To compute expected exit times from intervals of the form $(a,b)$ for one dimensional diffusions $X = (X_t)_{t \geq 0}$ started at $x \in (a,b)$, one can use that $$ E \tau_{a,b} = \int_a^b G_{a,b}(x,y) m(dy) \\ = - \int_a^x \left( p(x) - p(y) \right) m(dy) + \frac{p(x) - p(a)}{p(b) - p(a)} \int_a^b (p(b) - p(y)) m(dy), $$ where $p(x)$ is the scale function and $m(x)$ is the speed measure of the diffusion $X$, and $\tau_{a,b} = \inf\{ t \geq 0 :\: X_t \notin (a,b) \}$, with Green's function being defined as $$ G_{a,b}(x,y) = \frac {( p(x \wedge y) - p(a) ) ( p(b) - p(x \vee y )} { p(b) - p(a) }. $$ This is from [1] p. 343.
For example, if $X$ is a standard Brownian motion, then the above yields the well-known identity $E \tau_{a,b} = (b-x)(x-a)$. Here the scale function is $p(x) = x$ and the speed measure is $m(dx) = 2dx$.
I would like to use Green's function found in [2]. For the Brownian motion, using [2], p. 119, Green's function is $$ G_\alpha(x,y) = w_\alpha^{-1} e^{- \sqrt{2 \alpha} x} e^{\sqrt{2 \alpha y}}, \qquad x \geq y, $$ where $w_\alpha = 2 \sqrt{2 \alpha}$ is the Wronskian, and I can compute $$ \int_a^b G_\alpha(x,y) m(dy) = \frac {\left( -e^{\sqrt{2 \alpha} a} + e^{\sqrt{2 \alpha} b} \right) e^{- \sqrt{2 \alpha} x}} {2 \alpha}, $$ but this is not something I recognize.
What is the relationship between Green's function in [1] (which yields the desired result) and Green's function in [2]? How may I use Green's function in [2] to compute $E \tau_{a,b}$ and what is the role of $\alpha$?
[1] Karatzas & Shreve: Brownian Motion and Stochastic Calculus (1998).
[2] Borodin & Salminen: Handbook of Brownian Motion Facts and Formulae (2002).
Edit: According to John Dawkins, we should use the correct Green function below: $$ G_\alpha(x,y) = w_\alpha^{-1} \sinh ((b-x)\sqrt{2 \alpha}) \sinh ((y-a) \sqrt{2 \alpha}), \qquad b > x \geq y > a, $$ where $w_\alpha = \sqrt{2 \alpha} \sinh((b-a) \sqrt{2 \alpha})$ is the Wronskian. Then we let $$ G_0(x,y) := \lim_{\alpha \to 0} G_\alpha(x,y) = \frac{ab - ax - by + xy}{a-b}, \qquad b > x \geq y > a $$ and compute $$ \int_a^b 2 G_0(x,y) dy = \frac{(b-x)(a-x)^2}{-(a-b)}, $$ which is not quite $(b-x)(x-a)$. Where did I go wrong?
The formula you have taken from [2] is for the $\alpha$-order Green function, the density of the $\alpha$-resolvent with respect to $m$, for Brownian motion on all of $\Bbb R$. What you want is the formula in [2], in the later section 6, for "Brownian motion on $(a,b)$ killed at $a$ or $b$", on page 137 in the Second Edition. Let $\alpha$ go to $0$ in that formula and you'll get the same formula as in [1].