## anonymous one year ago what is the laplace transform of t^2u(t-1)?

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2. anonymous

$$\int_0^\infty e^{-st}t^2u(t-1)\, dt=\int_0^1e^{-st}t^2\cdot 0\, dt+\int_1^\infty e^{-st}t^2\cdot 1\, dt$$

3. anonymous

we can split the integral into one on $$(0,1)$$ and another on $$(1,\infty)$$ because we have that the step function $$u(t-1)=0$$ for $$t-1<0\Leftrightarrow t<1$$ and $$u(t-1)=1$$ for $$t-1>0\Leftrightarrow t>1$$

4. anonymous

the first integral is identically zero, while the second we evaluate using integration by parts: \begin{align*}\int_1^\infty e^{-st} t^2\, dt&=-\frac1s t^2e^{-st}\bigg|_1^\infty+\frac2s\int_1^\infty te^{-st}\, dt\\&=\left[-\frac1s t^2e^{-st}-\frac2{s^2}te^{-st}\right]_1^\infty +\frac2{s^2}\int_1^\infty e^{-st}\, dt\\&=\frac1se^{-st}+\frac2{s^2}e^{-st}+\frac2{s^2}\left[-\frac1se^{-st}\right]_1^\infty\\&=\left(\frac1s+\frac2{s^2}+\frac2{s^3}\right)e^{-s}\end{align*}

5. anonymous

which is the same thing you get after shifting $$t\to t+1$$ so that $$t^2\to(t+1)^2=t^2+2t+1$$ and pulling out the factor of $$e^{-s}$$ from $$e^{-st}\to e^{-s(t+1)}=e^{-s}e^{-st}$$, so: $$\mathcal{L}\{t^2 u(t-1)\}=e^{-s}\mathcal{L}\{t^2+2t+1\}=\left(\frac1s+\frac2{s^2}+\frac2{s^3}\right)e^{-s}$$ considering that $$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$

6. IrishBoy123

$$F(p) =\mathcal {L} \{ t^2 \; \; u(t-1)\}$$ [$$\bar t = t -1$$] $$F(p) = \mathcal {L} \; \left\{(\bar t+1)^2 \; u(\bar t) \right\}$$ $$= \mathcal {L} \; \left\{ ( \bar t^2 + 2 \bar t + 1 \; )u(\bar t) \right\}$$ $$= \left( \dfrac{2}{p^3} + \dfrac{2}{p^2} + \dfrac{1}{p} \right) e^{-p}$$ |dw:1443657391199:dw|