I have a piecewise function f(t), and I'm trying to get it's laplace transform. When I do it manually, i'm getting a different result than with Maple.
$$ f(t)=\!\cases{{t}^{2}&$0<t$ and $t<1$\cr 0&otherwise\cr}{ {\rm }}{} $$
My manual approach consists of the following steps: $$ F(s)=\int _{0}^{\infty }\!f \left( t \right) {{\rm e}^{-st}}{dt} $$ $$ =\int _{0}^{1}\!{t}^{2}{{\rm e}^{-st}}{dt}+\int _{1}^{\infty }\!0{dt} $$ Using integration by parts, with $u=t^2$, $u'=2t$, $v=e^{-st}/(-s)$, $v'=e^{-st}$ $$ \int _{}^{}\!u{\it v'}{}=uv-\int _{}^{}\!{\it u'}\,v{} $$ $$ (1)=-{\frac {{t}^{2}{{\rm e}^{-st}}}{s}}-\int _{0}^{1}\!-{\frac {2t{ {\rm e}^{-st}}}{s}}{dt} $$ $$ =-{\frac {{t}^{2}{{\rm e}^{-st}}}{s}}+2\,{\frac {\int _{0}^{1}\!t{ {\rm e}^{-st}}{dt}}{s}} $$ Again, integration by parts for the new integral we got from the previous integration by parts.
With $u=t$, $u'=1$, $v=e^{-st}/(-s)$, $v'=e^{-st}$ $$ (2)=2\,{\frac {\int _{0}^{1}\!t{ {\rm e}^{-st}}{dt}}{s}}=(2/s)(-{\frac {t{{\rm e}^{-st}}}{s}}-\int _{0}^{1}\!-{\frac {{{\rm e}^{-st}} }{s}}{dt}) $$ After solving (2), I substitute it into (1), which gives me: $$=-{\frac {{t}^{2}{{\rm e}^{-st}}}{s}}-2\,{\frac {t{{\rm e}^{-st}}}{{s}^ {2}}}+2\,{\frac {{{\rm e}^{-s}}}{{s}^{3}}}+2\,{s}^{-3}$$
$$ ={\it Eval} \left( -{\frac {{t}^{2}{{\rm e}^{-st}}}{s}}-2\,{\frac {t{ {\rm e}^{-st}}}{{s}^{2}}},t=1 \right) -{\it Eval} \left( -{\frac {{t}^ {2}{{\rm e}^{-st}}}{s}}-2\,{\frac {t{{\rm e}^{-st}}}{{s}^{2}}},t=0 \right) +2\,{\frac {{{\rm e}^{-s}}}{{s}^{3}}}+2\,{s}^{-3}$$ $$=-{\frac {{{\rm e}^{-s}}}{s}}+2\,{\frac {{{\rm e}^{-s}}}{{s}^{3}}}+2\,{ s}^{-3} $$
However, the laplace-transform function of Maple gives me the following: $${\frac {2-{{\rm e}^{-s}} \left( {s}^{2}+2\,s+2 \right) }{{s}^{3}}}$$ And I can't figure out what went wrong..
Thank you.