Find Laplace transform of a function multiplied by the Dirac delta function

Question

I know that the Laplace transform of the delta function $\delta(t-a)$ is $e^{-as}$.

Now my question is to find the Laplace transform of $te^{-3(t-2)}\delta(t-2) $.

Is there any theorem or easy thumb rule to solve these kinds of problems or should I integrate the function as we conventionally do, to find its Laplace transform?

Answer 1

Using the fact that,

$$\int^{\infty}_0f(t)\delta(t-a)dt=f(a)$$

$$F=\mathcal{L}\{te^{-3(t-2)}\delta(t-2)\}=-\frac{d}{ds}\mathcal{L}\{e^{-3(t-2)}\delta(t-2)\} = -\frac{d}{ds}\int^\infty_0e^{-st}e^{-3(t-2)}\delta(t-2)dt$$

$$F = -e^{6}\frac{d}{ds}\int^\infty_0e^{(-3-s)t}\delta(t-2)dt = -e^6\frac{d}{ds}\bigg[e^{(-3-s)2}\bigg] = -e^6\cdot(-2)e^{-6-2s} = +2e^{-2s}$$

$$F = 2e^{-2s}$$

Aliter

$$F = \mathcal{L}\{te^{-3(t-2)}\delta(t-2)\} = -\frac{d}{ds}\mathcal{L}\{e^{-3(t-2)}\delta(t-2)\} = -e^{-6}\frac{d}{ds}\mathcal{L}\{\delta(t-2)\}\bigg\vert_{s\to s+3}$$
$$F = -e^{-6}\frac{d}{ds}(e^{-2s})\bigg\vert_{s\to s+3} = -e^{-6}(-2)e^{-2(s+3)} = 2e^{-2s}$$

Answer 2

Use the fact that $f(t) \, \delta(t-a) = f(a) \, \delta(t-a)$:

$$ \mathcal{L}\{te^{-3(t-2)}\delta(t-2)\} = \mathcal{L}\{2e^{-3(2-2)}\delta(t-2)\} = 2 \mathcal{L}\{\delta(t-2)\} = 2 e^{-2s}. $$