Why the Laplace transform of u(-t) is 1/s? Yesterday I had my first contact with the Laplace transform, in an Electric Circuits class. $\mathcal{L} \left\{ u \left( -t \right) \right\}$ showed up. Our teacher said it was equal to $\frac{1}{s}$, but couldn't tell exactly why.
I want to know if it's possible to explain why $\mathcal{L} \left\{ u \left( -t \right) \right\} = \frac{1}{s}$ by using known properties and transforms, like those found in Wikipedia's page about Laplace transform, as if I was trying to do it myself, considering I learned the very basics.
No, it's not for an assignment.
Thanks in advance.
Edit: $u \left( t \right)$ is the unit step function. $u \left( -t \right)$ was used because it was needed reversed.
 A: I realize that this is an older question, but I'll answer it anyway in case other people are looking for the answer.
As an electrical engineer, a Laplace transform usually references the two-sided Laplace transform. Therefore the function being transformed does not have to be causal, meaning that the function can equal something other than $0$ for $t < 0$. (Here's a wiki reference to the two-sided Laplace transform: https://en.wikipedia.org/wiki/Two-sided_Laplace_transform )
Using this version of the Laplace transform, 
$$\mathcal{L}\{ u(-t)\}(s) = \int_{-\infty}^\infty u(-t)e^{-st} \ dt = \int_{-\infty}^0 e^{-st} \ dt = -\frac{1}{s}$$
This is true only when $Re[s] < 0$, which is this transform's region of convergence.
I hope this helps!
Sources: I'm an electrical engineering student.
A: This looks wrong to me. Assuming by $u$ you mean the unit step function $u(t) = 1$ if $t \geq 0$ and $u(t) = 0$ otherwise, then by definition of the Laplace transform
$$\mathcal{L}\{ u(-t)\}(s) = \int_0^\infty u(-t)e^{-st} \ dt = \int_0^\infty 0 \ dt = 0$$
as for all $t \in [0,\infty)$, $u(-t) = 0$.
It is however true by the same sort of calculation that $\displaystyle \mathcal{L}\{ u(t)\}(s) = \frac{1}{s}$, as
$$\mathcal{L}\{ u(t)\}(s) = \int_0^\infty u(t)e^{-st} \ dt = \int_0^\infty e^{-st} \ dt = \frac{1}{s}$$
A: I suspect you have a typo in what you posted since $u(-t)=0$ for $t>0$. Laplace transforms assume the underlying function is causal. Otherwise, computing the Laplace transform of the zero function is trivial.
Now, if you mean $u(t)$ (as I suspect) which is the unit step function, just compute its Laplace transform straightaway from the definition:
$$
\mathscr{L}\{u(t)\}=\int_0^\infty e^{-st}u(t)\,dt=\int_0^\infty e^{-st}\cdot 1\,dt={e^{-st}\over -s}\Bigg|_{t=0}^{t=\infty}=-{1\over s}(0-1)={1\over s}, \quad s>0.$$
A: Well he is wrong, the fact that he couldn't say why is a reason to mistrust his statement. You could come up with that result by using timescaling rule:
$\mathcal{L} f(a\cdot) = F(s/a)/a$
which with $a=-1$ would lead to the result that your teacher claims, but this rule assumes that $a>0$ so it can't be used. The reason the rule doesn't hold is that you in the process reverses the integration limits. If you put it into the formula you get:
$\mathcal{L} u(-\cdot) = \int_{-\infty}^{+\infty} u(-t)e^{-st}dt = \int_{-\infty}^0e^{-st}dt = [-s^{-1}e^{-st}]_{-\infty}^0 = -s^{-1}$ (if $\Re s < 0)$
