Can i generalize the results?

Question

g[z_, k_] := Exp[-z*(1/(k + 1))] *(1/(k + 1))
G[z_, k_] := 1 - Exp[-z*1/(k + 1)]

h[k_, f_, n_, y_, m_] := 
 Assuming[n ∈ Integers && n > 0, 
  Integrate[G[s, f]^(1)*F[s], {s, 0, y}]]
j[k_, f_, n_, m_] = 
 Assuming[k > 0 && f > 0 && k ∈ Reals && f ∈ Reals &&
      n ∈ Integers && n > 0, 
    Integrate[g[y, k]*h[k, f, n, y, m], {y, 0, Infinity}]] 
 // FullSimplify              

If i define F[s] as s or Sqrt[s] or log[s] the above code gives me back a pretty easy solution. But if i don't specify F[s] the code has problems. I assume the code is not aware, that F[s] should be an integrable function. Can i somehow specify this information? Or do you know a smart way to integrate thuis by hand?

Answer 1

You can use your own custom transformation rules for symbolic expressions with unevaluated integrals, if you know what you are doing. Mathematica won't do this automatically by itself and i think there is currently no way of specifying this information through Assumptions, so i actually think pattern based transformation rules are the canonical way of doing this (as basically everything in Mathematica under the hood).

As an example, let's start with something Mathematica won't integrate, like

Integrate[G[x] f[x], x]

. If we know that the indefinite integral of f[x] is F[x] we can write a rule that performs integration by parts for us, like

integrationByParts[f_, F_] := Integrate[G_*f, x] :> G*F - Integrate[D[G, x]*F, x]

And use it on our example to get

Integrate[G[x] f[x], x]
% /. integrationByParts[f[x], F[x]]
(* F[x] G[x] - Integrate[F[x] G'[x], x] *)

This works now also if G is known but f is not (except from the fact, that its integral is F).

Integrate[Sin[x] f[x], x]
% /. integrationByParts[f[x], F[x]]
(* -Integrate[Cos[x] F[x], x] + F[x] Sin[x] *)

Hope this helps to get the idea across, how you can apply the technique to your code.