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While working on my research, I ended up finding an infinite sum of the kind:

\begin{equation} S = \sum_{l=2}^{\infty} \frac{2l+1}{l^2 (l+1)^2} P_l (\cos(\theta)) \end{equation}

with $P_l$ the Legendre polynomial of order l and $\theta$ the polar angle in spherical coordinates, going from 0 to $\pi$.

I know that you can split the sum into two sums

\begin{equation} S = \sum_{l=2}^{\infty} \left( \frac{1}{l^2 }- \frac{1}{(l+1)^2} \right) P_l (\cos(\theta)) \end{equation}

I tried to look at the book by Prudnikov on integrals and series, but I couldn't find a relevant example. In addition, Wolfram Mathematica doesn't give me a closed-form expression. Could you help me find a closed-form expression?

Since the coefficients decay with the third power of $l$, only the first few terms are really relevant. Indeed, by plotting the first 20 or 40 terms with Mathematica, I can get a nice plot that shows that the sum converges over the entire interval of $\theta$.

For some context: This sum arises from looking at the Green's function of the biharmonic equation on a spherical surface $\vec \nabla^2(\vec \nabla^2 u)=0$, with $u(\theta)$ being a small deformation of the surface. The equation is complemented by the constraint that the volume of the sphere is constant and its center of mass is fixed. These two constraints eliminate the contributions of the modes $l=0$ and $l=1$ from the infinite sum, and this is why the sum starts from $l=2$.

SSC Napoli
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1 Answers1

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From book example: 6.11.2 page: 486 $$\sum _{l=2}^{\infty } \frac{(2 l+1) P_l(x)}{l^2 (l+1)^2}=1-\frac{\pi ^2}{6}+\text{Li}_2\left(\frac{1+x}{2}\right)-\frac{3 x}{4}$$ for: $-1\leq x\leq 1$

where: $\text{Li}_2$ is the polylogarithm function.

Mariusz Iwaniuk
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