LTI system output given input and frequency response The question I'm trying to understand is as follows: A linear time-invariant continuous-time system has the frequency response function $$H(\omega)=\frac{1}{j\omega+1} $$

Compute the output response $y(t)$ for $-\infty&lt;t&lt;\infty$ when the input $$x(t)=\cos(t),\quad -\infty&lt;t&lt;\infty$$

I know the answer, but I don't understand how to get there.

I know $H(\omega)=\frac{Y(\omega)}{X(\omega)}$, and from that $y(t)=h(t)*x(t)$. Taking the inverse Fourier transform of $H(\omega)$ I get $e^{-t}$. If I try to compute the convolution I just get a divergent integral. Is this wrong? The answer is supposedly $$y(t)=|H(1)|\cos(t+\angle H(1))$$ but this seems to come from nowhere. Can someone please explain?
Apply the eigenfunction property of the LTI system after decomposing the sinusoidal input by Euler identity. 

The eigenfunction property of the LTI system states that 

$$ x(t) = e^{j \omega_0 t} \implies y(t) = H(\omega_0) e^{j \omega_0 t} $$

where $H(w)$ is the frequency response of the LTI system.

Euler identity states that :

$$ x(t) = \cos(t) = 0.5 \{ e^{jt} + e^{-jt} \} $$

Then due to system being LTI, you can define the output as:

$$ y(t) = 0.5 \{ H(1) e^{jt} + H(-1) e^{-jt} \} .$$

Since the system impulse response was a real function, its frequency response $H(w)$ will be conjugate symmetric; i.e., $H(w) = H^*(-w)$.

Let's call $H(1) = A e^{j\phi} $ where $A = |H(1)|$ is the magnitude and $\phi = \angle{H(1)}$ is the phase of the frequency response at the frequency $\omega = 1$. 

Hence the output will be :
$$ y(t) = 0.5 \{ H(1) e^{jt} + H^*(1) e^{-jt} \} $$

consider the term $z = H(1)e^{jt}$  as a complex variable, then the second term will be $z^*$, hence the output is:

$$y(t) = 0.5 \{ z + z^* \} = \mathcal{Re} \{ z \} $$

$$y(t) = \mathcal{Re} \{ A e^{j\phi} e^{jt} \} = A \cos(t + \phi) = |H(1)| \cos(t + \angle{H(1)})$$