# Integral of a product of Laguerre polynomials

In order to estimate the non linear term in a particular PDE, I have to decompose $L_k^\alpha(x)^3\cdot x^{-\delta}$ (with $0<\delta<\alpha+1$) into a basis consisting of Laguerre polynomials $L_n^\alpha(x)$. To perform the decomposition I have to compute the following definite integral

$$T_n:=\int_0^\infty L_k^\alpha(x)^3 L_n^\alpha(x)x^{\alpha-\delta}e^{-x}dx\,.$$

Is there a closed form formula for the above integral or a way to obtain it?

EDIT:

I have found a paper by Artur Erdélyi where he gives a general formula for an integral of a product of $n$ Laguerre polynomials. Unfortunately, the result is presented in terms of Lauricella hypergeometric function and is not easily computable in the case of $T_n$. There are also some papers on a product of three Laguerre polynomials, which might be used to simplify $T_n$ but I won't be able to read them thoroughly for several days now.

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Note that the (weighted) integral of a product of Laguerre polynomias has a combinatorial meaning, (at least for $\alpha=0$); see this answer and the link therein: mathoverflow.net/questions/36824/… – Pietro Majer Dec 2 '13 at 22:30
The explicit answer is known for your case if $3\alpha=-\delta$. – Sergei Dec 28 '14 at 7:09
Is explicit formula exactly what you need for estimating? – Fedor Petrov Feb 26 '15 at 10:06

Maple finds the integral under consideration for concrete values of $n$ and $k$, producing huge outputs. For example, $$restart; with(orthopoly):VectorCalculus:-int(L(2, alpha, x)^3*L(3, alpha, x)*x^{alpha-delta}*exp(-x), x = 0 .. infinity) assuming\, alpha > delta-1$$ gives $$-1/48\,\Gamma \left( 4+\alpha-\delta \right) \left( {\delta}^{6}+12 \,\alpha\,{\delta}^{4}-27\,{\delta}^{5}+30\,{\alpha}^{2}{\delta}^{2}- 208\,\alpha\,{\delta}^{3}+319\,{\delta}^{4}+8\,{\alpha}^{3}-246\,{ \alpha}^{2}\delta+1422\,\alpha\,{\delta}^{2}-2081\,{\delta}^{3}+532\,{ \alpha}^{2}-4466\,\alpha\,\delta+7828\,{\delta}^{2}+5380\,\alpha-15976 \,\delta+13736 \right) +$$ $$1/48\,\Gamma \left( \alpha-\delta+1 \right) \left( \alpha+3 \right) \left( {\alpha}^{2}{\delta}^{6}-3\,\alpha\,{ \delta}^{7}+3\,{\delta}^{8}+12\,{\alpha}^{3}{\delta}^{4}-63\,{\alpha}^ {2}{\delta}^{5}+120\,\alpha\,{\delta}^{6}-66\,{\delta}^{7}+30\,{\alpha }^{4}{\delta}^{2}-298\,{\alpha}^{3}{\delta}^{3}+1069\,{\alpha}^{2}{ \delta}^{4}-1464\,\alpha\,{\delta}^{5}+698\,{\delta}^{6}+8\,{\alpha}^{ 5}-270\,{\alpha}^{4}\delta+2274\,{\alpha}^{3}{\delta}^{2}-7067\,{ \alpha}^{2}{\delta}^{3}+9342\,\alpha\,{\delta}^{4}-4356\,{\delta}^{5}+ 556\,{\alpha}^{4}-6356\,{\alpha}^{3}\delta+23458\,{\alpha}^{2}{\delta} ^{2}-34181\,\alpha\,{\delta}^{3}+17171\,{\delta}^{4}+6176\,{\alpha}^{3 }-38038\,{\alpha}^{2}\delta+72258\,\alpha\,{\delta}^{2}-42898\,{\delta }^{3}+23980\,{\alpha}^{2}-81016\,\alpha\,\delta+65264\,{\delta}^{2}+ 36800\,\alpha-54248\,\delta+18448 \right) .$$
Integrand is a polynomial in $x$ times $x^{\alpha-\delta}e^{-x}$. Such integral is always computable explicitly for any chosen $n$ and $k$. The question is, can we present the result in a compact form without fixing $n$ and $k$ beforehand. – pwl Dec 2 '13 at 14:22