TITLE: Dirac Delta function inverse Fourier transform
QUESTION [9 upvotes]: 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?

REPLY [5 votes]: You can't really compute that integral analytically, since the delta function isn't really a "function," and the integral is not really well-defined as you've written it down. What you CAN do is prove that when integrated against a test-function, the integral behaves like a delta function.
So let's test it. We want $\int\delta(t)f(t)dt = f(0)$ for any function $f$. We know, from your equation two, that we want
$$\int\delta(t)f(t)dt =\int\int f(t)e^{j\omega t} dt d\omega$$
Now we note that doing the t-integral merely gives us the fourier transform of f:
$$=\int \tilde{f}(-\omega) d\omega$$
Finally, we note that this final integral is just the reverse fourier transform of $f$, centered at $t=0$:
$$=\int \tilde{f}(-\omega)e^{j(-\omega)0} d\omega=f(0)$$
Thus, the integral you wrote down above behaves just like we want a $\delta$-fucntion to behave. Hope that helps!