Dirac Delta function inverse Fourier transform We know that the Fourier transform of the Dirac Delta function is defined as
$$\int_{-\infty}^{\infty} \delta(t) e^{-i\omega t} dt = 1,$$
and if I were to reconstruct the function back in time domain, the inverse Fourier transform is defined as
$$\delta(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega t} d\omega.$$
How do I compute this integral analytically?

You can view this as a limiting process. Start with the truncated integral:
$$
        \frac{1}{2\pi}\int_{R}^{R}e^{j\omega t}dt = \frac{1}{\pi}\frac{\sin(R\omega)}{\omega}
$$
If you integrate this against a function and take the limit as $R\rightarrow\infty$, then
$$
    \lim_{R\rightarrow\infty}\int_{-\infty}^{\infty}f(\omega)\frac{1}{2\pi}\int_{-R}^{R}e^{j\omega t}dt d\omega
    = \lim_{R\rightarrow\infty}\frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(\omega)e^{j\omega t}d\omega dt.
$$
The above is $(f^{\wedge})^{\vee}(0)=f(0)$ if $f$ has some smoothness at $0$.
