Proving that $L\{e^{at}\} = \frac{1}{s-a} s>a$

Question

The question states:
if $f(t)=e^{at}$ find $L\{f(t)\}$.

Solution
$$\begin{align} L\{e^{at}\} &= \int^\infty_0e^{-st}\cdot e^{at}\\ &=\lim_{B\rightarrow\infty}\int^B_0e^{-t(s-a)}dt\\ &=\lim_{B\rightarrow\infty}\left[\frac{e^{-t(s-a)}}{-(s-a)}\right]^B_0\\ &=\lim_{B\rightarrow\infty}\left(\frac{e^{-B(s-a)}-1}{-(s-a)}\right)\\ &=\frac{1}{s-a} &s>0 \end{align}$$

My questions are:
- How do we go from line 2 to line 3?
Where does the "$-(s-a)$" come from in the denominator in line 3? This must be some kind of lookup, no?
- And then, line 3 to line 4?
I see that we are substituting t for B but where does the "$-1$" come from in the numerator in line 4?

Answer 1

Hints:

What is the $\int e^{-w t} dt$?

Now, substitute $w = s-a$.

What is $e^{0}$ and the negative sign comes from the FTC using the lower limit of integration.

Is it clear now?