Let $F(s) = \dfrac{e^{-s}}{s-8}$
Using the residue method, we find that the residue at pole $s = 8$ for the function $F(s)e^{st}$ to be $e^{-8}e^{8t} =e^{8(t-1)} $
Since $f(t) = \sum\limits_k \text{Res}\{F(s)e^{st};p_k\}$ (see reference)
This means $f(t) = e^{8(t-1)}$
However, this is wrong. The correct answer is $f(t) = e^{8(t-1)}u(t-1)$, where $u$ is the heaviside/unit step function
How can I take into account of the unit step using the residue method?
Reference: 1. http://www.staff.city.ac.uk/~george1/laplace_residue.pdf
Similar problem: Inverse Laplace transformation using reside method of transfer function that contains time delay
If $F(s) = \int_{-\infty}^\infty f(t) e^{-st}dt$ converges for $Re(s) > \sigma_0$ with $f(t) e^{-\sigma_0 t} \in L^1$, then (by the Fourier inversion theorem) it is always true that for $\sigma > \sigma_0$ : $$f(t) = \frac{1}{2i\pi}\int_{\sigma-i \infty}^{\sigma+i\infty} F(s) e^{st}ds$$
Now you should look at the proof of $$\int_{\sigma-i \infty}^{\sigma+i\infty} F(s) e^{st}ds = 2i \pi \sum_\rho Res(F(s)e^{st},\rho) \tag{1}$$
It assumes that $F(s)e^{st}$ is holomorphic on $Re(s) \ge \sigma$ (always the case if $\sigma > \sigma_0$), and meromorphic on the rest of the complex plane.
It also requires that $F(s) e^{st} \to 0$ fast enough as $|Im(s)| \to \infty$ so that (by the residue theorem following from the Cauchy integral theorem) $$\int_{\sigma-i \infty}^{\sigma+i\infty}+\int_{\eta+i \infty}^{\eta-i\infty} F(s) e^{st}ds = 2i \pi \sum_{Re(\rho) \in (\eta,\sigma)} Res(F(s)e^{st},\rho)$$
Now if $F(s) e^{st} \to 0$ fast enough as $Re(s) \to -\infty$, you have $\lim_{\eta \to - \infty} \int_{\eta+i \infty}^{\eta-i\infty} F(s) e^{st}ds = 0$ and $(1)$ follows.
But in general this is true only for $t$ sufficiently large, namely $t > t_0$. For $t<t_0$, it is the opposite : if $F(s)e^{st} \to 0$ (fast enough) as $Re(s) \to +\infty$, you get that :
$$\int_{\sigma-i \infty}^{\sigma+i\infty} F(s) e^{st}ds = \lim_{\eta \to +\infty}\int_{\eta-i \infty}^{\eta+i\infty} F(s) e^{st}ds = 0$$