# Laplace-transform of $\delta''(t-1)\theta(t)$

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.

• Do you know the definition for the Laplace transform of a distribution? What happens when you apply that definition in this case? If you don't know a definition, that is where you should start. Jan 5, 2018 at 15:29
• @GEdgar Using the definition I get stuck at the integral $\int_{0}^{\infty}e^{-st}\cdot\delta''(t-1)dt$ Jan 5, 2018 at 15:37
• Use integrations by parts on that integral to "raise" $\delta''$ to $\delta$ and then use the definition of $\delta$. Jan 5, 2018 at 15:39
• The definition of $\delta''(t-1)$ is $\int f(t)\delta''(t-1)\,dt=(-1)^2 f''(1)$. Jan 5, 2018 at 15:57

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): $$\begin{eqnarray} \mathcal{L}\left(\delta''(t-1)\theta(t)\right)&=&\int _{0}^{\infty }e^{-st}\delta''(t-1)\,\mathrm {d} t=e^{-s}\int _{0}^{\infty }\delta''(x)e^{-sx}\,\mathrm {d}x=(-1)^2e^{-s}\left.\frac{\mathrm {d}^2e^{-sx}}{\mathrm {d}x^2}\right|_{x=0}=\\&=& s^2e^{-s} \end{eqnarray}$$