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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.

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    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jan 24 '17 at 16:15
  • $\begingroup$ 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 $\endgroup$ – SSC Napoli Jan 24 '17 at 17:11
  • $\begingroup$ They are not pleasant computations to carry on, would you add some context? What is the origin of this peculiar problem? $\endgroup$ – Jack D'Aurizio Jan 24 '17 at 17:19
  • $\begingroup$ 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. $\endgroup$ – SSC Napoli Jan 24 '17 at 17:59
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    $\begingroup$ 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. $\endgroup$ – Dr. Wolfgang Hintze Jan 27 '17 at 11:29
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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 $n=1$.

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

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

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

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

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    $\begingroup$ @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. $\endgroup$ – Dr. Wolfgang Hintze Jan 31 '17 at 0:05
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    $\begingroup$ @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. $\endgroup$ – Dr. Wolfgang Hintze Jan 31 '17 at 0:49
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    $\begingroup$ @SSC Napoli Yes, you are right in this case. $\endgroup$ – Dr. Wolfgang Hintze Feb 15 '17 at 16:06
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    $\begingroup$ @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. $\endgroup$ – Dr. Wolfgang Hintze Feb 19 '17 at 8:05
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    $\begingroup$ @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[7]). $\endgroup$ – Dr. Wolfgang Hintze Feb 22 '17 at 18:07

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