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Hi: I've been reading an introductory book on Fourier transforms. The author explains the $\delta$ function ( while noting that it's really a distribution ) in the following manner which makes a lot of sense to me:

$\delta(t) = \lim_{~a\to\infty } f_{a}(t)$

with $f_{a}(t) = \left\{ \begin{array}{r@{\quad\quad}l} a & for & -\frac{1}{2a} \le t \le \frac{1}{2a} \\ 0 & else & \end{array} \right. $

So, in this manner, the area underneath the function is always 1 even as the function value $f(t)$ approaches infinity at the value $t = 0$. That's clear. But then the author throws in the following as if it's obvious but it's not to me. The following is the exact statement:

Another representation for the $\delta$ function which we will frequently use is:

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

My question is why is the function also a delta function. It seems to me that in the integral defined, all the frequencies except $\omega = 0$, should average to $0$ because the function is periodic. So, the only frequency for which there is a non-zero value is at $\omega = 0$. At $\omega = 0$, the function evaluates to 1 so we have

$\frac{1}{2\pi} \int_{-\infty}^{\infty} 1dt $

This function integrates to T for $-\frac{T}{2} to \frac{T}{2}$ but I don't see how to relate that to the original definition above with the $f_{a}(t)$ because that has $a$ in the bottom and for this case the T is in the numerator. Or I may be totally barking up the wrong tree. He uses this concept of $\delta(\omega)$ a lot later in the chapter so I think that it's important to understand this definition. Thanks in advance for any wisdom concerning this.

mark leeds
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    The integral form is a consequence of the properties of the delta function and the use of the Fourier transform. Equations (31) and (32) demonstrate this on Weisstein's site: http://mathworld.wolfram.com/DeltaFunction.html – Leucippus May 16 '14 at 04:47
  • Hi Leucippus: I went through the link and I follow 31) and 32) but I don't see where the $\frac{1}{2\pi}$ comes from in the expression that I wrote. Wolfram I think is using the discrete version of the transform whereas the book I am using the continuous version so I'm not clear on the comparison. Thanks though for pointing the link out. I'm going to read it again and possibly purchase Bracewell text. – mark leeds May 16 '14 at 05:38
  • It falls more under "notational preference" in many cases. The Wiki page and presentation both use the $(2\pi)^{-1}$ factor. http://en.wikipedia.org/wiki/Dirac_delta_function , http://www.crystallography.fr/mathcryst/pdf/uberlandia/Estevez_Delta_Dirac.pdf – Leucippus May 16 '14 at 06:15
  • Whaaat, a "notational preference"? But one wants to compute things at the end, so the 2π factor cannot be a convention, can it? – Did May 16 '14 at 06:18
  • If using the symmetric Fourier transform, found http://hitoshi.berkeley.edu/221a/delta.pdf, for example then each integral gets a $(2\pi)^{-1/2}$ factor. One can also use the non-symmetric versions and either the transform integral gets a $(2\pi)^{-1}$ or the inverse transform integral gets the factor. With three possible ways to claim the same factor, it falls under notational preference. – Leucippus May 16 '14 at 06:23
  • I would say the equation "holds in the weak sense" (sense of distributions to be precise), i.e. "multiply by some $f(t)$ and integrate". – Dirk May 16 '14 at 06:30
  • @Leucippus In this case not since the relation involves both forward and inverse Fourier transform. – Dirk May 16 '14 at 06:31
  • this document at repository.ias.ac.in/1082/1/350.pdf gives a really nice explanation. On page 4 in equation 7, he shows how through the use of a limiting argument how one obtains the complex value. and if you use $\frac{1}{\epsilon}$ in place of $L$, then it maps to the original definition in terms of $a$. the only part I don't understand is how the sinc function can be defined that way. In other words, in equation (7), what justifies the second equality. Thanks again. – mark leeds May 16 '14 at 15:49

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In the following I use the non-unitary version of the Fourier transform with angular frequency $\omega$:

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

If you take the inverse Fourier transform of $\delta(\omega)$ you obtain by the definition of the delta impulse

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

and, consequently

$$\mathcal{F}\left\{\frac{1}{2\pi}\right\}=\delta(\omega)$$

And, by definition of the Fourier transform, we have

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

Of course, this integral must be interpreted as a distribution.

Matt L.
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