inverse laplace transformation (use convolution theorem)
example :L^(-1){s/(s^2+a^2)^2}
peter.vlasschaert@gmail.com,28/09/2017

kill(all)$

p: s/(s^2+a^2)^2;

method 1 : use menu ' Maxima ' =>
calculus -> Inverse Laplace Transformation.
------------------------------------------------------------------
first         field : expression -> p
second  field : old                -> s
thirdth    field : new              -> t 

1e) value 'a'

assume(a>0);

2e) use menu 'see above'

ilt(p, s, t);

method 2: use convolution theorem

1e) value 'a'   => 'see above'

2e) splitting  'p' two parts 'p5 and p6'

p1: part(p,1);
p2: part(p,2);
p3:part(p,2,1);
p4: part(p,2,2);

p5:p1/p3;
p6:1/p3;

3e)  use tables from books (online tables)
        use menu 'maxima'  : Calculus -> Inverse Laplace Transform    
for 'p5 and p6' -> 'p7 and p8'   

p7:ilt(p5,s,t);
p8:ilt(p6,s,t);

4e) use convolution theorem

definition : convolution theorem
 f star g = integrate(f(u)*g(t-u),u,0,t)

p9:FstarG ='integrate(f(u)*g(t-u),u,0,t);

select : g -> choose 'p7 or p8'
=> 'p8'  convert  g(t-u)
How :menu ' :simplify ->substitute (  result 'p10')
=> 'p7'   convert    f(u)
How :menu ' :simplify ->substitute  ( result  'p11')

p10:subst(t-u, t, p8);
p11:subst(u,t,p7);

find ::L^(-1){s/(s^2+a^2)^2} 'inverse laplace transform'

formula : 

p12:'integrate(p10*p11,u,0,t);

eval : above ,why assume (t=time >0),Maxima 

assume(t>0);

p13:integrate(p10*p11,u,0,t);

p14:ratsimp(p13);

Why use Maxima : same result direct,from method 1
=>  otherwise use convolution theorem to find Inverse Laplace Transform.