How to solve the following integral equation? $g(s) = \frac{\lambda A}{1-e^{-\lambda T}}s$ for $s\in [0, T]$. 
$g(s) = \int_{0}^{T} \lambda e^{-\lambda t} g(s-t) dt + A$. for $s \ge T$. 
$T, A, \lambda$ are constant. I want to get the closed-form of $g$ for $s > T$. 
 A: I think we have to solve it iteratively. First on [T,2T] and then on larger intervals with the same procedure. 
For $\epsilon\in [0,T] $ we have
$$
g(T+\epsilon) = \int_{\epsilon}^T \lambda e^{-\lambda t} g(T+\epsilon-t) dt + \int_0^\epsilon \lambda e^{-\lambda t} g(T+\epsilon-t) dt + A
$$
Since you have a explicit and nice representation for $g$ on $[0,T]$ that 
you can insert into the first integral,
the first integral can be explicitly computed, say it has the value $f(\epsilon)$. Its derivative can be also explicitly computed.
Writing $h(\epsilon) = g(T+\epsilon)$ and differentiating yields
$$
h'(\epsilon) = f'(\epsilon)+  \lambda e^{\lambda\epsilon }h(0) + \int_0^\epsilon 
\lambda e^{-\lambda t} h'(\epsilon-t) d t.
$$
Integration by parts yields
$$
h'(\epsilon) =f'(\epsilon)+  \lambda h(\epsilon) 
-\lambda \int_0^\epsilon 
\lambda e^{-\lambda t} h(\epsilon-t) dt.
$$
Inserting the first equation yields
$$
h'(\epsilon) = f'(\epsilon)
 + \lambda f(\epsilon) + \lambda A, h(0) =  g(T),
$$
so that the fundamental theorem of calculus implies
$$
g(T+\epsilon ) = h(\epsilon)= g(T) +  f(\epsilon) -f(0)  + \lambda \int_0^\epsilon f(s) d s +  \lambda A \epsilon.
$$
