What's the significance of negative exponents in Laplace transformations

[Integrate]

In the photo above why can't leave it as e^t(a-s)?

Why do we need to take out a negative to convert it to e^-t(s-a)

 

[MaxWong]

You can leave it as [imath]e^{t(a-s)}[/imath].

[imath]e^{t(a-s)}[/imath] and [imath]e^{-t(s-a)}[/imath] are just the same thing, and will surely give you the same result.

 

[mario99]

View attachment 37975

In the photo above why can't leave it as e^t(a-s)?

Why do we need to take out a negative to convert it to e^-t(s-a)

As MaxWong said, while solving the integral it does not matter which structure you use. But for convenience, it is better to make it look like [imath]e^{-(s - a)t}[/imath] for two reasons. The first reason is that the transformation answer will involve the expression [imath](s - a)[/imath] rather than [imath](a - s)[/imath]. The second reason is that it is some kind of a reminder to tell you that [imath]s > a[/imath].

Another way to see this is to look at this function, [imath]e^{-at}[/imath]. The convenient structure will be [imath]e^{-(s + a)t}[/imath] because the answer is [imath]\displaystyle \frac{1}{s + a}[/imath] which involves the expression [imath](s + a)[/imath].

 

[Integrate]

As MaxWong said, while solving the integral it does not matter which structure you use. But for convenience, it is better to make it look like [imath]e^{-(s - a)t}[/imath] for two reasons. The first reason is that the transformation answer will involve the expression [imath](s - a)[/imath] rather than [imath](a - s)[/imath]. The second reason is that it is some kind of a reminder to tell you that [imath]s > a[/imath].

Another way to see this is to look at this function, [imath]e^{-at}[/imath]. The convenient structure will be [imath]e^{-(s + a)t}[/imath] because the answer is [imath]\displaystyle \frac{1}{s + a}[/imath] which involves the expression [imath](s + a)[/imath].

This was a very helpful answer. Thank you.