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.
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.