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Wednesday, April 15, 2015

Laplace Transform of exponential

Now let's go slightly more complicated:


$ \begin{eqnarray*} \int\limits_{0}^{\infty}{e^{at}e^{-st}dt} &=& \int\limits_{0}^{\infty}{e^{(a-s)t}dt} \\ &=& \frac{{e^{(a-s)t}dt}}{(a-s)}|_{0}^{\infty} \\ &=& \frac{1}{s-a} \end{eqnarray*} $

Note that in the third equal sign above, we choose $ s \lt a $ so that we $ t \to \infty $, the exponential converge to $ 0 $.

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