In the laplace transform of 1, why is the s in the denominator?

Question

Ive been using the Laplace transform for a little while now for some electrical engineering differential equations, what I have never quite understood, is why is the $s$ in the denominator when you do the Laplace transform of 1. ie. this $\mathcal{L}\{1\}=\int_0^\infty{e^{-st}dt} = [-\frac{e^{-st}}{s}]_0^{\infty}$ how did the $s$ get into the denominator? Sorry if I am missing something obvious here. I thought it should be $[-s\cdot e^{-st}]_0^\infty$ using the chain rule.

Answer 1

The derivative of $e^{-st}$ is $-se^{-st}$. But what you want is the integral, and this is $\frac{e^{-st}}{-s}$.

Answer 2

Bear in mind that $s$ is a constant with respect to this integration, and we know that, for nonzero constants $a$,

$$\int e^{at}dt = \frac{e^{at}}{a} + C$$

Just differentiate with respect to $t$ to see that

$$\frac{d}{dt} \left( \frac{e^{at}}{a} \right) = a \cdot \frac 1 a \cdot e^{at} = e^{at}$$