How to prove that $\lim_{k\to+\infty}\frac{\sin(kx)}{\pi x}=\delta(x)$?

Question

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: $$\lim_{k\to+\infty}\int^{+\infty}_{-\infty}\frac{\sin(kx)}{\pi x}f(x)dx=f(0).$$

Answer 1

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.

Answer 2

$f_k(x)=\frac{\sin(k x)}{\pi x}$ 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.

Answer 3

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 $\epsilon→0$