Calculate the Laplace-transform of $f(t) = \delta''(t-1)\theta(t)$ where $\delta$ is the dirac-delta function and $\theta$ is the Heaviside-function.
My attempt:
Using the unilateral definition of laplace-transform $$\mathcal{L}\left(\delta''(t-1)\theta(t)\right) = s^2\mathcal{L}(\delta(t-1)\theta(t)) - sf(0) - s^0f'(0) = s^2\mathcal{L}(\delta(t-1)\theta(1))-sf(0) - f'(0) = s^2e^{-s} -sf(0) - f'(0).$$
However I can't compute $f(0), f'(0)$ since the delta function cannot be defined in a single point, and the Heaviside function isn't defined in $t=0.$ How do I proceed?
The answer is supposed to be $s^2e^{-s}.$
And another question, in $f(t) = \delta''(t-1)\theta(t)$, can I just set $\theta(t) = 1$ since, if i'm correct, this is the case where $\delta''(t-1)$ has any relevance.
First of all since you use the unilateral definition of Laplace-transform (as you have said) the Heaviside-function doesn't actually do anything.
Otherwise (using bilateral Laplace-transform) because of the definition of the Heaviside-function: $$\mathcal{L}\left(\delta''(t-1)\theta(t)\right)=\int _{-\infty }^{\infty }e^{-st}\delta''(t-1)\theta(t)\,\mathrm {d} t=\int _{0}^{\infty }e^{-st}\delta''(t-1)\,\mathrm {d} t$$ Then changing the variable (for easiness) to $x=t-1$ and using the definition of the derivative of Dirac delta (see eqn. 10): $$\begin{eqnarray} \mathcal{L}\left(\delta''(t-1)\theta(t)\right)&=&\int _{0}^{\infty }e^{-st}\delta''(t-1)\,\mathrm {d} t=e^{-s}\int _{0}^{\infty }\delta''(x)e^{-sx}\,\mathrm {d}x=(-1)^2e^{-s}\left.\frac{\mathrm {d}^2e^{-sx}}{\mathrm {d}x^2}\right|_{x=0}=\\&=& s^2e^{-s} \end{eqnarray}$$