As stated in title, how is the following true? $$\dfrac{1}{2\pi}\int_{\mathbb{R}} |\delta(x)|^2dx = \int_{\mathbb{R}}|\hat\delta(t)|^2dt$$.
I know that $\hat\delta =1$, hence the right hand side is infinity but I can't quite make how the left hand side is also the same.
This assertion is listed as an exercise in "Introduction to Applied Mathematics" by Gilbert Strang.
Compute $$ \int_{-\infty}^\infty |\delta(x)|^2\, dx = +\infty $$ by approximation: $$ \delta(x)=\lim_{\epsilon\to 0} \frac{1}{\epsilon} \zeta\left(\frac{x}{\epsilon}\right),$$ where $\zeta$ is a compactly-supported nonnegative function that integrates to one.
This is not rigorous but it is the usual way to think of the Dirac delta (generalized) function in applied mathematics, AFAIK.