Allora , ho applicato la definizione di trasformata.
$ int_(0)^(1)(1+t)^2e^(-st) dt $
$=[(1+t)^2e^(-st)/-s]_(0,1) - int_(0)^(1) 2(1+t)e^-(st)/-s dt$
$=[2^2 e^-s/-s -e^0/-s]+2/s int_(0)^(1) (1+t)e^-(st)dt$
$=[(1-4e^-s)/s]+2/s{[(1+t) e^-st/-s]_(0,1)- int_(0)^(1) e^(-st)/-s dt}$
$=(1-4e^-s)/s+2/s{[2e^-s/-s-e^0/-s]+1/s[e^-(st)/-s]_(0,1)}$
$=(1-4e^-s)/s+2/s{-1/s[2e^-s-1]-1/s^2[e^-(s)-e^0]}$
$=(1-4e^-s)/s+2/s{(1-2e^-s)/s+(1-e^-s)/s^2}$
$=(1-4e^-s)/s+(2-4e^-s)/s^2+(2-2e^-s)/s^3$
$ int_(0)^(oo ) (t^2+1)e^(-st)dt $
$=[(t^2+1)e^-(st)/-s]_(1,+ oo )- int_(1)^(+oo ) 2te^-(st) dt$
$=[(t^2+1)e^(-st)/-s]_(1,+oo )-{[2te^(-st)/s^2]_(1,+oo )-int_(1)^(+oo ) 2e^-(st)/s^2 dt}$
$=[(t^2+1)e^(-st)/-s]_(1,+oo )-[2te^(-st)/s^2]_(1,+oo )+2[e^(-st)/-s^3]_(1,+oo )$
$=-2e^-s/-s+2e^-s/s^2+2e^-s/s^3$
$=2e^-s/s+2e^-s/s^2+2e^-s/s^3$
Risultato:
$L[f](s)= (1-4e^-s)/s+(2-4e^-s)/s^2+(2-2e^-s)/s^3+2e^-s/s+2e^-s/s^2+2e^-s/s^3$
$=( s^2-2e^-s(s+1)s+2s+2)/s^3$
https://www.wolframalpha.com/input?i=LT+piecewise%5B%7B%7B%281%2Bt%29%5E2%2C0%3Ct%3C1%7D%2C%7B1%2Bt%5E2%2Ct%3E%3D1%7D%7D%5DProblema: troppi calcoli.Ho provato ad applicare la proprietà di linearità nel caso della trasformata della funzione scritta da gugo.
$L[t]=L[(1+t)^2u(t)]-2L[tu(t-1)]$
- I addendo:
$L[(1+t)^2u(t)]=L[(1+t)^2]$
perché moltiplicare per u(t) significa semplicemente considerare il segnale corrispondente a quella funzione
$=L[1+2t+t^2]=L[1]+2L[t]+L[t^2]= 1/s+2/s^2+2/s^3 =(s^2+2s+2)/s^3$
- II addendo:
$L[tu(t-1)](s)= ?$