Laplace Transform of $e^tu(-t)$

Question

What is the Laplace Transform of $e^tu(-t)$.

We got $1/(s-1)$, and we know its wrong.

Answer 1

$$e^{-t}u(t) + e^{t}u(-t)=e^{-|t|} $$ It seems the question is about the two-sided LT $\mathcal{L}(e^{-|t|})$.

$$\int_{-\infty}^{\infty}e^tu(-t)e^{-st}dt=\int_{-\infty}^{0}e^{(1-s)t}dt=-\frac{1}{s-1}$$

$$\int_{-\infty}^{\infty}e^{-t}u(t)e^{-st}dt=\int_{0}^{\infty}e^{-(s+1)t}dt=\frac{1}{s+1}$$

Therefore, $$\mathcal{L}(e^{-a|t|})=\frac{1}{s+1}-\frac{1}{s-1}=\frac{2}{1-s^2}$$