how they prove this Fourier Transform of unit impulse function

Question

In my text book , to prove $\ \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega t}d\omega $ behaves like unit impulse function they evaluate the integral :

$\ \int_{-\infty}^{\infty}\bigg[\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega t}d\omega \bigg]g(t)dt$

where $\ g(t)$ is an arbitrary well behaved signal which is continuous at $\ t=0 $ and whose Fourier Transform $\ G(\omega)$ . So

$\ \frac{1}{2\pi}\int_{-\infty}^{\infty}\bigg[\int_{-\infty}^{\infty}g(t)e^{j\omega t}dt \bigg]d\omega= \frac{1}{2\pi }\int_{-\infty}^{\infty}G(-\omega)d\omega $

and the inversion formula it follows that:$\ \frac{1}{2\pi }\int_{-\infty}^{\infty}G(-\omega)d\omega = \frac{1}{2\pi }\int_{-\infty}^{\infty}G(\omega)d\omega =g(0) $

How they get this last line. There is no additional info is given.How they write this two integrals are equal or there value is the value of $\ g$ at $\ t=0$ ?

Answer 1

Recall that

$$g(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)e^{i\omega t}d\omega$$

Therefore,

$$g(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)e^{i\omega 0}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}G(\omega)d\omega$$

Now substitute $\omega \to -\omega$. The limits flip, but by absorbing the minus sign on the "differential", we have

$$\int_{-\infty}^{\infty}G(\omega)d\omega=\int_{\infty}^{-\infty}G(-\omega)(-d\omega)=\int_{-\infty}^{\infty}G(-\omega)d\omega$$

Answer 2

Hint, for the first equality, note that $f(x)|_{a} = f(-x)|_{-a}$ and that you're integrating over a symmetric interval. For the second equality, look at the formula for the inverse Fourier transform. What happens to the exponential term in the integrand when $t=0$?