2
$\begingroup$

By definition, I know the Laplace transform of $e^{at}$ is $\int_0^\infty e^{at} e^{-st}\ dt = \frac{1}{s-a}$. I understand this fully. Now the book I'm reading derives the Laplace transform of $te^{at}$ as follows:

Taking the derivative of both sides with respect to $a$ of the previous integral:

$\int_0^\infty te^{at}e^{-st}\ dt = \frac{d}{da}(\frac{1}{s-a}) = \frac{1}{(s-a)^2}$.

It seems that on the left hand side, the derivative of the integral is equal to the integral of the derivative. This, however, does not seem to be a case of the Fundamental Theorem of Calculus. Is this the case because integrals, like derivatives, are linear operators and the integral is with respect to $t$ and not $a$?

I'm trying to teach myself differential equations and any help and understanding would be much appreciated.

$\endgroup$
4
  • 3
    $\begingroup$ See Leibniz's rule en.wikipedia.org/wiki/Leibniz_integral_rule. $\endgroup$
    – copper.hat
    Commented May 13, 2017 at 3:44
  • 1
    $\begingroup$ Thanks ... that was quick. $\endgroup$
    – lhoernle
    Commented May 13, 2017 at 3:49
  • 1
    $\begingroup$ @copper.hat Joe, that version of Leibniz's Rule does not apply to an improper integral. A sufficient condition is that the integral of the derivative uniformly converges. We can easily show that $\int_0^\infty te^{at}e^{-st}\,dt$ converges uniformly for $s-a\ge \delta>0$. $\endgroup$
    – Mark Viola
    Commented May 13, 2017 at 4:34
  • $\begingroup$ @Dr.MV: Thanks for catching that Mark, that was sloppy of me (a frequent occurence!). I will add an elaboration in the morning. $\endgroup$
    – copper.hat
    Commented May 13, 2017 at 4:46

2 Answers 2

3
$\begingroup$

Let $F(s)=\int_0^\infty e^{at}e^{-st}\,dt$. Then, we have

$$\begin{align} \left|\frac{F(s+\Delta s)-F(s)}{\Delta s}-\int_0^\infty -te^{at}e^{-st}\,dt\right|&=\left|\int_0^\infty te^{at}\left(\frac{e^{-(s+\Delta s)t}-e^{-st}}{(\Delta s)t}+1\right)\right|\\\\ &\le \int_0^\infty te^{-(s-a)t}\left|\frac{e^{-(\Delta s)t}-1+(\Delta s)t}{(\Delta s)t}\right|\,dt\\\\ &\le \int_0^\infty te^{-(s-a)t}\left(\frac12 (\Delta s)t\right)\,dt\\\\ &=(\Delta s)\left(\frac12\int_0^\infty t^2\,e^{-(s-a)t}\,dt\right)\\\\ &\to 0\,\,\text{as}\,\,\Delta s\to 0 \end{align}$$

Hence, we have

$$F'(s)=-\int_0^\infty te^{at}e^{-st}\,dt=\int_0^\infty e^{at}\frac{d}{ds}e^{-st}\,dt$$

as was to be shown!

$\endgroup$
1
  • $\begingroup$ @copper.hat Thank you Joe! Much appreciative. I could have appealed to a Theorem on differentiating under the integral for improper Riemann integrals. But this is clean and more elementary. $\endgroup$
    – Mark Viola
    Commented May 13, 2017 at 5:10
2
$\begingroup$

If you also know that

  1. $\mathcal{L}\{c_1y_1+c_2y_2\}=c_1\mathcal{L}\{y_1\}+c_2\mathcal{L}\{y_2\}$ and
  2. $ \mathcal{L}\{y^\prime\}=s\mathcal{L}\{y\}-y(0)$

then you may find $\mathcal{L}\{te^{at}\}$ as follows:

\begin{eqnarray} \mathcal{L}\{(te^{at})^\prime\}&=&s\mathcal{L}\{te^{at}\}-0\\ \mathcal{L}\{e^{at}+ate^{at}\}&=&s\mathcal{L}\{te^{at}\}\\ \mathcal{L}\{e^{at}\}+a\mathcal{L}\{te^{at}\}&=&s\mathcal{L}\{te^{at}\}\\ \mathcal{L}\{e^{at}\}&=&s\mathcal{L}\{te^{at}\}-a\mathcal{L}\{te^{at}\}\\ \frac{1}{s-a}&=&(s-a)\mathcal{L}\{te^{at}\}\\ \mathcal{L}\{te^{at}\}&=&\frac{1}{(s-a)^2} \end{eqnarray}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .