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
$ \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
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