We expand in small $T$.
Note that $e^x = 1+x+O(x^2)$.
We have
\begin{align*}
\cdots &= \exp\left(-\int_0^T (r_t+h_t) dt\right)
+ \int_0^T RR \cdot h_t
\exp\left(-\int_0^t(r_s+h_s) ds\right) dt \\
&=
\left(1 - \int_0^T (r_t+h_t) dt + O(T^2)\right)
+ \int_0^T RR\cdot h_t
\left(
1 - \underbrace{\int_0^t(r_s+h_s)ds}_{O(t)} + O(t^2)
\right)dt \\
&= 1 - \int_0^T(r_t+\underbrace{(1-RR)h_t)}_{s_t}dt
+ O(T^2) \\
&= \exp\left(
- \int_0^T(r_t+s_t)dt
\right) + O(T^2)
\end{align*}
How well this approximation works depends on the specific form for $r_t$ and $h_t$.
We assume that $r_t$ and $h_t$ are analytic on some open interval centered on $t=0$ and containing $[0,T]$.
Generally corrections are $O(T^2)$.
If $h_0=0$, corrections are $O(T^3)$.
If $r_0=h_0=0$, corrections are $O(T^4)$.
Define $ G(x) $ with derivative $g(x)$ such that (assume $\zeta\phi\rho=b$)
$$
G(x)=
\begin{array}{cc}
\left\{
\begin{array}{cc}
0 & x< 0 \\
1-F(x) &x\ge0
\end{array}
\right.
\end{array}
$$
$$ \implies g(x)= -f(x) $$
Now $\log_2(1+x)=\log_2(e)\cdot\log_e(1+x)$
Taking away $ \frac{-\log_2(e)}{2} $ the integral reduces to
$$ I = \int_0^\infty \log (1+x) g(x) \,dx $$
We evaluate the improper integral, by taking the limit of the definite integral
$$ I = \lim \limits_{y\to\infty} \left[ \int_0^y \log (1+x) g(x) \,dx \right] $$
Simplifying by parts
[EDIT: This is where the utility of introducing an auxiliary function is most useful. Due to the definition of $G(x)$, the indefinite integral of its derivative $g(x)$ is identical to it. Hence the integral evaluation by parts is elegant.]
$$ I = \lim \limits_{y\to\infty} \left[ G(y)\log(1+y)-G(0)log(1) - \int_0^y \frac{G(x)}{1+x} \,dx \right] $$
$$ I = \lim \limits_{y\to\infty} G(y)\log(1+y) - \int_0^\infty \frac{G(x)}{1+x} \,dx $$
By application of L'hospital rule, we can evaluate the limit to be zero.
Hence, $I$ reduces to
$$ I =-\int_0^\infty \frac{G(x)}{1+x} \,dx = -\int_0^\infty \frac{e^{-\frac{x}{b}}}{(1+x)(1+ax)} \,dx $$
$$ = \frac{1}{a-1} \int_0^\infty \frac{(1-a)e^{-\frac{x}{b}}}{(1+x)(1+ax)} \,dx = \frac{1}{a-1} \int_0^\infty \left(\frac{1}{1+x}-\frac{a}{1+ax}\right) e^{-\frac{x}{b}} \,dx $$
Multiplying with $ \frac{-\log_2(e)}{2} $, we get result (2)
Best Answer
You need to consider the domain of integration. This is a triangle in the $z-s$ plane, bounded by $s=t$, $z=T$ and $z=s$. When you exchange integrals you need to re-describe this region so the outer integral limits do not depend on the inner integral variable of integration. Specifically: $$\int_t^T\Big(\int_s^T f(z, s) dz\Big) ds=\int_t^T\Big(\int_t^z f(z, s) ds\Big) dz.$$ I hope this is enough for you to complete the integral yourself.