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Solution for summation involving associated Laguerre polynomials

I want the expression for the following:

Sum[(1/Factorial[(l + p)])*(x y)^l*LaguerreL[p, l, x^2]*
  LaguerreL[p, l, y^2], {l, 0, Infinity}]

When I write this in Mathematica, it just gives back the same thing written in mathematical form, not the solution.

If I specify the value of p, then I get closed form expression, but I am not getting for a general p.
POSTED BY: Radhika Prasad
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can you explain how can you solve this Integration

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POSTED BY: Anuj Malik

I believe there probably will be a closed form. I saw the Hardy-Hille formula on Wikipedia:https://en.wikipedia.org/wiki/Laguerre_polynomials which is of the similar form, just with an extra factorial in the numerator. Further, when specifying the p values from 0 to 10, Mathematcia does give closed form expressions for those particular p values.

I need closed form expression because this is a small part of my project.
POSTED BY: Radhika Prasad

From here we have:

$L_p^l\left(a x^2\right)=\sum _{k=0}^p \frac{(-1)^k \binom{p+l}{p-k} \left(a x^2\right)^k}{k!}$

Then we have:

.

$\sum _{l=0}^{\infty } \frac{(c x y)^l L_p^l\left(a x^2\right) L_p^l\left(b y^2\right)}{(l+p)!}=\\\sum _{l=0}^{\infty } \frac{(c x y)^l \left(\sum _{k=0}^p \frac{(-1)^k \binom{p+l}{p-k} \left(a x^2\right)^k}{k!}\right) \sum _{j=0}^p \frac{(-1)^j \binom{p+l}{p-j} \left(b y^2\right)^j}{j!}}{(l+p)!}=\\\sum _{k=0}^p \sum _{j=0}^p \left(\sum _{l=0}^{\infty } \frac{(-1)^{j+k} \left(a x^2\right)^k (c x y)^l \left(b y^2\right)^j \binom{l+p}{-j+p} \binom{l+p}{-k+p}}{j! k! (l+p)!}\right)=\\\sum _{k=0}^p \sum _{j=0}^p \frac{(-1)^{j+k} \left(a x^2\right)^k \left(b y^2\right)^j \binom{p}{-j+p} \binom{p}{-k+p} \, _2F_2(1,1+p;1+j,1+k;c x y)}{j! k! p!}$
POSTED BY: Mariusz Iwaniuk

Do you have any reason to think there is a closed form?

Questions like this should always indicate why it is reasonable to expect a solution.

Most integrals, series don't have one.

Why do you want closed-form expression ?
POSTED BY: Mariusz Iwaniuk

How did you get this form?
POSTED BY: Radhika Prasad

Going from a single sum to a triple sum will probably not help me. I would need a closed form solution because this will then go inside an integral.
POSTED BY: Radhika Prasad

Another solution:

POSTED BY: Mariusz Iwaniuk
Jim Baldwin
Jim Baldwin
Posted 4 months ago

The result for any positive integer n is the sum of two terms: (1) an Exp[x y] term multiplied by a polynomial in x and y and (2) a polynomial in x and `y'. The first term is

Exp[x  y] LaguerreL[n, (x - y)^2]/n!

but I haven't figured out a general form for the second polynomial (assuming there is a closed form for that polynomial). So far for that second polynomial the sum of the x and y exponents is always even and the coefficient for x^i y^j is the same as the coefficient for x^j y^i. That's all I've determined. Maybe someone else can see the pattern.
POSTED BY: Jim Baldwin

It can be expressed by The Srivastava and Daoust multivariable hypergeometric function. I don't know if this triple sum can be simplified.

$\sum _{l=0}^{\infty } \frac{(c x y)^l L_p^l\left(a x^2\right) L_p^l\left(b y^2\right)}{(l+p)!}=\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \sum _{u=0}^{\infty } \frac{\left((1)_u (-p)_m (-p)_n (1+p)_u\right) \left(\left(a x^2\right)^m (c x y)^u \left(b y^2\right)^n\right)}{\Gamma (1+p) \left((1)_{m+u} (1)_{n+u}\right) (m! n! u!)}$
POSTED BY: Mariusz Iwaniuk
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