How to find Laplace Transform

Question

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

Answer 1

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 $F$.

Answer 2

$$ \text{Unit Step Fuction : } \begin{eqnarray} u(t-c) \begin{cases} 1 & \text{if } t > c \\ 0 & \text{otherwise}\end{cases} \end{eqnarray} $$ $$ \mathcal{L} \left[ f(t-c) u(t-c) \right]=e^{-cs}F(s)$$ $$ \mathcal{L} \left[ f(t) u(t-c) \right]=e^{-cs} \mathcal{L} \left[ f(t+c)\right]$$ $$ \int_t^\infty e^{-sx}f(x)dx \space \triangleq \space \mathcal{L} \left[ f(t) u(t-c) \right] $$