TITLE: Why $\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{iw(t-x)}dw$ is the Dirac delta function?
QUESTION [12 upvotes]: I'm reading a book that says:
$$f(x) = \int_{-\infty}^{\infty}f(t)\left\{\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{iw(t-x)}dw\right\}dt$$
and then says that the term in curly brackets can be seen as the Dirac Delta Function. 
As I understand, the Dirac Delta Function should be $0$ when $t\neq x$, right? 
$$\delta(t-x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{iw(t-x)}dw$$
why this thing is $0$ when $t\neq x$?

REPLY [2 votes]: Let $$\phi_N(x) = \frac{1}{2\pi}\int_{-N}^N e^{i \omega x}d\omega = \frac{\sin(Nx)}{2\pi x} =  \frac{N}{2\pi}\text{sinc}(Nx)$$
Then for $f \in C^0 \cap L^1,f' \in L^1$ such that $\lim_{x \to \pm\infty} f(x) = 0$, we can use integration by parts to obtain
$$\int_{-\infty}^\infty f(x) \phi_N(x)dx = \frac{-1}{2\pi}\int_{-\infty}^\infty f'(x) \left(\int_{-\infty}^{Nx} \text{sinc}(y)dy\right) dx $$
And hence, with $A = \int_{-\infty}^\infty \text{sinc}(y)dy = 2\pi$
$$\lim_{N \to \infty} \int_{-\infty}^\infty f(x) \phi_N(x)dx = -\frac{1}{2\pi}\int_0^\infty f'(x) A dx = f(0)$$
Which is the definition of $\phi_N \to \delta$ in the sense of distributions
$\scriptstyle \text{(also note how this proves the Fourier inversion theorem)}$