What is the relation between Fourier's Inversion theorem and the Dirac-Delta function?

Question

This is a direct quote from page 472 of this book:

From Fourier's Inversion theorem $$f(t)= \int_{-\infty}^\infty f(u) \, \mathrm{d}{u} \left( \frac{1}{2\pi}\int_{-\infty}^\infty e^{-i\omega(t-{u})} \,\mathrm{d}\omega \right) \tag{1}$$ comparison of $(1)$ with the Dirac-Delta property: $$f(a)= \int f(x) \, \mathrm{d}x \, \delta(x-a)$$ shows we may write the $\delta$ function as $$\delta(t-u)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{i\omega(t-{u})} \, \mathrm{d}\omega$$

My question is what is the part in the large parentheses of $(1)$ got to do with $\delta(t-u)$?

Many thanks.

Answer 1

If you have $$ f(t)= \int_{-\infty}^\infty f(u) \Big(\cdots\cdots\text{blah blah} \cdots\cdots\Big) \, \mathrm{d}{u} $$ then you can conclude that $$ \Big(\cdots\cdots\text{blah blah} \cdots\cdots\Big) = \delta(u-t) $$

Answer 2

You can interpret this classically instead. Assuming Fourier conditions on $f$, the inverse Fourier transform applied to the Fourier transform gives you back $f$: $$ f(t)=\lim_{R\rightarrow\infty}\frac{1}{2\pi} \int_{-R}^R e^{i\omega t} \int_{-\infty}^\infty f(u)e^{-i\omega u} \, du \, d\omega. $$ Interchanging orders of integration, $$ f(t) = \lim_{R\rightarrow\infty}\int_{-\infty}^\infty f(u)\left(\frac{1}{2\pi} \int_{-R}^R e^{i\omega(t-u)} \, d\omega\right) \, du. $$