Laplace transform of the derivative of the dirac delta function times another function

Question

I'm trying to solve a DE involving terms of the form $\dot{\delta}(t-k)f(t)$ and $\ddot{\delta}(t-k)f(t)$, where $k>0$. I would therefore like to find the Laplace transform of these terms. My approach so far has been to simply integrate by parts, ie in the case of $\dot{\delta}(t-k)f(t)$ I get

$\mathcal{L}\left\{ \dot{\delta}\left(t-k\right)f\left(t\right)\right\} =\int_{0}^{\infty}\dot{\delta}\left(t-k\right)f\left(t\right)\exp\left(-ts\right)dt$

$=\left[\delta\left(t-k\right)f\left(t\right)\exp\left(-ts\right)\right]_{0}^{\infty}-\int_{0}^{\infty}\delta\left(t-k\right)\left[\dot{f}\left(t\right)\exp\left(-ts\right)-sf\left(t\right)\exp\left(-ts\right)\right]dt$

Since $\delta\left(t\right)=0$ for all $t\neq k$ and since $k>0$, it must hold that $\left[\delta\left(t-k\right)f\left(t\right)\exp\left(-ts\right)\right]_{0}^{\infty}=0$, so by the sifting property of $\delta(t)$ I get

$\mathcal{L}\left\{ \dot{\delta}\left(t-k\right)f\left(t\right)\right\} = \left(sf\left(k\right)-\dot{f}\left(k\right)\right)\exp\left(-ks\right)$

Using the same approach, I get

$\mathcal{L}\left\{ \ddot{\delta}\left(t-k\right)f\left(t\right)\right\} =\left(\ddot{f}\left(k\right)-2s\dot{f}\left(k\right)+s^{2}f\left(k\right)\right)\exp\left(-ks\right)$

My question has two parts:

1) Is this the correct approach?

2) What happens if $k=0$?

Answer 1

The approach is correct in a heuristic sense. But the symbol $\int_0^\infty \delta^{(n)}\phi(t)\,dt$ is not and integral.

The Distributional Derivative of the Dirac Delta is defined such that for every suitable test function $\phi$, we have

$$\langle \delta_a',\phi \rangle =(-1)\phi'(a)\tag 1$$

The $n$'the derivative is similarly defined by

$$\langle \delta_a^{(n)},\phi \rangle =(-1)^n\phi^{(n)}(a)\tag 2$$

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Note that both $(1)$ and $(2)$ can be heuristically derived by formally applying integration by parts, proceeding as if the Dirac Delta was an ordinary function rather than a generalized function.

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Therefore, if $\phi(t) =f(t)e^{-st}$ in $(1)$, then

$$\int_0^\infty \delta'(t-k)f(t)e^{-st}\,dt=-\left.\frac{d(f(t)e^{-st})}{dt}\right|_{t=k}=(-f'(k)+sf(k))e^{-sk}$$

And if $\phi(t) =f(t)e^{-st}$ in $(2)$ with $n=2$, then

$$\int_0^\infty \delta''(t-k)f(t)e^{-st}\,dt=\left.\frac{d^2(f(t)e^{-st})}{dt^2}\right|_{t=k}=(f''(k)-2sf'(k)+s^2f(k))e^{-sk}$$