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This is a follow-up of this question/answer.

I'm working on a recursive integral more complicated than the one on the linked question, but this one can be used as an example.

I would like to obtain a closed expression for the dependence of the integral on the dimension $d$, that is, on the recursive parameter. In other words, I want to know what the general term of the recurrence relation is. RSolve is not able to find it (in my example).

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1 Answer

Using @ubpdqn's answer for getting a closed expression for the dependence of the integral on the dimension n.

vol[n_] := vol[n] = FullSimplify[Nest[Integrate[# /. r -> Sqrt[r^2 - x^2], {x, -r, r}] &, 2 r, n], 
                                                                      Element[r, Reals] && r > 0]

FindSequenceFunction[Table[vol[n], {n, 1, 6}], n]
(*
 (π^((1 + n)/2) r (r^2)^(1/2 (-1 + n)) (r - (-1)^n r + (1 + (-1)^n) Sqrt[r^2]))/(2 Gamma[(3 + n)/2])
*)

I don't quite understand your sentence "I want to know what the general term of the recurrence relation is", since that is exactly what @ubpdqn used.

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Thanks. The general term is the one you wrote between (* *), that is, the term as an explicit function of $n$. Unfortunately, FindSequenceFunction does not work for my recurrence relation. Any idea? Should I write my specific recurrence relation? –  drake Nov 23 '13 at 0:59
    
@drake 1) Keep in mind that it isn't always possible to find a closed form 2) Post it in another question, as this one would radically change , and that isn't convenient –  belisarius Nov 23 '13 at 1:08
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