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How to prove $$\lim_{\alpha\rightarrow\infty} \frac{\sin^2\alpha x}{\alpha x^2} ~=~\pi\delta(x)~?$$

I have encountered this limit while learning time dependent perturbation and transition probability in Sakurai. How to show this limit? I tried to integrate around $x=0$ but didn't get anything useful? I am out of ideas. Any help is appreciated.

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    $\begingroup$ There's a clear presentation in the quantum mechanics textbook by R. Shankar. $\endgroup$ – Andrew Steane Sep 26 '19 at 8:59
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  1. More generally, for an arbitrary integrable function $\eta \in {\cal L}^1(\mathbb{R})$ with unit integral $ \int_{\mathbb{R}} \! \mathrm{d}y~\eta(y)=1$, one has the following representation $$ \delta(x)~=~\lim_{\varepsilon\to 0^+} \frac{1}{\varepsilon}\eta\left(\frac{x}{\varepsilon}\right) \tag{1} $$ of the Dirac delta distribution $\delta(x)$. In more detail, eq. (1) states that $$\begin{align} \int_{\mathbb{R}}\frac{\mathrm{d}x}{\varepsilon}f(x)~\eta\left(\frac{x}{\varepsilon}\right)~\stackrel{x=\varepsilon y}{=}~& \int_{\mathbb{R}}\mathrm{d}y~f(\varepsilon y) \eta(y)\cr ~\longrightarrow~&f(0)\int_{\mathbb{R}} \! \mathrm{d}y~\eta(y)~=~f(0)\quad\text{for}\quad \varepsilon\to 0^+\quad \end{align}\tag{1'}$$ for all continuous test functions $f\in C_c(\mathbb{R})$ with compact support. Note that each test function $f\in C_c(\mathbb{R})$ necessarily is bounded $|f|\leq C$. The proof of eq. (1') uses Lebesgue's Dominated Convergence Theorem with $C|\eta|$ as integrable majorant function.

  2. OP's formula corresponds to the case $$\eta(x)~=~\frac{\sin^2x}{\pi x^2}.\tag{2}$$

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  • $\begingroup$ Hello. Is there any reference you could provide, were the second statement is proved, using the first of course? Thank you. $\endgroup$ – Constantine Black Feb 28 '16 at 9:43
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    $\begingroup$ @ConstantineBlack The second is not a consequence of the first. The first says you can take any Lebesgue integrable function with unit integral and take scaled versions of it and the family of distributions associated with those scaled functions has a limit that is the dirac delta distribution. The second says that the particular function in question is Lebesgue integrable and has unit integral. So it says the particular function is an example of one of the many things you can use for the first statement. As would a Gaussian, or any Lebesgue integrable function with unit integral. $\endgroup$ – Timaeus Feb 29 '16 at 0:36
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You want to show that a limit of some functions is a particular distribution. Specifically:

$$\lim_{\alpha\rightarrow\infty} \frac{\sin^2\alpha x}{\alpha x^2} =\pi\delta(x)$$

So first you can ask what a distribution is. A distribution $F$ is a functional. And specifically it takes a test function $g$ and gives you a scalar. An example is if we start with a function $f:x\mapsto f(x)$ we can make the functional $F_f$ by the following map

$$F_f:g\mapsto F_f[g]=\int f(x)g(x) dx.$$ Or in abbreviated form $$F_f:g\mapsto \int f(x)g(x) dx.$$

So in a sense we can imagine functions as distributions since $f:x\mapsto f(x)$ and $F_f:g\mapsto \int f(x)g(x) dx$ are related to each other.

Great. So now we can talk about the limit of some distributions and ask what it means for one distribution to be the limit of some other distributions. The technical details depend on the norm we put on the space of distributions, but we do want that if $F=\lim F_f$ (one distribution is the limit of some other distributions) then $F[g]=\lim F_f[g]$ (the action on a test function, a scalar, is the limit of the actions on the test function, which are also scalars).

So now we can check to see if $\lim_{\alpha\rightarrow\infty} \frac{\sin^2\alpha x}{\alpha x^2} =\pi\delta(x).$

So pick a random test function. A test function is smooth, and it and all of its derivatives are bounded, and go to zero fast enough that integration by parts gives zero on the boundary. So let's let $g$ be an arbitrary test function. First note that $\int \pi\delta(x) g(x)dx=\pi g(0).$ So we want to show that $$\pi g(0)=\lim_{\alpha\rightarrow\infty} \int\frac{\sin^2\alpha x}{\alpha x^2}g(x)dx.$$

And now this is a regular calculus problem.

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    $\begingroup$ I think the OP's question is where the factor of $\pi$ comes from? This is nontrivial; I believe it requires complex analysis. $\endgroup$ – knzhou Feb 27 '16 at 21:07
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    $\begingroup$ Hello. Although I find your answer correct i think this is not just a regular calculus problem, at least concerning the tools I possess. For example, the Dirichlet kernel converges to a Dirac delta function. This integral breaks exactly the same dominated convergence schemes that we are trying to use, and the limit - in whatever sense it exists (which it does) - is outside of the function space we are looking in. This is why, unless we have fancier tools available, we need to work it from the ground up. Copied from a link I have gave in my answer, to be right. Thank you. $\endgroup$ – Constantine Black Feb 28 '16 at 9:37
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I shall try here to prove the statement posted in your question, but I have to admit that this is an effort with one particular danger; since I'm not a mathematician I can only rely on others with more knowledge from me, so that, any mistakes in the formality of my proof or any inconsistencies on the mathematics involved may be found and corrected. So, having said that:

We can define the δ Dirac function as follows:

$$δ(x)= \lim_{n\to \infty} D_n (x) .$$

Some examples or D functions are ${{\sin(nx)} \over {πx}} $ or $ \sqrt{{{π} \over {n}}e^{-nx^2}} .$

We want to prove the same for: $$D_n (x)= {{\sin(ax)} \over {πax^2}}=δ(x). $$

So from the definition of Dirac function: $\int_{- \infty}^{\infty} f(x) δ(x) = f(0), $ we shall deal with this relationship:

$$ I= \lim_{a\to \infty}{ \int_{- \infty}^{\infty} (f(x) – f(0)){{1} \over {π}} {{\sin^2 (ax) } \over {ax^2}} dx } =0 ,$$ where we brought $f(0)$ in the integral by noticing:

$$\int_{- \infty}^{\infty}{{1} \over {π}} { {\sin^2 (ax)} \over {ax^2}} =1 , $$ for positive a.

For this problem to have meaning, $f$ must be a continuous function at 0. So, it must hold:

$$|x|<δ \rightarrow |f(x) – f(0)| < ε $$, were δ,ε some arbitrary numbers for now. Now we make the change $ax=t $. Taking the above integral, we have:

$$A= \int_{- \infty}^{\infty} (f(t/a) – f(0)){{1} \over {π}} {{\sin^2 (t) } \over {t^2}} dt $$ so that $$ A \leqslant { {1} \over {π} } \left[ \int_{- \infty}^{\infty} |f(t/a) - f(0) |^2 dt \right]^{1/2} ~~~\left[ \int_{- \infty}^{\infty} \left|{{\sin^2 (t)} \over {t^2}} \right|^2 dt \right]^{1/2} .$$

The inequality came from using the Cauchy-Schwarz inequality. We set $ A_1$ the first integral term and $A_2 $ the second. So:

$$A_1 = \sqrt { \int_{- \infty}^{\infty} |f(t/a) - f(0) |^2 dt } \leqslant ε \sqrt{δa}, $$ where this result came from the continuity considerations of $f$. Also,

$$ [ \int_{- \infty}^{\infty} \left|{{\sin^2 (t)} \over {t^2}} \right|^2 dt \leqslant \prod \left[2 \int_0 ^{ \infty} {{\sin t} \over {t}} dt \right]^{1/2} = 4π, $$ where we again used the Cauchy-Schwarz inequality if I have understood correctly (see also http://arxiv.org/abs/1405.1265).

So, we have, in the end:

$$A=A_1 A_2 \rightarrow A \leqslant ε \sqrt{δa} .$$

For certain, fixed values of δ,α we may choose ε so that the term is close to zero, thus having: $$I= \lim_{a\to \infty} A =0 .$$

I am sure there might exist mistakes but if anyone with more knowledge is interested he/she 's welcome to correct them.

Hope this helps.

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Consider a real function $\;f(x)\;$ of the real variable $\;x\in\mathbb{R}\;$ for which \begin{align} f(x)\boldsymbol{=}0 \quad & \text{for any} \quad x\boldsymbol{\ne} x_{0} \quad \textbf{and} \tag{01a}\label{01a}\\ \mathcal{I}\boldsymbol{=}\!\!\!\!\int\limits_{\boldsymbol{x_{0}-\varepsilon}}^{\boldsymbol{x_{0}+\varepsilon}}\!\!\!f(x)\mathrm dx\boldsymbol{=}1\quad & \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0 \tag{01b}\label{01b} \end{align} Under these conditions it seems that this function is not well-defined at $\;x_{0}$, may be because of a singularity at this point. But we have good reasons to $^{\prime\prime}$believe$^{\prime\prime}$ that
\begin{equation} f(x)\boldsymbol{\equiv}\delta\left(x\boldsymbol{-}x_{0}\right) \tag{02}\label{02} \end{equation} since equations \eqref{01a},\eqref{01b} remind us the defining properties of Dirac delta function on the real axis $\;\mathbb{R}$.

For the function in the question we have at first \begin{equation} f(x)\boldsymbol{=}\lim_{\alpha\rightarrow\infty} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2}\boldsymbol{=}0 \quad\text{for any} \quad x\boldsymbol{\ne} 0 \boldsymbol{\equiv}x_{0} \tag{03}\label{03} \end{equation} and second for the integral \begin{equation} \mathcal{I}\boldsymbol{=}\!\!\!\int\limits_{\boldsymbol{x_{0}-\varepsilon}}^{\boldsymbol{x_{0}+\varepsilon}}\!\!\!f(x)\mathrm dx\boldsymbol{=}\!\!\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\!\!\lim_{\alpha\rightarrow\infty} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2}\mathrm dx \tag{04}\label{04} \end{equation} we have
\begin{equation} \mathcal{I}\boldsymbol{=}\lim_{\alpha\rightarrow\infty}\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2}\mathrm dx \tag{05}\label{05} \end{equation} leaving to mathematicians the proof of commutability of the limit and integral operators.

Now, we have the following indefinite integral(1) \begin{equation} \int \frac{\sin^2 y}{y^2}\mathrm dy \boldsymbol{=}\mathrm{Si}\left(2y\right) \boldsymbol{-}\frac{\sin^2 y}{y}\boldsymbol{+}\text{constant} \tag{06}\label{06} \end{equation} where \begin{equation} \mathrm{Si}\left(z\right) \boldsymbol{=}\int\limits_{0}^{z} \frac{\sin t}{t}\mathrm dt \tag{07}\label{07} \end{equation} the Sine Integral Function(2). This function has between others the following properties(3) to be used in the next \begin{align} \mathrm{Si}\left(z\right) & \boldsymbol{=}\boldsymbol{-}\mathrm{Si}\left(\boldsymbol{-}z\right) \tag{08a}\label{08a}\\ \lim_{z\rightarrow\boldsymbol{+}\infty}\mathrm{Si}\left(z\right) & \boldsymbol{=}\int\limits_{0}^{\boldsymbol{+}\infty} \frac{\sin t}{t}\mathrm dt \boldsymbol{=}\frac{\pi}{2} \tag{08b}\label{08b} \end{align} Replacing $y\boldsymbol{=}\alpha x$ in equation \eqref{06} and integrating from $\boldsymbol{-}\varepsilon $ to $\boldsymbol{+}\varepsilon $ \begin{equation} \int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2}\mathrm dx \boldsymbol{=}\:\Biggl|\mathrm{Si}\left(2\alpha x\right) \boldsymbol{-}\frac{\sin^2 \left(\alpha x\right)}{\alpha x}\Biggr|_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\stackrel{\eqref{08a}}{\boldsymbol{=\!=}}2\cdot\mathrm{Si}\left(2\alpha \varepsilon\right) \boldsymbol{-}2\cdot\frac{\sin^2 \left(\alpha \varepsilon\right)}{\alpha \varepsilon} \tag{09}\label{09} \end{equation} so \begin{equation} \lim_{\alpha\rightarrow +\infty}\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2}\mathrm dx \stackrel{\eqref{08b}}{\boldsymbol{=\!=}}2\cdot\frac{\pi}{2} \boldsymbol{-}2\cdot0 \boldsymbol{=}\pi \tag{10}\label{10} \end{equation} that is \begin{equation} \boxed{\:\:\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\!\!\lim_{\alpha\rightarrow +\infty} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2}\mathrm dx \boldsymbol{=}\pi \quad \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0\:\:} \tag{11}\label{11} \end{equation} From equations \eqref{03},\eqref{11} we have according to equations \eqref{01a},\eqref{01b} and \eqref{02} \begin{equation} \boxed{\:\:\lim_{\alpha\rightarrow +\infty} \frac{\sin^2\left(\alpha x\right)}{\alpha x^2} \boldsymbol{=}\pi\delta\left(x\right)\vphantom{\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}} \:\:} \tag{12}\label{12} \end{equation}


(1) $\textbf{Proof of equation}$ \eqref{06} \begin{align} \int \frac{\sin^2\! y}{y^2}\mathrm dy & \boldsymbol{=} \boldsymbol{-}\int \sin^2\! y\,\mathrm d\left(\frac{1}{y}\right)\boldsymbol{=} \boldsymbol{-}\left[\frac{\sin^2\! y}{y}\boldsymbol{-}\int\frac{1}{y}\,\mathrm d\left(\sin^2\! y\right)\right] \nonumber\\ & \boldsymbol{=}\int\frac{2\sin y \cos y}{2y}\,\mathrm d\left(2 y\right)\boldsymbol{-}\frac{\sin^2\! y}{y}\boldsymbol{=}\underbrace{\:\:\int\frac{\sin\left(2y\right)}{2y}\,\mathrm d\left(2 y\right)\:\:}_{\mathrm{Si}\left(2y\right)}\boldsymbol{-}\frac{\sin^2\! y}{y} \nonumber\\ & \boldsymbol{=}\mathrm{Si}\left(2y\right)\boldsymbol{-}\frac{\sin^2\! y}{y} \nonumber \end{align} Adding a constant, equation \eqref{06} is proved \begin{equation} \int \frac{\sin^2 y}{y^2}\mathrm dy \boldsymbol{=}\mathrm{Si}\left(2y\right) \boldsymbol{-}\frac{\sin^2 y}{y}\boldsymbol{+}\text{constant} \nonumber \end{equation}


(2) $\textbf{The Sine Integral Function } \mathrm{Si}\left(z\right)$ enter image description here

A graph of the Sine Integral Function is shown in above Figure. Since \begin{equation} \dfrac{\mathrm d\mathrm{Si}\left(z\right)}{\mathrm dz}\boldsymbol{=}\dfrac{\sin z}{z}\,, \quad \dfrac{\mathrm d^2\mathrm{Si}\left(z\right)}{\mathrm dz^2}\boldsymbol{=}\dfrac{z\cos z\boldsymbol{-}\sin z}{z^2} \tag{ftnt2-1}\label{ftnt2-1} \end{equation} we have for the extrema \begin{align} \textbf{maxima at : } z &\boldsymbol{=}\left(2k\boldsymbol{+}1\right)\pi \qquad k=0,1,2,3,\cdots \tag{ftnt2-2a}\label{ftnt2-2a}\\ \textbf{minima at : } z & \boldsymbol{=}\:\:\:\: 2k\pi \qquad k=1,2,3,\cdots \tag{ftnt2-2b}\label{ftnt2-2b} \end{align}


(3) "Handbook of Mathematical Functions" by M.Abramowitz & I.Stegun, pages 231,232

$\textbf{Note on equation}$ \eqref{08b}

This equation is proved via "The Residue Theorem" of complex analysis see for example here "Complex Variables - Schaum's Outline Series [Spiegel-Lipschutz S.-Schiller-Spellman], 2nd edition 2009 - page 221..

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The short answer is that $\sin^2 (ax) /ax^2 $ becomes increasingly localized at zero. The effective domain shrinks like $1/a$ while its value at zero is $a$. Moreover, $\int_{-\infty}^\infty \sin^2 (ax) /ax^2 = \pi$. The rest is math.

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