Deriving the Laplace transform of $te^{at}$ 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.
 A: 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!
A: If you also know that


*

*$\mathcal{L}\{c_1y_1+c_2y_2\}=c_1\mathcal{L}\{y_1\}+c_2\mathcal{L}\{y_2\}$ and

*$ \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}
