Green Function for $y''(t)-y(t)=f(t)$ I have to solve this diff equation using Green function where f is a particular function. The first step is :
$(d_{t}^2-1)G(t-t_0)=\delta(t-t_0) $
so i put 
$G(t)=(2\pi)^{-1}\int_{R} G(w) e^{iwt} dw $
so i obtain 
$G(w)=- \frac{1}{w^2+1}$
using res theorem we have
$G(t)= -\frac{e^{-|t|}}{2}$
Unfortunately it is not the Green function of $(d_{t}^2-1)$ which is
$G(t)= H(t)\sinh(t)$ where H is Heaviside function.
From the homogeneus solution using the continuity and the jump discontinuity of derivative of Green function i get the correct result.
There is another point not clear to me: in this case the Green function (the form from residue theorem) is defined also for negative value of t because for the negative i have the residue in -i. In other situations i could choose the path of integration so that for negative value of t i don't have singularities so the G(t) is equal to zero.
 A: I disagree that it should be $\sinh(t)H(t)$ and it also appears you are missing a parameter in your solution. I will assume we are looking for solutions which decay at $\pm \infty$ and provide a solution. 
In finding a solution to 
$$
(\partial_{t}^2-1)G(t,t_0)=\delta(t-t_0)
$$
we solve the homogeneous ode's to the right and left of the discontinuity of the delta function. 
Namely, if $t<t_0$ we have 
$$
(\partial_{t}^2-1)G(t,t_0)=0\implies G(t,t_0)=a(t_0)e^t+b(t_0)e^{-t}
$$
identically, if $t>t_0$ we have
$$
G(t,t_0)=c(t_0)e^t+d(t_0)e^{-t}
$$
Imposing the decay conditions we find $b(t_0)=0=c(t_0)$ and the Green's function is given by 
$$
G(t,t_0)=\begin{cases} a(t_0)e^{t},&t<t_0\\
d(t_0)e^{-t},&t>t_0 \end{cases}
$$
Imposing the continuity condition at $t=t_0$ we have 
$$
a(t_0)e^{t_0}=d(t_0)e^{-t_0}
$$
Imposing the Dirac delta behavior at $t=t_0$ of $LG$ we find 
$$
\lim_{\epsilon \to 0}\int_{t_0-\epsilon}^{t_0+\epsilon} \partial_{t}^2G(t,t_0)-G(t,t_0)\mathrm dt=1\\
\implies \lim_{\epsilon\to 0}\int_{t_0-\epsilon}^{t_0+\epsilon}\partial_t^2G(t,t_0)\mathrm dt=1\\
\implies \lim_{\epsilon\to 0}(G_t(t_0+\epsilon,t_0)-G_t(t_0-\epsilon,t_0))=1\\
\implies -d(t_0)e^{-t_0}-a(t_0)e^{t_0}=1
$$
Solving the system yields
$$
d(t_0)=-\frac12 e^{t_0}\\
a(t_0)=-\frac12 e^{-t_0}
$$
and our Green's function is 
$$
G(t,t_0)=\begin{cases} -\frac12e^{t-t_0},&t<t_0\\
-\frac12e^{t_0-t},&t>t_0 \end{cases}\\
=-\frac12e^{-|t_0-t|}
$$
Note that $G(t,t_0)=G(t_0,t)$ and $G$ in fact a convolution operator, as you would expect since your operator is translation invariant.
