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.