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

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@Litum: Do you have any information about $t$? –  Babak S. Oct 10 '12 at 12:53
    
Do you mean you want to express that integral in terms of the laplace transform of $f$? –  fgp Oct 10 '12 at 14:38
    
@fgp: I think if $t$ be the period of $f(x)$ then we could write some calculations. –  Babak S. Oct 10 '12 at 15:15
    
@BabakSorouh Who says that $f$ is periodic? It think what you might be looking for is the relationship between the laplace transform of $f$ and that of $f$ shifted by $t$. –  fgp Oct 10 '12 at 15:47
    
@fgp: Nobody says. It seems that we can only find a relationship between the laplace transform of f and ...as you noted. Doesn't it? –  Babak S. Oct 10 '12 at 16:08

2 Answers 2

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 $F$.

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What are the differences between LST and LT? –  Litun Oct 12 '12 at 9:28

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

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