How to calculate the energy in the dirac delta function signal?
\begin{align}
&amp;\int_{\omega_1}^{\omega_2}|\delta (\omega-\omega^*)|^2.d\omega \\
&amp;= \int_{\omega_1}^{\omega_2}|\delta (\omega-\omega^*)|.|\delta (\omega-\omega^*)|.d\omega \\
&amp;= \int_{\omega_1}^{\omega_2}\delta (\omega-\omega^*).\delta (\omega-\omega^*).d\omega \\
&amp;= \int_{\omega_1}^{\omega_2}\delta (\omega^*-\omega^*).d\omega \\
&amp;= \int_{\omega_1}^{\omega_2}\delta (0).d\omega \\
&amp;= \delta (0).\Delta \omega 
\end{align}
Which is undefined! See this for better clarification: https://en.wikipedia.org/wiki/Distribution_(mathematics)#Adding_and_multiplying_distributions

What you have to understand is that $\delta$ is not a function, it is a distribution. What you learn in introductory DSP courses is just an "engineering trick" so that things will work the way they should (you will understand what that means once you study sampling and analog to digital conversion).

EDIT: Sorry I misinterpretation the question. (Thanks Jazzmaniac for pointing that out)

As far as I know the $\delta$ "function" in the continuous domain is not in $L^2$, it isn't a square-integrable function.

Think about your question in the time domain. If you have 2 symmetric $\delta$'s in the frequency domain, you will have a INFINITLY lasting sinusoid. It's energy is infinite, i.e, you can't associate a value to it (undefined, remember that $\infty$ is not a number). If you have a single $\delta(f)$, centered at zero, you will have a complex exponential in the time domain, and it's energy is also undefined (unbounded since $|e^x|^2=1$.)