Finding the Laplace transform of a piecewise function

[Another]

Homework Statement

$f(t) = -e^{-t}$ ; $t ≤ 0$ and $f(t) = 0$ ;$t > 0$ find Laplace transform this function.

Homework Equations

Laplace transform
$F(s) = \int_{[-∞<r<+∞]} f(t) e^{-st} dt$

The Attempt at a Solution

$F(s) = \int_{[-∞<r<0]} -e^{-t} e^{-st} dt +\int_{[0<r<+∞]} (0) e^{-st} dt$
$F(s) = \int- e^{-(s+1)t} dt + 0$ ,[-∞<r<0]
$F(s) = \frac{1}{s+1}[e^{-(s+1)t}]$

$e^{-(s+1)(0)}=1$ when t = 0
$e^{-(s+1)t}= 0$ when t = -∞ I'm not sure because t≤0 . when take -∞ to $e^{-(s+1)t} = e^{-∞} = 0$

please check my solution

 

[BvU]

from http://mathworld.wolfram.com/LaplaceTransform.html

The unilateral Laplace transform is almost always what is meant by "the" Laplace transform

 

[Ray Vickson]

Homework Statement

$f(t) = -e^{-t}$ ; $t ≤ 0$ and $f(t) = 0$ ;$t > 0$ find Laplace transform this function.

Homework Equations

Laplace transform
$F(s) = \int_{[-∞<r<+∞]} f(t) e^{-st} dt$

The Attempt at a Solution

$F(s) = \int_{[-∞<r<0]} -e^{-t} e^{-st} dt +\int_{[0<r<+∞]} (0) e^{-st} dt$
$F(s) = \int- e^{-(s+1)t} dt + 0$ ,[-∞<r<0]
$F(s) = \frac{1}{s+1}[e^{-(s+1)t}]$

$e^{-(s+1)(0)}=1$ when t = 0
$e^{-(s+1)t}= 0$ when t = -∞ I'm not sure because t≤0 . when take -∞ to $e^{-(s+1)t} = e^{-∞} = 0$

please check my solution

If $s < -1$ you get $F(s) = 1/(s+1)$. When $s \geq -1$ the Laplace integral diverges.

 

[Another]

from http://mathworld.wolfram.com/LaplaceTransform.html

yes. this problem Forced to use the equation (2) .