Prove that $\lim_{n\to\infty}\sin(nx)/ \pi x$ is a delta function

Question

I need to prove that $\delta_n(x)=\sin (nx)/\pi x$ is a delta function. That is, to prove that: $$\underset{n\rightarrow \infty}{\lim}\int^\infty_{-\infty}f(x)\frac{\sin( nx)}{\pi x}dx=f(0).$$ For that I made $y=x/n$ and took the limit under the integral since $f(x)$ is supposed to be analytic in $\mathbb{R}$. I have a fundamental limit for the $\sin$ leaving me with $$\int^\infty_{-\infty}f(0)\cdot1 \;dy.$$ But this is equal to $f(0)$ $\iff$ $f(x)$ have contribuition only at $x=0$, but here the only assumption is that $f(x)=0$ when $x\rightarrow \pm \infty$.

What is wrong here?

Answer 1

I thought that it might be instructive to present a way forward that uses a regularization of the Dirac Delta. To that end we proceed.

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PRELIMINARIES:

Let $\displaystyle \delta_n(x)=\frac1{2\pi}\int_{-n}^ne^{ikx}\,dk$. Then, we can write

$$\delta_n(x)=\begin{cases}\frac{\sin(nx)}{\pi x}&,x\ne0\\\\\frac n\pi&,x=0\tag1\end{cases}$$

The function $\delta_n(x)$ has the following properties:

1. For each $n$, $\delta_n(x)$ is an analytic function of $x$.
2. $\lim_{n\to \infty} \delta_n(0)= \infty$
3. $\left|\int_{-\infty}^x \delta_n(x')\,dx'\right|$ is uniformly bounded.
4. $\lim_{n\to \infty}\int_{-\infty}^{x}\delta_n(x')\,dx'=u(x)$, where $u$ is the unit step funciton.
5. For each $n>0$, $\int_{-\infty}^\infty \delta_n(x)\,dx=1$

While heuristically $\delta_n(x)$ "approximates" a Dirac Delta when $n$ is "large," the limit of $\delta_n(x)$ fails to exist. However, if we interpret this limit in the distributional sense, then $\lim_{n\to\infty}\delta_n(x)\sim\delta(x)$. We will now show that this is indeed that case.

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ANALYSIS:

Let $\phi(x)\in S$ where $S$ is the Schwarz Space of functions.

We will now evaluate the limit

$$\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\lim_{n\to \infty}\int_{-\infty}^\infty \frac{\sin(nx)}{\pi x}\phi(x)\,dx\tag1 \end{align}$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=\phi(x)$ and $v=\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'$ reveals

$$\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx&=-\lim_{n\to \infty}\int_{-\infty}^\infty \phi'(x)\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'\,dx\tag2 \end{align}$$

Using Property 3 in the Preliminaries section, there exists a number $C$ such that $\left|\phi'(x)\int_{-\infty}^{nx}\frac{\sin(x')}{x'}\,dx'\right|\le C\,|\phi'(x)|$. Inasmuch as $C|\phi'(x)|$ is integrable, the Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx&=-\lim_{n\to \infty}\int_{-\infty}^\infty \phi'(x)\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'\,dk\tag3\\\\ &=-\int_{-\infty}^\infty \phi'(x)\lim_{n\to \infty}\left(\int_{-\infty}^{nx}\frac{\sin(x')}{\pi x'}\,dx'\right)\,dx\\\\ &=- \int_{-\infty}^\infty \phi'(x)\underbrace{u(x)}_{\text{Unit Step}}\,dx\\\\ &=-\int_0^\infty \phi'(x)\,dx\\\\ &=\phi(0) \end{align}$$

Therefore, in the sense of distributions as given by $(3)$, we assert that $\lim_{n\to\infty}\delta_n(x)\sim \delta(x)$ as was to be shown!

Answer 2

Since $\delta_n(x)=n\delta_1(nx)$,$$\begin{align}\int_{\Bbb R}\delta_n(x)f(x)dx&=\int_{\Bbb R}\delta_1(y)f(y/n)dy\\\implies\lim_{n\to\infty}\int_{\Bbb R}\delta_n(x)f(x)dx&=\lim_{n\to\infty}\int_{\Bbb R}\delta_1(y)f(y/n)dy\\&\stackrel{*}{=}\int_{\Bbb R}\delta_1(y)\lim_{n\to\infty}f(y/n)dy\\&=f(0)\underbrace{\int_{\Bbb R}\delta_1(y)dy}_{1},\end{align}$$provided $f$ has the right behaviour for $\stackrel{*}{=}$ to be legal. For example, while $f$ being a Schwartz function suffices, $f$ being analytic doesn't.

Answer 3

Make your substitution, then break up your

$$\int_{-\infty}^\infty f\left(\frac{u}{n}\right) \frac{\sin u}{\pi u} d u$$

as $I_1 + I_2,$ where $I_1$ is the integral from $-\sqrt{n}$ to $\sqrt{n}$ and $I_2$ is the rest. The point is that $I_1$ converges to $$\int_{-\infty}^\infty f(0)\frac{\sin u}{\pi u} du = f(0),$$ and $I_2$ converges to $0.$

Answer 4

There are indeed many ways to understand the conclusion, and they are all worthwhile. But some may be less reassuring or persuasive than others... of course depending on taste and prior experience.

For me, at this moment, ... let $\langle,\rangle$ be the pairing of Schwartz functions and tempered distributions. By an extended form of Plancherel, $$ \langle f,u\rangle \;=\; \langle \widehat{f},\widehat{u}\rangle $$ for Schwartz $f$ and tempered $u$. If we get things normalized properly, for a suitable version $u_n(x)={\sin nx\over x}$, $\widehat{u}_n$ is the characteristic function of $[-n,n]$. This computation is most easily done by invoking Fourier inversion after computing the Fourier transform of that characteristic function. These converge as tempered distributions to $1$, which is the Fourier transform of $\delta$...