I am trying to find a closed form for the summation:
$\sum_{l=2}^{\infty} \frac{(2l+1)(l+1)}{K+(l+1)(l+2)(l-1)} P_l\left(\cos(\theta)\right)$
Where $K$ is a positive real number.
I managed to find a closed form when $K=0$ by using the formulas 5.10.4; 5.10.5 and 5.10.6 of the Book "Integrals and Series Vol 2" by Proudnikov.
Indeed, when $K=0$, reducing the above fraction to simpler fractions, and using the fact that:
$\sum_{k=1}^{\infty} \frac{1}{k}P_k\left(x\right)=\ln{(\frac{2}{1-x +\sqrt{2-2x} })}$
and
$\sum_{k=0}^{\infty} \frac{1}{k+1}P_k\left(x\right)=\ln{(1+\frac{\sqrt{2}}{\sqrt{2-2x} })}$
you can get a closed form for the summation.
Could you give me some hint if it is possible to find a closed form even when the constant $K$ is not zero? and maybe where to start from? Or if it is possible to obtain approximated solutions.
EDIT: First steps in working out the solution, starting from the answer of Dr. Wolfgang Hinze:
we start by casting the original fraction into a sum of its complex partial fractions:
$\frac{(2l+1)(l+1)}{K+(l+1)(l+2)(l-1)} = \frac{A}{l-q1}+\frac{B}{l-q2}+\frac{C}{l-q3}$
where $q1$, $q2$, and $q3$ are the three solutions of the equation $K+(l+1)(l+2)(l-1)=0$ and the complex coefficients $A$, $B$, and $C$ are obtained matching the coefficients of the powers of $l$ in the numerator.
Now for each of the fractions $\frac{A}{l-q1}$ it is possible to apply the answer of Dr. Wolfgang Hintze which starts by writing the fraction $\frac{A}{l-q1}= A \int_0^{\infty} \exp{\left(-z (l-q1) \right)}dz $ and then using the generating function, as explained below.
However this integral does not converge for arbitrary values of $q1$; indeed it is necessary that $l > Re(q1)$.
Unfortunately, by choosing $K >0$ in the equation $K+(l+1)(l+2)(l-1)=0$ two solutions always have real part bigger than zero, meaning that the above integral representation of the fraction doesn't apply for all the $l$ in the summation.
Is it possible to use a different integral representation for this fractions?
Is there a workaround for it?
EDIT2: To successfully complete the bounty I expect the solution as a closed form (even if not pleasant), or if not possible the demonstration why it is not possible
if you feel like you need more details, I would be glad to provide them, as well as collaborating towards the solution.
Best Answer
This is not a complete solution but it shows, for a simplified version of the problem, how a frequenty used approach works.
Consider the simplified problem to calculate the sum
$$s(k,x) = \sum _{n=0}^{\infty } \frac{P_n(x)}{k+n}$$
Where $k>0$ is a real parameter.
The generating function for the Legendre polynomials is
$$g(t,x) = \frac{1}{\sqrt{t^2-2 t x+1}} = \sum _{n=0}^{\infty } t^n P_n(x)$$
Now writing
$$\frac{1}{k+n}=\int_0^{\infty } \exp (-z (k+n)) \, dz$$
inserting this into the expression for $s$ we get
$$\sum _{n=0}^{\infty } P_n(x) \int_0^{\infty } \exp (-z (k+n)) \, dz$$
Interchanging summation and integration gives
$$\int_0^{\infty } \left(\sum _{n=0}^{\infty } P_n(x) \exp (-z (k+n))\right) \, dz$$
Extracting the factor $ \exp (- k z)$ from the sum this can be written as
$$\int_0^{\infty } \exp (- k z) \sum _{n=0}^{\infty } \exp (- n z) P_n(x) \, dz$$
Now the sum can be done using the formula for the generating function of the Legendre polynomials (with $t = \exp(- z))$ with the result
$$\sum _{n=0}^{\infty } \exp (-n z) P_n(x) = \frac{1}{\sqrt{-2 x e^{-z}+e^{-2 z}+1}}$$
Hence we arrive at this integral representation of $s$:
$$s1(k,x) = \int_0^{\infty } \frac{e^{-k z}}{\sqrt{-2 e^{-z} x+e^{-2 z}+1}} \, dz$$
Mathematica gives for this integral the closed form expression:
$$s1(k,x) = \frac{1}{k (k+1) (k+2)}\left( (k+1) (k+2) F_1\left(k;-\frac{1}{2},-\frac{1}{2};k+1;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)+2 k (k+2) x F_1\left(k+1;\frac{1}{2},\frac{1}{2};k+2;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)-k (k+1) F_1\left(k+2;\frac{1}{2},\frac{1}{2};k+3;x+i \sqrt{1-x^2},x-i \sqrt{1-x^2}\right)\right)$$
Here $F_1$ is the Appell1 hypergeometric function (http://mathworld.wolfram.com/AppellHypergeometricFunction.html).
Admittedly, this is not a very "pleasant" expression, but the result for the complete problem of the OP can be expected to be even uglier, if it has a closed form at all, which I doubt.
EDIT
For integer $k$ we get closed form expressions (always for the simplified Problem). The first few are
$$s(0,x) = \log (2)-\log \left(-x+\sqrt{2-2 x}+1\right)$$
Notice that for $k=0$ the n-sum starts at $n=1$.
$$s(1,x) = \log \left(\frac{x-\sqrt{2-2 x}-1}{x-1}\right)$$
$$s(2,x) = \sqrt{2-2 x}+2 x \coth ^{-1}\left(\sqrt{2-2 x}+1\right)-1$$
$$s(3,x) = \frac{1}{2} \left(\left(6 x^2-2\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)$$
$$s(4,x) = \frac{1}{6} \left(6 x \left(5 x^2-3\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+5 x \left(3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)-2 \sqrt{2-2 x}+4\right)$$