Dirac delta function defined in Zee's Quantum Field Theory book This is from Appendix 1 of the first chapter of Zee's Quantum Field Theory in a Nutshell:

I am not sure whether it is correct to call this the Dirac delta function. Sure, the integral over all space is 1, and it is sharply peaked at $x=0$. But its width doesn't approach $0$ when $K \to \infty$. $d_k(\Delta x) \neq 0$ for small $\Delta x$, and hence the integral $ \lim_{k \to \infty} \int_{-\Delta x}^{\Delta x} d_k(x) \neq 1$.
 A: Your concern is that$$\lim_{K\to\infty}d_K(x)=\left\{ \begin{array}{rl} \infty & x=0\\ 0 & x\ne0 \end{array}\right.\implies\int_{\Bbb R}\lim_{K\to\infty}d_K(x)dx=0\ne\lim_{K\to\infty}\int_{\Bbb R}d_K(x)dx=1.$$This shows $\delta$ cannot be a function defined as a pointwise limit of $d_K$. Indeed, $\delta$ is not a function at all, which is all such a pointwise limit can obtain. Instead $\delta$ is a distribution with a measure, and is recovered as a distributional limit viz.$$\int_{\Bbb R}\delta(x)f(x)dx=\lim_{K\to\infty}\int_{\Bbb R}d_K(x)f(x)dx=f(0)$$for sufficiently nice $f$ (in particular, $f$ can be any Schwartz function).
A: This derivation is based on the Fourier integral.
Indeed you may know that the Fourier transform of the Dirac's delta is 1, and therefore the inverse Fourier transform of 1 is the Dirac's delta.
Indeed you can easily see that
\begin{equation}
\mathcal{F}[\delta(x)](k) = \int_{-\infty}^{+\infty}\delta(x)e^{-ikx}\,\mathrm{d}x=e^0=1
\end{equation}
If we make the inverse Fourier transform of the above
\begin{equation}
\delta(x)=\mathcal{F}^{-1}[1](x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}1\cdot e^{ikx}\,\mathrm{d}k = \lim_{K\to\infty}\int_{-K/2}^{K/2}\frac{e^{ikx}}{2\pi}\,\mathrm{d}k
\end{equation}
which is what you wanted to prove.
A: To say the same thing in a dumbed down way: If you had no idea what a delta function was, but wanted to mimic
$$f_n = \sum_{n'} \delta_{n,n'} f_{n'} $$
and write $f(x)$ as a sum (i.e. now an integral) of something like this that was zero everywhere except at $x$ giving $f(x)$, how could one do it?
Although it seems immediate what to do, we can make the generalization explicit by re-writing the delta above as a function of only one argument $n-n'$ via
$$f_n = \sum_{n'} \delta_{n-n',0} f_{n'}$$
or even more explicitly by setting $f_n = f(n)$ and $\delta_{n-n',0} = \delta(n-n')$ to get
$$f(n) = \sum_{n'} \delta(n-n') f(n') $$
so that the generalization becomes
$$f(x) = \int dx' \delta(x - x') f(x').$$
To find an expression for $\delta$ we can recall the Fourier integral form of $f$
$$f(x) = \frac{1}{2 \pi} \int f(k) e^{i kx} dk = \frac{1}{2 \pi} \int [ \int dx' f(x') e^{-ikx'}] e^{ikx} dk = \int dx' [\frac{1}{2 \pi} \int dk e^{ik(x-x')}] f(x')  $$
Thus we find
$$\delta(x - x') = \frac{1}{2 \pi} \int dk e^{ik(x-x')} = \lim_{K \to \infty} \frac{1}{2 \pi} \int_{-K/2}^{K/2} dk e^{ik(x-x')}  $$
Now you can rationalize this to make it work, e.g. $\delta$ satisfies property X etc... for example $1 = \int dx' \delta(x - x')$ follows immediately. You might have to give up the very idea of what a function is to make all this work, but so be it.
A: *

*When considering a nascent delta function $\delta_{\epsilon}:\mathbb{R}\to \mathbb{C}$ with a regularization parameter$^1$ $\epsilon>0$, it is not necessary that (the Lebeque measure of) the support ${\rm supp}(\delta_{\epsilon})$ vanishes for $\epsilon\to 0^+$. There are many counterexamples. E.g. the heat kernel or the Poisson kernel representation of $\delta$.


*The Dirac delta distribution $\delta$ satisfies by definition that$^2$ $$\delta[f]~=~f(0)$$ for test functions $f$.


*The nascent delta function $\delta_{\epsilon}$ satisfies
$$\lim_{\epsilon\to 0^+}\int_{\mathbb{R}}\! \mathrm{d}x~\delta_{\epsilon}(x)f(x)~=~f(0).$$
--
$^1$ Zee's regularization parameter can be viewed as $\epsilon=1/K$.
$^2$ $\delta[f]$ is often written with the notation $\int_{\mathbb{R}}\! \mathrm{d}x~\delta(x)f(x)$.
A: One way to directly see why this is true is to recall:
$$\delta(x) = \int_{-\infty}^{\infty} \frac{dk}{2 \pi} e ^ {ik.x} \ \ \ \ \ \ \ \  \  \ \ [the \ standard \ definition] \\ = \underset{K\to \infty}{lim} \int_{- \frac{K}{2}}^{\frac{K}{2}} \frac{dk}{2\pi} e^{ik.x} =\underset{K\to \infty}{lim} (\frac{1}{\pi x} sin(\frac{Kx}{2}))$$
