ \frac{1}{2\pi} \quad &|t|=1. \end{cases} \]

What we can learn from this integral

Since \(\psi(x)=\left(\frac{\sin{x}}{x}\right)\) is even, dividing \(I_A\) by \(\sqrt{\frac{1}{2\pi}}\), we actually obtain the Fourier transform of \(\psi\) by abuse of language. We also get

\[ \hat\psi(t)= \begin{cases} \sqrt{\frac{\pi}{2}}\quad & |t|<1, \\ 0 \quad & |t|>1, \\ \frac{1}{2\pi\sqrt{2\pi}} & |t|=1. \end{cases} \]

Note that \(\hat\psi(t)\) is not continuous, let alone being uniformly continuous. Therefore, \(\psi(x) \notin L^1\). The reason is, if \(f \in L^1\), then \(\hat{f}\) is uniformly continuous (proof). Another interesting fact is, this also gives us the value of the Dirichlet integral since we have

\[ \begin{aligned} \int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)dx&=\int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)e^{0\cdot ix}dx \\ &=\sqrt{2\pi}\hat\psi(0) \\ &=\pi. \end{aligned} \]

We end this section by evaluating the inverse of \(\hat\psi(t)\). The calculation is not that difficult. Now you can see why we put \(\sqrt\frac{1}{2\pi}\).

\[ \begin{aligned} \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty}\hat\psi(t)e^{itx}dt &= \sqrt{\frac{1}{2\pi}}\int_{-1}^{1}\sqrt{\frac{\pi}{2}}e^{itx}dt \\ &=\frac{1}{2}\cdot\frac{1}{ix}(e^{ix}-e^{-ix}) \\ &=\frac{\sin{x}}{x}. \end{aligned} \]

Problem 2

For real \(t\), compute

\[ J=\int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)^2e^{itx}dx. \]

Now since \(h(x)=\frac{\sin^2{x}}{x^2} \in L^1\), we are able to say with ease that the integral above is the Fourier transform of \(h(x)\) (multiplied by \(\sqrt{2\pi}\)). But still we will be using the limit form

\[ J(t)=\lim_{A \to \infty}J_A(t) \]


\[ J_A(t)=\int_{-A}^{A}\left(\frac{\sin{x}}{x}\right)^2e^{itx}dx. \]

And we are still using the contour integration as above (keep \(\Gamma_A\), \(\Gamma_U\) and \(\Gamma_L\) in mind!). For this we get

\[ \left(\frac{\sin z}{z}\right)^2e^{itz}=\frac{e^{i(t+2)z}+e^{i(t-2)z}-2e^{itz}}{-4z^2}. \]

Therefore it suffices to discuss the function

\[ \mu_A(z)=\int_{\Gamma_A}\frac{e^{itz}}{2z^2}dz \]

since we have

\[ J_A(t)=\mu_A(t)-\frac{1}{2}(\mu_A(t+2)-\mu_A(t-2)). \]

Dividing \(\mu_A(z)\) by \(\frac{1}{\pi i}\), we see

\[ \frac{1}{\pi i}\mu_A(t)=\frac{1}{2\pi i}\int_{\Gamma_A}\frac{e^{itz}}{z^2}dz. \]

An integration of \(\frac{e^{itz}}{z^2}\) over \(\Gamma_L\) gives

\[ \begin{aligned} \frac{1}{\pi i}\mu_A(z)&=\frac{1}{2\pi i}\int_{-\pi}^{0}\frac{\exp(itAe^{i\theta})}{A^2e^{2i\theta}}dAe^{i\theta} \\ &=\frac{1}{2\pi}\int_{-\pi}^{0}\frac{\exp(itAe^{i\theta})}{Ae^{i\theta}}d\theta. \end{aligned} \]

Since we still have

\[ \left|\frac{\exp(itAe^{i\theta})}{Ae^{i\theta}}\right|=\frac{1}{A}\exp(-At\sin\theta), \]

if \(t<0\) in this case, \(\frac{1}{\pi i}\mu_A(z) \to 0\) as \(A \to \infty\). For \(t>0\), integrating along \(\Gamma_U\), we have

\[ \frac{1}{\pi i}\mu_A(t)=it-\frac{1}{2\pi}\int_{0}^{\pi}\frac{\exp(itAe^{i\theta})}{Ae^{i\theta}}d\theta \to it \quad (A \to \infty) \]

We can also evaluate \(\mu_A(0)\) by computing the integral but we are not doing that. To conclude,

\[ \lim_{A \to\infty}\mu_A(t)=\begin{cases} 0 \quad &t>0, \\ -\pi t \quad &t<0. \end{cases} \]

Therefore for \(J_A\) we have

\[ J(t)=\lim_{A \to\infty}J_A(t)=\begin{cases} 0 \quad &|t| \geq 2, \\ \pi(1+\frac{t}{2}) \quad &-2<t \leq 0, \\ \pi(1-\frac{t}{2}) \quad & 0<t <2. \end{cases} \]

Now you may ask, how to find the value of \(J(t)\) at \(0\), \(2\) or \(-2\)? \(\mu_A(0)\) is not even evaluated. But \(h(t) \in L^1\), \(\hat{h}(t)=\sqrt{\frac{1}{2\pi}}J(t)\) is uniformly continuous (!), thus continuous, and the values at these points follows from continuity.

What we can learn from this integral

Again, we get the value of a classic improper integral by

\[ \int_{-\infty}^{\infty}\left(\frac{\sin{x}}{x}\right)^2dx = J(0)=\pi. \]

And this time it's not hard to find the Fourier inverse:

\[ \begin{aligned} \sqrt{\frac{1}{2\pi}}\int_{-\infty}^{\infty}\hat{h}(t)e^{itx}dt&=\frac{1}{2\pi}\int_{-\infty}^{\infty}J(t)e^{itx}dt \\ &=\frac{1}{2\pi}\int_{-2}^{2}\pi(1-\frac{1}{2}|t|)e^{itx}dt \\ &=\frac{e^{2ix}+e^{-2ix}-2}{-4x^2} \\ &=\frac{(e^{ix}-e^{-ix})^2}{-4x^2} \\ &=\left(\frac{\sin{x}}{x}\right)^2. \end{aligned} \]


问题的引入: 最大值、最小值的局限性

我们先考虑一个简单的函数: \[ f(x)=x\quad x\in(0,1) \]

那么问题来了: 这个函数有没有最小值或者最大值? 答案是没有. 定义域中的严格不等式给我们制造了无法得到严格解的麻烦. 可能读者会想到, 这是个单调函数, 可以使用极限解决这个问题. 但是这样做局限性太大了. 如果我们连函数的表达式都不知道, 又要到哪个极限找“最值”呢? 我们要做的是, 找到一个数, 它描述集合的范围时, 不依赖于最大值最小值的存在, 而且它不依赖于极限.

理论基础: 实数是有序的

我们一直以来接触各种不等式, 最基本的理论依据就是, 实数的顺序. 这个顺序的意思就是大小关系. 没有顺序关系的数也是有的, 例如复数. 实数的顺序其实可以叙述成以下两点:

  1. 如果\(x\in\mathbb{R}\)而且\(y\in\mathbb{R}\), 那么下面三个关系有且仅有一个成立 \[ x<y\quad x=y\quad y<x \]
  2. 如果\(x,y,z\in\mathbb{R}\), 而且有\(x<y\)\(y<z\), 那么\(x<z\)

至于“\(\leq\)”,应该理解成小于或等于, “或”的意思在这里是并没有强调哪个一定成立.


我们接下来只讨论上界. 也就是说, 我们只讨论小于的情况. 至于另一侧, 将小于号换成大于号即可.


取集合\(E\subset\mathbb{R}\), 如果存在\(\beta\in\mathbb{R}\)使得对任意\(x\in{E}\)满足不等式\(x\leq\beta\), 那么就称\(E\)有上界, \(\beta\)就是\(E\)的一个上界.

这其实就是一个存在命题. 只要满足就够了. 满足之后呢? 没有然后了. 剩下的细节在这里不重要. 这和极限很类似. 一个很简单的例子: \((-\infty,1]\)就是有上界的. 而且任意的上界\(\beta\)都满足\(\beta\in[1,+\infty)\).

上确界: 最小的上界

有的时候, 我们需要知道一个严格最小的上界, 这就是上确界. 首先, 我们给出上确界的定义.

对于有上界的集合\(E\), 假设存在一个\(\alpha\in\mathbb{R}\)满足以下条件, 那么就称它为\(E\)的上确界, 记作\(\alpha=\sup{E}\):

  • \(\alpha\)\(E\)的一个上界

  • 如果\(\gamma<\alpha\), 那么\(\gamma\)不是\(E\)的上界

下确界可以通过相反的不等式进行确定, 记作\(\alpha=\inf{E}\). 当然\(\sup\)\(\inf\)后面也不一定要跟一个集合, 也可以跟一个函数或者数列. 回到开头的那个函数, 我们能很容易得到 \[ \sup{f(x)}=1\quad\inf{f(x)}=0 \]

这时我们不再依赖于极限, 也可以发现, 如果函数的最大值或最小值存在, 那么它一定等于上确界或下确界, 但是上下确界存在时, 最值不一定存在. 在函数最值不存在时, 我们仍然能利用确界对函数的范围进行严格的分析, 而且不一定需要求极限, 这体现了上下确界的价值所在. 在这里建议读者写几个函数, 再试着求以下上下确界.



\(\{a_n\}\)\(\mathbb{R}\)里的一个数列, 并且定义 \[ b_k=\sup\{a_k,a_{k+1},a_{k+2},\cdots\} \]

再定义 \[ \beta=\inf\{b_1,b_2,b_3,\cdots\} \]

那么就有 \[ \beta=\limsup_{n\to\infty}a_n \]

这就完成了上极限的定义. 而下极限的求法就是将上面的上确界、下确界交换顺序即可. 可是这里为什么出现了极限符号? 这是因为 \[ b_1\geq b_2\geq b_3\geq\cdots\geq\beta \]

所以有 \[ \lim\limits_{k\to\infty}b_k=\beta \]

此外,\(\{a_n\}\)中有一个子列\(\{a_{n_i}\}\)收敛于\(\beta\), 而且\(\beta\)是具有这个性质最大的数.

这其实还是解决了一件事情, 如果一个数列发散, 我们还是有可能利用上下确界的极限形式来研究极限的一些性质. 一个很简单的例子比如 \[ \limsup_{n\to\infty}{(-1)^n}=1\quad\liminf_{n\to\infty}{(-1)^n}=-1 \]


函数中上下极限的求法其实和数列形式是非常类似的. 我们先举一个上极限的例子, 理解构建过程时也可以利用这个例子: \[ \limsup_{x\to{0}}\sin\frac{1}{x}=1 \]

对于函数, 值域可能是连续集合. 求\(x\to{a}\)时的上极限, 我们先求出\(x\to{a}\)时, \(f(x)\)的上确界构成的集合 \[ U=\{y|y=\sup\{f(x)|x\in(a-\varepsilon,a+\varepsilon)\}\} \]

那么上极限就是 \[ \limsup_{x\to{a}}f(x)=\inf{U} \]


我们单纯利用不等式, 建立起一个非常严格,也非常有应用价值的概念: 确界. 这个概念源于最值, 又高于最值. 当然确界和最值什么时候相等又是一个拓扑学问题. 这又是另外一个话题了.



洛必达法则我想甚至不少高中生甚至初中生都听说过,知道怎么进行简单的应用。简单点说,处理\(\frac{0}{0}\)的函数时,对上下进行求导,可能会很大程度上简化计算。但是洛必达法则为什么能奏效? 能不能用严格的数理语言进行论证? 这是这篇文章需要解决的.


假设有定义在\((a,b)\)可导的实函数\(f\)\(g\),且\(g’(x)\neq0\)对所有\(x\in(a,b)\)恒成立,其中\(a\)\(b\)满足 \[ -\infty\leq{a}<{b}\leq+\infty.\]
若有\[\lim_{x\to a}\frac{f’(x)}{g’(x)}=A,\]且如果\[\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0,\]\[\lim_{x\to a}g(x)=+\infty,\]那么\[\lim_{x\to a}\frac{f(x)}{g(x)}=A\]类似的结论对\(x\to{b}\)或者\(g(x)\to-\infty\)也成立。



洛必达法则首次出现于1696年洛必达的 Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes 一书中。这本书当然以”洛必达法则”闻名于世。证明是这样完成的: \[ \frac{f(a+dx)}{g(a+dx)}=\frac{f(a)+f’(a)dx}{g(a)+g’(a)dx}=\frac{f’(a)dx}{g’(a)dx}=\frac{f’(a)}{g’(a)} \]

这个证明很好理解,线性近似展开,再考虑到\(f(a)=g(a)=0\)就得到结果。但是这个做法肯定是不合适的,\(dx\)在这里非常模糊,也不方便表达\(x\to\infty\)的情况。关于历史内容可以参见 The Historical Development of the Calculus 一书。



对函数导数有 \[ f’(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}, \]

我们可以写成 \[ f’(x) = \frac{f(x+h)-f(x)}{h} + r(h) \]

其中\(\lim\limits_{h\to0}r(h)=0\),且\(r(h)\)为连续函数。进行代数变形(这里\(r(h)\)的正负进行了调整),我们的得到线性近似 \[ f(x+h)=f(x)+f’(x)h+r(h)h \]

同样可以写出\(g(x)\)的线性近似 \[ g(x+h)=g(x)+g’(x)h+s(h)h \]

那么就能得到 \[ \frac{f(a+h)}{g(a+h)}=\frac{f(a)+f’(a)h+r(h)h}{g(a)+g’(a)h+s(h)h}=\frac{f’(a)h+r(h)h}{g’(a)h+s(h)h}=\frac{f’(a)+r(h)}{g’(a)+s(h)} \]



这个证明中,我们会利用柯西中值定理(GMVT)对所有的情况进行完整的证明,这期间涉及到一些不等式运算技巧。证明来自W. Rudin的 Principles Of Mathematical Analysis,我会在其中加上一些额外的解释。

情况1: \(-\infty\leq{A}<+\infty\)

选取实数\(\varepsilon>0\)\(q\)使得\(A<A+\varepsilon<q\)。因为\(\frac{f(x)}{g(x)}\to{A}\),必定有实数\(\delta\in(0,b-a)\)使得对于所有\(a<x<a+\delta\),始终有\(-\varepsilon<\frac{f’(x)}{g’(x)}-A<\varepsilon\)。也就是说 \[\frac{f’(x)}{g’(x)}<A+\varepsilon.\]

\(a<x<y<c\),由GMVT可知,存在\(t\in(x,y)\)使得不等式(A)成立: \[ \frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f’(t)}{g’(t)}<A+\varepsilon \] 最后一个不等式成立是因为\(t\in(x,y)\subset(a,b)\),而\((a,b)\)中这个不等式成立。

情况1.1: \(g(x)\to0\)


也就是说,对任意实数\(\varepsilon>0\),有\(\delta>0\),使得\(a<y<a+\delta\)时,满足不等式(B): \[ \frac{f(y)}{g(y)}\leq\varepsilon+A<q \]


情况1.2: \(g(x)\to+\infty\)

\(r=A+\varepsilon\)。固定不等式(A)中的\(y\),因为\(g(x)\to+\infty\),能找到一个值\(c\in(a,b)\)使得\(g(x)>g(y)\)\(g(x)>0\)对所有\(x\in(a,c)\)同时成立。那么不等式(A)两边同时乘以\([g(x)-g(y)]/g(x)\),能得到不等式(C) \[ \frac{f(x)}{g(x)}<r-r\frac{g(y)}{g(x)}+\frac{f(y)}{g(x)}\quad(a<x<c) \]

\(x\to{a}\),因为\(g(x)\to+\infty\),有点\(c_1\in(a,c)\)使得不等式(D)成立: \[ \frac{f(x)}{g(x)}<q\quad(a<x<c_1) \]



这里要注意,不等式(B)和(D)都只在\(q>A\)时成立,也就是说,如果\(q=A\),那么有\(\frac{f(x)}{g(x)}\geq{q}=A\)。也就是说,对于所有\(q>A\),都存在\(c\in(a,b)\),使得对于所有\(x\in(a,c)\),满足 \(A\leq\frac{f(x)}{g(x)}<q\),若令\(q\to{A}\),就能得到\(\frac{f(x)}{g(x)}\to{A}\)

情况2: \(-\infty<{A}\leq+\infty\)





假设它无意义。如果有\(g(x)=g(y)\),那么有\({x}<t<y\)使得\(g’(t)=0\),此时不满足 \[ f’(t)/g’(t)<A+\varepsilon \] > 不等式(B)中为什么变成小于等于?