Dirac delta and fourier transform

[aaaa202]

In my book the dirac delta is described by the equation on the attached picture. This realtion is derived from the Fourier transform, but I'm not sure that I understand what it says. If u=t it is clear that one gets f(u) in the Fourier inversion theorem. But why wouldn't u=t? In the derivation of the Fourier transform from the discrete Fourier series t was just changed to u in the expression of the coefficients to avoid confusion.
Can anyone try to picture what this expression fundamentally says? I should suspect that it is like the analogue of the ortogonality relation of the discrete Fourier series, but I can't quite understand it.
And what would the situation u≠t represent?

 

[jbunniii]

What it is saying is that

$$\delta(t - u)$$
and
$$\exp(-iu\omega)$$

are a Fourier transform pair. A complex exponential with "frequency" equal to $u$ has a Fourier transform with all of its energy concentrated at $u$.

 

[jbunniii]

P.S. The equation

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

is mathematically nonrigorous. The integral on the right hand side does not actually exist for any values of $u$ and $t$. What is true is that if $\mathcal{F}$ denotes the Fourier transform operator, then

$$\mathcal{F}(\delta(t-u)) = \exp(-i\omega u)$$
and
$$\delta(t-u) = \mathcal{F}^{-1}(\exp(-i\omega u))$$

in the sense of distributions. See here for more details:

http://en.wikipedia.org/wiki/Distribution_(mathematics)