$L_{s} [t f(t) f'(t)] $

Question

Is there any simplification by the way of standard properties of Laplace Transform for this? $$L_{s} [t f(t) f'(t)]$$

where $f'(t)$ is $\frac{d}{dt} f(t)$

Alternately, is there a simplification of $L_{s} [f(t) f'(t)]$, so that I can compute my original expression by computing the derivative $-\frac{d}{ds} L_{s} [f(t) f'(t)]$ ?

Answer 1

$$\int_0^{+\infty} f'(t)f(t) e^{-st} \, \mathrm d t = I(s)$$

Define:

$$u(t)=f(t)e^{-st}$$ $$u'(t)=f'(t)e^{-st}-s e^{-st} f(t)$$ $$v'(t)=f'(t)$$ $$v(t)=f(t)$$

Integrate by parts: $\int v'u=vu-\int vu'$:

$$I(s)=f^2(t)e^{-st}\vert_{t=0}^{t \to + \infty} - I(s) + s \int_0^{+\infty} f^2(t)e^{-st} \, \mathrm d t$$

Solve for $I(s)$.

Answer 2

Since: $$\mathcal {L}\{tg(t)\}=-\frac {d}{ds}\mathcal {L}\{g(t)\}$$ We have: $$\mathcal {L}[tf(t)f′(t)]=-\frac {d}{ds}\mathcal {L}\{f(t)f′(t)\}$$ $$\mathcal {L}[tf(t)f′(t)]=-\frac 1 2\frac {d}{ds}\mathcal {L}\{(f^2(t))′\}$$ Use the fact that : $$ \mathcal{L}{(g'(t))}=sG(s)-g(0) $$ $$ \begin{align} \mathcal {L}[tf(t)f′(t)]&=-\frac 1 2\frac {d}{ds}(s\mathcal {L}\{f^2(t)\}(s)-f^2(0)) \\ \mathcal {L}[tf(t)f′(t)]&=-\frac 1 2\frac {d}{ds}(s\mathcal {L}\{f^2(t)\}(s)) \\ \end{align} $$