# Summation of Legendre polynomials $\sum_{l=2}^{\infty} \frac{(2l+1)(l+1)}{K+(l+1)(l+2)(l-1)} P_l\left(\cos(\theta)\right)$

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.

• I suggest to write the $l$-th coefficient as an integral (of a function times $x^K$) then exploit the generating function for Legendre polynomials. It is not granted that the final outcome is a "pleasant" function, but that is the same approach one may use to prove the stated (known) identities. Jan 24, 2017 at 16:15
• Many thanks for the comment. Do you have any reference on how to proceed? Or maybe could you elaborate your comment into an answer? I would really appreciate any hint on how to proceed Jan 24, 2017 at 17:11
• They are not pleasant computations to carry on, would you add some context? What is the origin of this peculiar problem? Jan 24, 2017 at 17:19
• This summation comes from a spherical harmonics expansion of an axysimmetric problem on the unit sphere. It describes the deformation of the surface under a point forcing located at the north pole. Indeed I am expecting that the above summation diverges for $\theta=0$ (the north pole) as $P_l(1)=1$ and for large $l$ the summation is proportional to the harmonic series. But I believe it is finite everywhere else. I am looking for an indication on how to proceed, something like a starting kick and a good reference (also an example might work) i can work out unpleasant calculations myself. Jan 24, 2017 at 17:59
• Following the hint of Jack D'Aurizio I find with Mathematica for the much simpler case Sum[P(n,x)/(k+n),{n,0,oo}] a combination of AppellF1 hypergeometric functions. Jan 27, 2017 at 11:29

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 $n=1$.

$$s(1,x) = \log \left(\frac{x-\sqrt{2-2 x}-1}{x-1}\right)$$

$$s(2,x) = \sqrt{2-2 x}+2 x \coth ^{-1}\left(\sqrt{2-2 x}+1\right)-1$$

$$s(3,x) = \frac{1}{2} \left(\left(6 x^2-2\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)$$

$$s(4,x) = \frac{1}{6} \left(6 x \left(5 x^2-3\right) \coth ^{-1}\left(\sqrt{2-2 x}+1\right)+5 x \left(3 x \left(\sqrt{2-2 x}-1\right)+\sqrt{2-2 x}\right)-2 \sqrt{2-2 x}+4\right)$$

• @user3810266 This case can be done with similar methods: 1) decompose 1/(k+n^2) into (complex) partial fractions. These can be represented via a z-integral as before 2) generate the factor n derivating Exp(-a n) with respect to a 3) now you have this geometric sum: $\exp ((i k) (-z)) \sum _{n=1}^{\infty } \exp (-a n) \exp (n (-z)) P_n(x)$ 4) this gives the sqrt expression of the g.f. 5) take the real part, derive with respect to a and let a ->0 and you get the Integrand of the z-integral. Jan 31, 2017 at 0:05
• @user3810266 It is even better: from the hints of my last comment you can gather how to transform your original L-sum into a sum of z-integrals. Jan 31, 2017 at 0:49
• @SSC Napoli Yes, you are right in this case. Feb 15, 2017 at 16:06
• @SSC Napoli I have now done the complete calculation. I have found that your sum can be expressed through the Appell1-function and its derivatives. The remaining problem is now for me to find the time to check the result thoroughly and to write it down here as an answer. Feb 19, 2017 at 8:05
• @SSCNapoli Very good. I have the mathematica notebook ready. If you give me your email adress I can send it to you, and we can continue the discussion via email. The result is very long and not suitable to be simply presented in this forum. I have found that the case of a double root in the denominator leads to an integral which I (and Mathematica) could not solve. This happens if K = 2/27 (10 +/- 7 Sqrt). Feb 22, 2017 at 18:07