If asked to find the Laplace transform of the derivative of the Dirac delta function, I would naively integrate by parts and conclude that $$ \begin{align}\int_{0}^{\infty} \delta'(t) e^{-st} \, dt &= \delta(t)e^{-st} \Big|_{0}^{\infty} + s \int_{0}^{\infty} \delta(t) e^{-st} \, dt \\ &= 0 – \lim_{t \to 0^{+}} \delta(t)e^{-st} + s \, \mathcal{L} \{\delta(t)\}(s) \\ &= – \lim_{t \to 0^{+}} \delta(t)e^{-st} + s. \end{align}$$
But I don't know why $\delta(t) e^{-st}$ would tend to zero as $t$ appraches $0$ from the right, nor do I know if integrating by parts was even an appropriate thing to do in the first place.
Best Answer
There's a simpler way to do that. For any distribution $\Delta$ you have that $$ \int \Delta'\,f \:dt = -\int \Delta\,f' \:dt \text{.} $$ And for the $\delta$-distribution you have that $$ \int \delta\,f \:dt = f(0) \text{.} $$ This immediatly gives $$ \int \delta'\,e^{-st} \:d(t) = -\int \delta\,(-s)e^{-st} \:d(t) = s \text{.} $$ Don't take this notation too serious, btw. A distribution is not a function, so whatever $\int \delta\,f \:dt$ is, it's neither a Riemann nor a Lebesgue integral. For some distributions ($\delta$ for example), you can interpret it as a lebesgue integral, but one with a measure other than the normal lebesgue measure. But that doesn't work for all distributions, $\delta'$ is an example of one where it doesn't I think. In reality, distributions are objects which map functions to values. So what $\int \Delta\,f \:dt$ really means is $\Delta(f)$, which is some real (or complex) number. These things get written as integraly mainly because that makes it easier to remember some identities, like $\Delta(f+g) = \Delta(f)+\Delta(g)$.