Question: How to evaluate $\int\_t^\infty e^{-sx}f(x)dx$ using laplace transform properties?

Answer: 
You can interpret that integral as the laplace transform of $f$ multiplied with a shifted step function $H$. Since a multiplication in the time domain maps to a convolution in the frequency domain, and since the laplace transform of $H(t-a)$ is $\frac{e^{-as}}{s}$ you get $$
\int\_t^{\infty} e^{-sx}f(x)dx = \int\_0^\infty e^{-sx} f(x)H(x-t) dx = \frac{1}{2\pi i}\int\_{\sigma-i\cdot\infty}^{\sigma+i\cdot\infty}F(u)\frac{e^{-t(s-u)}}{s-u} du 
$$
where $$\begin{eqnarray}
 F(s) &=& \int\_0^\infty e^{-sx}f(x)dx \\
 H(x) &=& \begin{cases} 1 & \text{if } x \geq 0 \\ 0 & \text{otherwise}\end{cases} 
\end{eqnarray}$$
and $\sigma$ lies within convergence region of $F$.
$$ \text{Unit Step Fuction : } \begin{eqnarray}
 u(t-c) \begin{cases} 1 & \text{if } t > c \\ 0 & \text{otherwise}\end{cases} 
\end{eqnarray} $$
$$ \mathcal{L} \left[ f(t-c) u(t-c) \right]=e^{-cs}F(s)$$
$$ \mathcal{L} \left[ f(t) u(t-c) \right]=e^{-cs} \mathcal{L} \left[ f(t+c)\right]$$
$$ \int\_t^\infty e^{-sx}f(x)dx \space \triangleq \space \mathcal{L} \left[ f(t) u(t-c) \right] $$