Replacing s with jw in transfer function. When the input function is oscillatory and has exponential decay
\$y(t)=\int_0^\infty h(\tau) u(t-\tau)d\tau  &lt;=&gt; Y(s) = H(s)U(s)\$,

\$H(s)\$ = \$\frac{1}{s^2+2s+2} = \frac{1}{(s+1+i)(s+1-i)}\$ 

Modifing the input \$u(t)\$ so that it is actually a real function (as suggested by Chu).

$$ u(t) = \frac{e^{(-1+i)t} + e^{(-1-i)t}}{2}, $$

$$U(s)= \frac{1/2}{s+1-i} + \frac{1/2}{s+1+i}$$

$$Y(s) = \frac{1}{(s+1+i)(s+1-i)}  \left( \frac{1}{s+1+i}+\frac{1}{s+1-i} \right), $$

$$ =  \frac{1/2}{(s+1+i)(s+1-i)^2} + \frac{1/2}{(s+1+i)^2(s+1-i)}, $$

$$ =  \frac{ \tfrac{1}{2} (s+1+i) }{(s+1+i)^2(s+1-i)^2} + \frac{ \tfrac{1}{2}(s+1-i)}{(s+1+i)^2(s+1-i)^2}, $$

$$ =  \frac{\tfrac{1}{2}/2 (s+1+i+s+1-i) }{(s+1+i)^2(s+1-i)^2}, $$

$$ =  \frac{\tfrac{1}{2} (2s+2) }{(s+1+i)^2(s+1-i)^2}, $$

$$ =  \frac{(s+1) }{(s+1+i)^2(s+1-i)^2}, $$

The denominator of H(s) is \$ (s+1+i)^2(s+1-i)^2 \$, it is not zero.