finding energy content of signal from "energy spectral density" function
The so called auto-correlation function that's used in the example (which was found in a previous example) is not correct.

First note that it's the deterministic auto-correlation computation, compared to its theoretical counterpart which requires that the signals be random for their ACF to be defined at all.

Anyway, so the following integral signifies one definition of deterministic ACF of a given deterministic signal $x(t)$ as :

$$R_{xx}(\tau)  = \int_{-\infty}^{\infty} x(t)x(t-\tau) dt$$

applying this definition to the signal $x(t) = e^{-a t} u(t)$ with $a &gt;0$ we have: 

$$
\begin{align}
R_{xx}(\tau)  &amp;= \int_{-\infty}^{\infty} x(t)x(t-\tau) dt \\ \\
&amp;= \int_{-\infty}^{\infty} e^{-a t} u(t) e^{-a (t-\tau)} u(t-\tau) dt \\ \\
&amp;= e^{a \tau} \int_{-\infty}^{\infty} e^{-2 a t} u(t) u(t-\tau) dt \\ \\
&amp;= e^{a \tau} \int_{\max\{0,\tau\}}^{\infty} e^{-2 a t} dt \\ \\
&amp;= \begin{cases}{ e^{a\tau} \int_{\tau}^{\infty} e^{-2at} dt ~~~,~~~\tau \geq 0 \\ \\ \\
e^{a\tau} \int_{0}^{\infty} e^{-2at} dt ~~~,~~~\tau &lt; 0 }\end{cases}\\ \\
&amp;= \begin{cases}{ e^{a\tau} \frac{e^{-2a\tau}}{2a} ~~~,~~~\tau \geq 0 \\ \\ \\
e^{a\tau} \frac{1}{2a} dt ~~~,~~~\tau &lt; 0 }\end{cases}\\ \\
&amp;= \begin{cases}{ \frac{e^{-a\tau}}{2a} ~~~,~~~\tau \geq 0 \\ \\ \\
\frac{e^{a\tau}}{2a} dt ~~~,~~~\tau &lt; 0 }\end{cases}\\ \\
R_{xx}(\tau) &amp;=   \frac{e^{-a |\tau|}}{2a}  \\ \\
\end{align}
$$