Find the inverse Laplace Tranformation by using convolution theorem for the function frac{1}{s^3(s-5)}

Step 1
According to the given information, it is required to find the inverse of Laplace transform using convolution theorem.
$\frac{1}{s^3(s-5)}$
it is required to find $L^{-1}\left(\frac{1}{s^3(s-5)}\right)$
Step 2
Solving further:
$\text{let }G(s)=\frac{1}{s^3},H(s)=\frac{1}{(s-5)}$
the inverse laplace of G(s) and H(s) are
$L^{-1}(G(s))=L^{-1}\left(\frac{1}{s^3}\right)=\frac{t^2}{2}=g(t)$
$L^{-1}(H(s))=L^{-1}\left(\frac{1}{(s-5)}\right)=e^{5t}=f(t)$
Step 3
The convolution theorem states that if G(s) and H(s) are the Laplace transform of g(t) and h(t) then,
$L((g\cdot h)(t))=G(s)H(s)$
$\Rightarrow (g\cdot t)(t)=L^{-1}(G(s)H(s))$
$\Rightarrow g(t)h(t)=L^{-1}(F(s))={\int }_0^{t}f(t-u)g(u)du$
Step 4
Now solve the integral to find the inverse.
${\int }_0^t{e}^{5(t-u)}\frac{u^2}{2}du=\frac{1}{2}{e}^{5t}{\int }_0^t(e^{-5u}\cdot u^2)du$
$=\frac{1}{2}{e}^{5t}[-\frac{1}{125}(25t^2{e}^{-5t}-2(-5t{e}^{-5t}-e^{-5t})-2)]$
$=-\frac{1}{250}{e}^{5t}[25t^2{e}^{-5t}+10t{e}^{-5t}+2e^{-5t}-2]$
${\int }_0^t{e}^{5(t-u)}\frac{u^2}{2}du=L^{-1}(\frac{1}{s^3(s-5)})=-\frac{1}{10}{t}^2-\frac{1}{25}t-\frac{1}{125}+\frac{1}{125}{e}^{5t}$