Lets pass a pure frequency in continuous-time through the convolution integral:

\begin{align*}
y(t) &= \int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau \\
y(t) &= \int_{-\infty}^{\infty}h(\tau)e^{i \omega (t-\tau)}d\tau \\
y(t) &= \int_{-\infty}^{\infty}h(\tau)e^{i \omega t}e^{-i \omega \tau}d\tau \\
y(t) &= \left( \int_{-\infty}^{\infty}h(\tau)e^{-i \omega \tau}d\tau\right)e^{i \omega t} \\
y(t) &= H(\omega)e^{i \omega t} \\
\end{align*}

In CT (and DT), an LTI system cannot create new frequencies. The golden rule still holds:

\begin{align*}
x & \to \fbox{h} \to y \\
e^{i\omega t} & \to \fbox{h} \to H(\omega)e^{i\omega t} 
\end{align*}

Thus, \emph{the continuous-time frequency response} is defined as $$H(\omega) = \int_{-\infty}^{\infty}h(\tau)e^{-i\omega \tau}d\tau$$

Some important distinctions between the DT and CT frequency response:

\begin{enumerate}
\item In discrete-time, we have the $2\pi$ ambiguity:
$$H(\omega + 2\pi)=\sum \limits_{n\in Z} h(n) e^{-i\omega n} e^{-i2\pi n} = \sum \limits_{n\in Z} h(n) e^{-i\omega n} = H(\omega)$$
because $n$ is an element of $\Z$.
\item In continuous-time, we have a different story:
$$ H(\omega +2\pi) = \int_{-\infty}^{\infty}h(t)e^{-i\omega t}e^{-i2\pi t}dt \neq H(\omega)$$
because $t$ is an element of $\R$.
\end{enumerate}

\begin{example}

Given an impulse response $h(t) = e^{-\alpha t} u(t)$, where $\alpha \gt 0$, graph the magnitude response and solve for the frequency response.

\begin{align*}
h(t) &= e^{-\alpha t} u(t)\mbox{} \quad \alpha>0 \\
H(\omega) &= \int_{-\infty}^{\infty} h(t)x(t)dt \\
H(\omega) &= \int_{-\infty}^{\infty} \left( e^{-\alpha t} u(t) \right)e^{i \omega t}dt \\
H(\omega) &= \int_{0}^{\infty} e^{-\left(i \omega + \alpha\right)t}dt \\
H(\omega) &= \left. \frac{e^{-\left(i \omega + \alpha\right)t}}{-\left(i \omega + \alpha\right)} \right|_{0}^{\infty} \\
\end{align*}

Since $\alpha \gt 0$, as $t$ approaches $\infty$, $e^{-\left( i\omega + \alpha\right)t} $ approaches $0$.
\begin{align*}
H(\omega) &= \lim_{t \to \infty} \left( \frac{e^{-\left(i \omega + \alpha\right)t}}{-\left(i \omega + \alpha\right)}\right) - \left( \frac{e^{0}}{-\left(i \omega + \alpha\right)} \right) \\
H(\omega) &= 0 - \left( \frac{1}{-\left(i \omega + \alpha\right)} \right) \\
H(\omega) &= \frac{1}{i \omega + \alpha} \\
\end{align*}

\end{example}

A good way to graph this is to write the equation as $\frac{1}{i\omega - (-a)}$, and analyze the graph of $i \omega - (-a)$

CT Frequency Response