Laplace Transform of a polynomial

Now we are getting to the fun, here is a key result that we will need to know about differentiating a power multiplied with an exponential.

$ \begin{eqnarray*} \frac{d}{dt}t^n e^{-at} &=& nt^{n-1}e^{-at} - at^ne^{-at} \end{eqnarray*} $

The fun begins with integrating both sides:

$ \begin{eqnarray*} \int{\frac{d}{dt}t^n e^{-at}dt} &=& \int{(nt^{n-1}e^{-at} - at^ne^{-at})dt} \end{eqnarray*} $

To make life simple we denote $ I(n) = \int{t^ne^{-at}dt} $, so we have:

$ \begin{eqnarray*} t^n e^{at} &=& nI(n-1) - aI(n) \\ I(n) &=& \frac{n}{a}I(n-1) -\frac{1}{a}t^n e^{-at} \end{eqnarray*} $

We could have just expanded this, but the notation is so cubersome, let consider a simpler problem below like this:

$ \begin{eqnarray*} A(n) &=& npA(n-1) + qt^n \\ &=& \frac{n!p}{(n-1)!} A(n-1) + q\sum\limits_{k=0}^{0} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& \frac{n!p}{(n-1)!} ((n-1)pA(n-2) + qt^{n-1}) + q\sum\limits_{k=0}^{0} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& \frac{n!p}{(n-1)!} (n-1)pA(n-2) + q\frac{n!p t^{n-1}}{(n-1)!} + q\sum\limits_{k=0}^{0} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& \frac{n!p^2}{(n-2)!} A(n-2) + q\sum\limits_{k=0}^{1} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& ... \\ &=& \frac{n!p^n }{(n-n)!} A(n-n) + q\sum\limits_{k=0}^{n-1} \frac{n!p^k t^{n-k}}{(n-k)!} \\ &=& p^n n! A(0) + q\sum\limits_{k=0}^{n-1} \frac{n!p^k t^{n-k}}{(n-k)!} \end{eqnarray*} $

Imagine how much more complicated if we have not used the substitution. Now substituting back we have this:

$ \begin{eqnarray*} I(n) &=& \frac{n}{a}I(n-1) -\frac{1}{a}t^n e^{-at} \\ &=& (\frac{1}{a})^n n! I(0) -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!(\frac{1}{a})^kt^{n-k}}{(n-k)!} \\ &=& (\frac{1}{a})^n n! \frac{e^{-at}}{-a} -\frac{1}{a}e^{-at}\sum\limits_{k=0}^{n-1} \frac{n!t^{n-k}}{a^k(n-k)!} \\ \end{eqnarray*} $

I know, the last summation term is ugly, bear with me, we will kill it, by taking $ t \to \infty $ and $ t \to 0 $. Using l'hopital's rule, we can write

$ \begin{eqnarray*} \lim_{t \to \infty}{t^n e^{-at}} &=& \lim_{t \to \infty}{\frac{t^n}{e^{at}}} \\ &=& \lim_{t \to \infty}{\frac{nt^{n-1}}{ae^{at}}} \\ &=& ... \\ &=& \lim_{t \to \infty}{\frac{n!}{a^ne^{at}}} \\ &=& 0 \end{eqnarray*} $

So as promised, the ugly terms goes away, and therefore:

$ \begin{eqnarray*} \mathcal{L}(t^n) &=& \int\limits_{0}^{\infty}{t^ne^{-st}dt} \\ &=& I(n)|_{0}^{\infty} \\ &=& \frac{n!}{s^{n+1}} \end{eqnarray*} $

QED

Laplace Transform of complex exponential and its consequences

There is nothing blocking us from making the $ a $ in the previous post complex, so we get this Laplace transform pair for free!


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

This, of course, imply $ |a - s| < 0 $. Interesting things happen when we use the Euler's formula. Let $ a = r + i \theta $, we have $ e^{at} = e^{rt}(\cos \theta t + i \sin \theta t) $. Plugging this into the formula above we have

$ \begin{eqnarray*} \mathcal{L}(e^{rt}(\cos \theta t + i \sin \theta t)) &=& \frac{1}{s-a} \\ \mathcal{L}(e^{rt}(\cos \theta t)) &=& \operatorname{Re}(\frac{1}{s-a}) \\ &=& \operatorname{Re}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Re}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Re}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{s-r}{(s-r)^2 + \theta^2} \\ \mathcal{L}(e^{rt}(\sin \theta t)) &=& \operatorname{Im}(\frac{1}{s-a}) \\ &=& \operatorname{Im}(\frac{1}{s-(r + i \theta))}) \\ &=& \operatorname{Im}(\frac{1}{s - r - i \theta}) \\ &=& \operatorname{Im}(\frac{s - r + i \theta}{(s - r - i \theta)(s - r + i \theta)}) \\ &=& \frac{\theta}{(s-r)^2 + \theta^2} \\ \end{eqnarray*} $

Of course we can further specialize so that $ r = 0 $, and in that case we will have

$ \begin{eqnarray*} \mathcal{L}(\cos \theta t) &=& \frac{s}{s^2 + \theta^2} \\ \mathcal{L}(\sin \theta t) &=& \frac{\theta}{s^2 + \theta^2} \end{eqnarray*} $

See now little we need to come up with so many results!

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 $.

Laplace Transform of 1

This is the easiest Laplace Transform I can think of

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

Introduction to Laplace Transform

I have been working on a few problems around Laplace Transform on Yahoo Answer (in fact, the one in Hong Kong/Taiwan). It would be a waste for those content to only post there, so I am posting them here. Feel free to checkout my profile there if you are interested

This post is about Laplace Transform, the Laplace Transform of a function can be simply written as $ Y(s) = \mathcal{L}(y) = \int\limits_{0}^{\infty}{y(t)e^{-st}dt} $.

There are a few things to note:

1. $ s $ is in general a complex number, $ t $ is usually real.
   
2. The integral is improper, in particular, by writing $\int\limits_{0}^{\infty}{y(t)e^{-st}dt} $, we really mean $ \lim_{T \to \infty}{\int\limits_{0}^{\infty}{y(t)e^{-st}dt}} $.
   
3. When necessary, we pick ranges of $ s $ such that the limit converges and define $ Y(s) $ there only.

That's it for this post. Next we will talk about some common Laplace Transform pair and properties.