Energy of a Signal A signal is given by
$$x(t)=\begin{cases} e^{-t} &t\geq 0\\ 0 &t < 0 \\\end{cases}$$


*

*Find the Fourier transform of the signal.


$$X(\omega)=\int_{0}^{\infty}e^{-t}e^{-j\omega t} dt$$
and I get 
$$X(\omega)=\left(\frac{1}{1+j\omega}\right)$$


*Find total energy of the signal using time domain representation


$$X(\omega)=\int_{0}^{\infty}|e^{-t}|^2 dt$$
and I get
$$E = \frac{1}{2}$$ Is this correct?


*Find total energy of the signal using its frequency domain representation
$$E(\omega)=\int_{-\infty}^{\infty}\left|\frac{1}{1+j\omega}\right|^2 d\omega$$


I think this is the approach I have to use in order to calculate the energy right? And since this contains an imaginary part, how can I integrate this? :)


*Find the percentage of energy contained in the frequency range 200 - 500 Hz
If I take the magnitude in the 3rd Q in order to integrate it, I don't get a function of $\omega$

 A: The Laplace transform
$$
               \mathscr{L}\{f\}(\omega)=\int_{0}^{\infty}e^{-\omega t}f(t)dt
$$
may be examined on a vertical line in the complex plane $\omega = u+iv$ for $-\infty < v < \infty$ for any fixed $u$ for which $\int_{0}^{\infty}e^{-ut}|f(t)|dt < \infty$. This function of $u$
$$
                    \mathscr{L}\{f\}(u+iv)=\int_{0}^{\infty}e^{-ivt}(e^{-ut}f(t))dt
$$
is related to the Fourier transform of $e^{-ut}f(t)$ where $f$ is extended to be $0$ on $(-\infty,0)$. By Parseval's equality for the Fourier transform,
$$
   \int_{0}^{\infty}|f(t)e^{-ut}|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}
    |\mathscr{L}\{f\}(u+iv)|^2 dv
$$
This is true for all $u$ for which $F_{u}(t) = f(t)e^{-ut}$ is square integrable in $t$ on $[0,\infty)$.
In your case,
$$
              \mathscr{L}\{f\}(u+iv) = \int_{0}^{\infty}e^{-t}e^{-(u+iv)t}dt
    = \frac{1}{1+u+iv}
$$
Your function is square integrable for $u=0$, which gives energy
\begin{align}
           \frac{1}{2}=\int_{0}^{\infty}e^{-2t}dt 
  & = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{1}{1+iv}\right|^2 dv  \\
  & =\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{1}{1+v^2}dv \\
  & = \left.\frac{1}{2\pi}\tan^{-1}(v)\right|_{v=-\infty}^{\infty} \\
  & =\frac{1}{2\pi}\pi=\frac{1}{2}.
\end{align}
The reconstruction of $f$ is
$$
         f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathscr{L}\{f\}(iv)e^{ivt}dv
$$
The signal $e^{ivt}$ is a periodic function which goes through one complete cycle every $2\pi/|v|$ seconds. So the frequency of $e^{ivt}$ is
$$
  \frac{1\mbox{cycle}}{(2\pi/|v|) \mbox{seconds}}=\frac{|v|}{2\pi}\frac{\mbox{cycles}}{\mbox{second}}
$$
You want the components of the signal for which
$$
     200 < \frac{|v|}{2\pi} < 500.
$$
The energy corresponding to these components is obtained by integrating the energy density function over these frequencies.
$$
           \frac{1}{2\pi}\int_{-500(2\pi)}^{-200(2\pi)}\frac{1}{1+v^2}dv
      + \frac{1}{2\pi}\int_{200(2\pi)}^{500(2\pi)}\frac{1}{1+v^2}dv \\
        = \frac{1}{\pi}\int_{400\pi}^{1000\pi}\frac{1}{1+v^2}dv \\
        = \frac{1}{\pi}\{\tan^{-1}(1000\pi)-\tan^{-1}(400\pi)\}
$$
