calculating laplace transform of t^3*delta(t - 4)

Question

How can I calculate Laplace transform of $t^3\delta(t - 4)$

I used the formula for calculating L-transform of $t^n$ => $\frac{n!}{s^{n+1}}$ and $\delta(t-n)$ => $e^{-ns}$ so the answer I calculated is

$\frac{6e^{-4s}}{s^4}$

but the correct answer in the book is $4^3*e^{-4s}$. How does this answer come, or the answer is incorrect? Please explain.

Answer 1

First, note: $$\mathcal{L}\big(f(t) \times g(t)\big) \neq \mathcal{L}\big(f(t)\big) \mathcal{L}\big(g(t)\big).$$ The simplest example would be: take $f(t) = t$, $g(t) = t$. Then the LHS reads $\frac{2}{s^3}$ and RHS would be $\frac{1}{s^4}$.

To answer your question lets just write out the integral:

$$\mathcal{L}\big(f(t)\big) = \int t^3 \delta(t-4)e^{-st} dt = \int h(t)\delta(t-4) dt $$

where I defined a handy function $h$ by $h(t) := t^3 e^{-st}.$ Now, do you know how an integral over a Dirac delta works?