Use the Inverse Fourier transform to show the Dirac-Delta function as a limit of the sinc function

Question

This question is not the same as Dirac delta function as a limit of sinc function because I am asking about the inverse Fourier transform and more specifically the relation of equation $(1)$ below to the expression in $\color{purple}{\mathrm{purple}}$ below.

From this previous question asked by myself with some help I became convinced that the Dirac-Delta function can be written as $$\delta(t-u)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega(t-{u})} \, \mathrm{d}\omega \tag{1}$$

Consider the rectangular distribution of frequencies in $\displaystyle\mathrm{(a)}$, taking the inverse Fourier transform

$$f_{\Omega}=\frac{1}{\sqrt{2\pi}}\int_{-\Omega}^{\Omega}1\times e^{i\omega t} \,\mathrm{d}\omega$$

$$=\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t}$$ My text tells me that as $\Omega \to \infty \implies f_{\Omega} \to \sqrt{2\pi}\delta(t)$ by virtue of $(1)$.

Could someone please explain this to me as I don't see why $$\color{purple}{\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \to \sqrt{2\pi}\delta(t)}$$ as $\Omega \to \infty$?

Thank you.

Answer 1

First of all, I would like to emphasize once again, that I think you should learn distribution theory properly. It will increase your understanding. You should not be too upset when we say this over and over again. We have walked the path, and our intentions are good. Let me sketch how to do this problem for you, using distribution theory. This is the last time I answer questions like this from you, though.

You want to show that $$ \sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \to \sqrt{2\pi}\delta(t) $$ First of all, and as I mentioned in a comment, one has to explain what is meant by such a limit. We have a function in the left-hand side, and a distribution in the right hand side. Let us give a definition of what it means for a distribution to converge. (Below I will write $\mathcal D=C_0^{+\infty}(\mathbf R)$ and $\mathcal D'$ for the set of distributions with $\mathcal D$ as test functions.)

Definition The family of distributions $T_\Omega\in\mathcal D'$, $\Omega>0$ converge to the distribution $T\in D'$ precicely when $$ T_\Omega(\phi)\to T(\phi) $$ all test function $\phi$.

Note that the limit $T_\Omega(\phi)\to T(\phi)$ is a limit of real numbers.

To interpret the function $\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t}$ we have to say how it acts on functinos in $\mathcal D$. This was discussed in another of my answers to you. We let $T_\Omega$ be the distribution associated with the function, i.e. $$ T_\Omega(\phi)=\int_{-\infty}^{+\infty}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \phi(t)\,dt. $$ Also, to have notations that fit with the limit definition above, we let $T$ be the distribution associated with the dirac distribution you have in the right-hand side. $$ T(\phi)=\sqrt{2\pi}\phi(0). $$ Hence, what you have to show is the following:

What we need to show Let $\phi\in\mathcal D$. Then $$ \lim_{\Omega\to+\infty} \int_{-\infty}^{+\infty}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \phi(t)\,dt=\sqrt{2\pi}\phi(0). $$

Note that, when stated like this, everything makes sense! We have no divergent integrals or something that could confuse us.

Here, we could continue in essentially two ways. Either we use properties of Fourier transform of distributions to work on the Fourier side, or we try to work to show the limit directly.

To prove it directly

First, divide the integral as $$ \int_{-\infty}^{+\infty}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \phi(t)\,dt =\int_{|t|\geq 1}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \phi(t)\,dt+ \int_{|t|\leq 1}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \phi(t)\,dt. $$ In the first integral, integrate by parts to see that it goes to zero as $\Omega\to+\infty$. Then write the second integral as $$ \int_{|t|\leq 1}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \phi(t)\,dt =\int_{|t|\leq 1}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \bigl(\phi(t)-\phi(0)\bigr)\,dt+\phi(0)\int_{|t|\leq 1}\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \,dt. $$ For the first integral in the right-hand side, you note that $t\mapsto (\phi(t)-\phi(0))/t$ is smooth, and hence that integral goes to zero as $\Omega\to+\infty$ (this can be seen via Riemann-Lebesgue, or just by integrating by parts). In the second integral in the right-hand side, you do the substitution $t\mapsto t/\Omega$ and use the Calculus fact that $$ \lim_{\Omega\to+\infty}\int_{-\Omega}^{+\Omega}\sqrt{\frac{2}{\pi}}\frac{\sin(t)}{t}\,dt=\sqrt{2\pi} $$ to conclude.

Using Fourier transforms

For a distribution $T$, its Fourier transform $\hat T$ is defined via $$ \hat T(\phi)=T(\hat \phi), $$ i.e. it acts on a test function as $T$ acts on the Fourier transform of the test function. Note that $\phi$ is in $\mathcal D$, so it has a nice Fourier transform, and all formulas and properties for Fourier transforms one could expect is possessed by $\hat\phi$. I leave it to you to try to show the limit, using the nice Fourier transform of the sinc function.

Answer 2

As $\Omega \to \infty$, the rectangular distribution tends to a constant function. And the Fourier of a constant is (by definition?) the dirac delta function.

Edit:

There are two conventions for the Fourier Transform.

The first convention:

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega) e^{i\omega t} d\omega \iff \int_{-\infty}^{\infty}f(t) e^{-i\omega t} dt$$

the dirac delta function, shifted in time by $u$, is then defined by,

$$\delta(t - u) := \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega (t - u)}d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)e^{i\omega t} d\omega$$ where $F_u(\omega) = e^{-i\omega u}$. (i.e. modulation)

When there is no time shift, i.e. $u = 0$. Then $F_u(\omega) = 1$, the constant function. And we have $$\delta(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega t} d\omega$$

The second conventions, which assures a "symmetry" between Time and frequency

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(\omega) e^{i\omega t} d\omega \iff \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t) e^{-i\omega t} dt $$

Then the inverse Fourier of the constant function is related to delta delta (keeping its definition as above) by:

$$\frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty} e^{i\omega t} d\omega = \sqrt{2\pi}\delta(t)$$

Conclusion: Knowing that the (unit) rectangular function has inverse Fourier $\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t}$. When $\Omega \to \infty$, the rectangular function tends to $1$. And then $\sqrt{\frac{2}{\pi}}\frac{\sin(\Omega t)}{t} \to \sqrt{2\pi}\delta(t)$.