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I know the following recurrence relation


with $a_0=1$ can be represented alternatively as an integral


Verifying this is easy, but is there any general technique to do this kind of transformations?

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Do you have a specific "class" or recurrence relations in mind? Otherwise the answer might be "No". – Aryabhata Mar 5 '12 at 0:49
@Aryabhata: What conclusions can you draw in general? Even for $$a_n=a\int_0^1{x^{a-n-1}(b-cx)^ndx}$$? – littleEinstein Mar 5 '12 at 0:55
I don't understand. Are you looking for general techniques which take you from a recurrence relation to integral, or are you asking if getting an integral representation is any useful? – Aryabhata Mar 5 '12 at 0:58

If a polynomial formula for the $a_n$ term is known, then the numerator of each term can be multipled by $x^{y-1}$ where $y$ is the denominator of that term.

$$ \textstyle a_n=\frac{a}{a-n}+\frac{ n(n-1)a}{(a-n)(a-(n-1))}+\frac{n(n-1)(n-2)a}{(a-n)(a-(n-1))(a-(n-2))}+ ... +\frac{n!(a+a_0)(a-n)!}{(a-1)!} $$

$$\begin{eqnarray} a_n&=&\int_0^1 ( ax^{a-n-1}+nax^{(a-n)(a-(n-1))-1}\\ &&+n(n-1)ax^{(a-n)(a-(n-1))(a-(n-2))-1}+\dots\\ &&+n!(a+a_0)x^{\frac{(a-1)!}{(a-n)!}-1})dx\\ &=&\int_0^1 a_0 n!x^{\frac{(a-1)!}{(a-n)!}-1} + \sum_{k=0}^n k! {n\choose k} x^{k! {a-1\choose k}-1 } \text dx \end{eqnarray} $$

This is not the simplest integral form, but at least it allows $a_n$ to be calculated when $n$ is not an integer.

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