Laplace of $e^{-s+3}$?

Question

I understand that the $e^{-ks}$ will time shift the function in the time domain by $k$ and will result in a time function of $u(t-3)$, but what does it mean when you have $e^{-s-k}$? How will that affect the function?

Answer 1

$$\mathcal{L}\left(e^{ax+k}\right)=e^k\mathcal{L}\left(e^{ax}\right)=e^k\frac{1}{s-a}$$ because the constant $e^k$ factors out by linearity of the transform.

Answer 2

If you meant find the Laplace of $e^{-t+3}$, instead of $e^{-s+3}$, then hope this helps. \begin{align} \mathcal{L}\{e^{-t+3}\}&=\int_{0}^{\infty} e^{-st}(e^{-t+3}) \, dt \\ &= \int_{0}^{\infty} e^{-t(s+1)}e^3 \, dt \\ &=e^3 \int_{0}^{\infty} e^{-t(s+1)} \cdot (1) \, dt & e^3 \text{ is a constant value; move it outside the integral} \\ &=e^3 \mathcal{L}\{1\}_{s \rightarrow s+1} \\ &=e^3 \frac{1}{s+1} \end{align} Normally, $\mathcal{L}\{1\}=\frac{1}{s}$, but note the fact that we are replacing the variable $s$ with $s+1$. Hence, $\mathcal{L}\{1\}_{s \rightarrow s+1}=\frac{1}{s+1}$.