Could anyone explicity show the following step? How did the writer change from one integration to a double integration, and when can i do this?

 A: This is not at all a typical move. As @TheChaz commented, the inner integral produces the integrand of the outer integral, by direct integration. It's not about double integrals, just about how to integrate $e^{-tx}$ from $a$ to $b$.
The point of doing this is surely less obvious... Context?
EDIT: recall that $\int_a^b e^{tx}\;dt = {e^{bx}-e^{ax}\over x}$.
Also, the peculiar feature of this general type of integral has a search-able name, "Frullani integral". It is not entirely trivial to justify the conclusion, but what you give is the standard heuristic, indeed.
A: To answer the question,
$$
\int_{a}^{b}e^{(-x)t}dt = \frac{e^{(-x)t}}{-x} \Bigl|^{b}_{a} = \frac{e^{-x(b)} - e^{(-x)a}}{-x} = \frac{e^{-ax} - e^{-bx}}{x} 
$$
So the author just rewrote the right-hand side of the last equation as an integral. As noted, this is a very unusual move.
A: This is a standard form of Frullani's integral.(You can find a proof here and here). Basically, it says that -
$$\int_{0}^{\infty} \frac {f(ax) - f(bx)}{x} dx = \left(f(0) - \lim_{n \to \infty}  f(n)\right) \cdot \ln\left(\frac {b}{a}\right)$$
In your case, $f(x) = e^{-x}.$ Taking the limits we can see that the answer is $\ln(\frac {b}{a}).$
