the time shifting property of Laplace transform:
f(t−a)u(t−a)=L−1{e−asF(s)},a>0 where u(t) is the Heaviside step function and
F(s)=L{f(t)} hence
L−1{e−s/s}=[a=1>0]=u(t−1)f(t−1) where
F(s)=s1=[for p=−21]=Γ(p+1)sp+1Γ(p+1)=Γ(21)sΓ(21) The Gamma function is an extension of the normal factorial function and
Γ(21)=π for pth power of Laplace transform:
L−1{sp+1Γ(p+1)}=[Re(p)>−1]=tp So
f(t)=L−1{Γ(1/2)sΓ(1/2)}=π1L−1{sΓ(1/2)}=πt1L−1{e−s/s}=u(t−1)f(t−1)=π(t−1)u(t−1) Answer: L−1{e−s/s}=π(t−1)u(t−1)
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