It is well-known that: $$\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x).$$ This can also be written as $$ 2\pi\delta(x)=\int^{+\infty}_{-\infty}e^{ikx}\,\mathrm dk.$$ However, I don't know how to prove this without using Fourier Transform. I have already searched google and looked for some books, but I just get nothing.

In short, I want to know the proof of this equation: $$\lim_{k\to+\infty}\int^{+\infty}_{-\infty}\frac{\sin(kx)}{\pi x}f(x)dx=f(0).$$

  • $\begingroup$ The antiderivative of the function in the LHS is the Sine Integral, which tends to a step. 2dcurves.com/trig/h17sinix.gif $\endgroup$
    – user65203
    Jun 25, 2016 at 9:23
  • $\begingroup$ The limit doesn't even exist unless $x$ is a multiple of $\pi$! $\endgroup$ Jun 25, 2016 at 10:11

4 Answers 4


I will prove that the limit is true in distribution sense: for all $\varphi \in C_{c}^{\infty}(\Bbb{R})$ we have

$$ \lim_{k\to\infty} \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = \varphi(0). $$

The standard approximation-to-the-identity argument splits the integral into near-zero part and away-from-zero part, and estimate each part separately. In this case, however, this approach is a bit painful due to the fact that $\sin(kx)/\pi x$ is not absolutely integrable. So we adopt an indirect but easier approach.

Proof. Define $F(x) = \frac{1}{\pi}\int_{-\infty}^{x} \frac{\sin t}{t} \, \mathrm{d}t$. It is easy to check that $F$ is bounded, $F(-\infty) = 0$ and $F(+\infty) = 1$. Then performing integration by parts, we can write

$$ \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = - \int_{\Bbb{R}} F(k x) \varphi'(x) \, \mathrm{d} x. $$

Now notice that the integrand of the RHS is uniformly bounded by $\| F \|_{\mathrm{sup}} |\varphi'| $, which is integrable. Thus taking $k \to \infty$, the dominated convergence theorem shows that

$$ \lim_{k\to\infty} \int_{\Bbb{R}} \frac{\sin (kx)}{\pi x} \varphi(x) \, \mathrm{d}x = - \int_{\Bbb{R}} \lim_{k\to\infty} F(k x) \varphi'(x) \, \mathrm{d} x = - \int_{\Bbb{R}} H(x) \varphi'(x) \, \mathrm{d} x, $$

where $H(x)$ is the Heaviside step function. Now manually computing the last integral gives

$$ - \int_{\Bbb{R}} H(x) \varphi'(x) \, \mathrm{d} x = - \int_{0}^{\infty} \varphi'(x) \, \mathrm{d} x = \varphi(0) $$

and therefore the claim follows.

  • 1
    $\begingroup$ Hi Sangchul, my friend! I hope that you're enjoying the weekend. A big (+1) for this answer. Would the following work? Fix $\delta >0$$$\begin{align} \int_{-\infty}^\infty \frac{\sin(kx)}{x}\phi(x)\,dx=\pi \phi(0)\text{sgn}(k)&+\int_{|x|\le \delta} \frac{\sin(kx)}{x}(\phi(x)-\phi(0))\,dx\\&+\int_{|x|\ge \delta} \frac{\sin(kx)}{x}\phi(x)\,dx\\\\ &+\phi(0)\int_{|x|\ge k\delta} \frac{\sin(x)}{x}\,dx \end{align}$$The first integral is $O(\delta)$, the second approaches $0$ by the Riemann Lebesgue Lemma, and the third goes to $0$ as the improper integral converges. $\endgroup$
    – Mark Viola
    Sep 6, 2020 at 0:01
  • $\begingroup$ @MarkViola, Hi, and sorry for the late reply! I was busy preparing the teaching this quarter. As for your solution, it seems correct to me. I definitively love this one. $\endgroup$ Sep 14, 2020 at 19:39
  • $\begingroup$ I liked your solution. Very cool. Where do you teach? $\endgroup$
    – Mark Viola
    Sep 14, 2020 at 20:04
  • $\begingroup$ @MarkViola, I am teaching at Stanford now. The teaching is done remotely, though, and I am still living in LA. :) $\endgroup$ Sep 14, 2020 at 20:59
  • $\begingroup$ Congratulations Sangchul on your new (I presume) appointment. Well done! $\endgroup$
    – Mark Viola
    Sep 14, 2020 at 21:03

$f_k(x)=\frac{\sin(k x)}{\pi x}$ is a function with a unit integral over the whole real line. It is an entire function, its value at $x=0$ equals $\frac{k}{\pi}$ and the integral over the whole real line is concentrated in a neighbourhood of the origin that shrinks to zero as $k\to +\infty$:

$$ \lim_{k\to +\infty}\int_{-\frac{\pi}{2k}}^{\frac{\pi}{2k}}\frac{\sin(kx)}{\pi x}\,dx =\lim_{k\to +\infty}\int_{-\frac{\pi}{2k}}^{\frac{\pi}{2k}}\frac{k}{\pi}\,dx=1$$ hence: $$ \lim_{k\to +\infty}\int_{|x|\geq\frac{\pi}{2k}}\frac{\sin(kx)}{\pi x}\,dx = 0$$ by the regularity (analiticity) of the $\text{sinc}(kx)$ in a neighbourhood of the origin. It can be seen as a consequence of the Paley-Wiener theorem, too, since the Fourier transform of $\text{sinc}(x)$ is compact-supported.

  • $\begingroup$ I don't think this is a complete proof. Just because $\lim_{k\to \infty}\int_{\mathbb R\setminus U_\epsilon(0)} f_k(x) dx = 0$, how can you conclude that also $\lim_{k\to \infty}\int_{\mathbb R\setminus U_\epsilon(0)} f_k(x) \varphi(x) dx = 0$ for all $\varphi$? $\endgroup$
    – Hyperplane
    Jul 21, 2020 at 15:38
  • $\begingroup$ Isn't the following a counter example that the proof is wrong: let $g_k(x) = 2k(1-\operatorname{abs}(x))\mathbf{1}_{[-\frac{1}{k}, +\frac{1}{k}]}(x)$. Then add a positive bump to the left and a negative bump to the right. (for instance $f_k(x) = g_k(x) + (x-1)x(x+1)\mathbf{1}_{[-1,+1]}(x)$) The resulting function satisfies the same integral properties, but does not converge against dirac delta. $\endgroup$
    – Hyperplane
    Jul 21, 2020 at 15:48

k=100 Intuitively, as $k\to\infty$, the function oscillates faster and faster, and so it "locally averages out to zero". We note that for all $a<b$:

$$\lim_{k\to\infty}\int_a^b \sin(kx) dx = \lim_{k\to\infty}\Big[\frac{\cos(kx)}{k}\Big]_a^b = 0$$

Secondly, take a test function $\varphi\in C_c^\infty(\mathbb R$). Then, for all $\epsilon>0$, we have $\frac{\varphi(x)}{\pi x}\in C^\infty_c(\mathbb R\setminus(-\epsilon,+\epsilon))$. Since this function is continuous and compactly supported, it is even uniformly continuous. In particular, it can be approximated arbitrarily well by step functions (locally constant function with finitely many discontinuities). So let $\operatorname{supp}(\varphi)\subset [-C, C]$, and take a step function $f_\delta$ such that $|\frac{\varphi(x)}{\pi x} - f_\delta(x)|\le \delta$. Then it follows

$$ \int_\epsilon^C \frac{\sin(kx)}{\pi x}\varphi(x)dx = \int_\epsilon^C \sin(kx) \underbrace{\Big(\frac{\varphi(x)}{\pi x} - f_\delta(x)\Big)}_{\in(-\delta, +\delta)\forall x} dx + \underbrace{\int_\epsilon^C \sin(kx)f_\delta(x) dx}_{\to 0 \text{ since $f_\delta$ is step function}} $$

Hence, since $f$ can be chosen to approximate arbitrarily well, $\forall \delta>0: \lim\limits_{k\to\infty}|\int_\epsilon^C \frac{\sin(kx)}{\pi x}\varphi(x)dx| \le C\delta$; i.e. in the limit this integral must be zero. And in the neighborhood of the origin we find:

$$ \int_{-\epsilon}^{+\epsilon} \frac{\sin(kx)}{\pi x}\varphi(x) dx \sim \int_{-\epsilon}^{+\epsilon} \frac{\sin(kx)}{\pi x}\Big(\varphi(0)+\varphi'(0)x\Big) dx \to \varphi(0) $$


Hint:You just need to prove that for any $f(x)$ which is continuous at $x=a$, the following result holds:

  • $\int_{-\infty}^{\infty} f(x)$.$\delta(x-a)dx=f(a)$.

    Added-Substituting $\delta(x-a)$ using its definition, one gets

$\int_{a}^{a+\epsilon}$$f(x).\frac{1}{\epsilon}dx$$=\frac{1}{\epsilon}\int_{a}^{a+\epsilon}$$f(x)dx=\frac{1}{\epsilon}.(a+\epsilon-a).f(c)$ (where $a\le c\le a+\epsilon$)$=f(a)$ for $\epsilon→0$

  • $\begingroup$ I dnt understand...why downvote? $\endgroup$ Jun 26, 2016 at 1:39
  • 2
    $\begingroup$ I didn't downvote, but I really don't see how this has much at all to do with the question. How does this show that $\sin(kx) / (kx)$ converges distributionally to the delta? $\endgroup$
    – user296602
    Jun 27, 2016 at 2:44

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