What does the following integral evaluate to? $$ \int_{-1}^1 \mathrm dx\,\sqrt{1-x^2}P_m^n(x)P_l^n(x)? $$


I have developed a condition that may be written as $$ \sum_{m=0}^\infty \sum_{n=-m}^m A_{m,n}P_m^n(\cos\theta)e^{in\phi}=\sum_{m=0}^\infty \sum_{n=-m}^m B_{m,n}\left[\alpha\sin\theta + \beta\cos\theta\right]P_m^n(\cos\theta)e^{in\phi}, $$ where $A_{m,n}$ and $B_{m,n}$ are constant coefficients, $\alpha$ and $\beta$ are constants, $\theta$ and $\phi$ are angles in spherical coordinates, and $P_m^n(x)$ are the associated Legendre polynomials. I am interested in using orthogonality to get rid of the summations. Orthogonality over $\phi$ simply yields $$ \sum_{m=0}^\infty A_{m,n}P_m^n(\cos\theta)=\sum_{m=0}^\infty B_{m,n}\left[\alpha\sin\theta + \beta\cos\theta\right]P_m^n(\cos\theta). $$ In order to implement orthogonality with respect to $\theta$, multiply by $P_l^n(\cos\theta)\sin\theta$ and integrate over $\theta$ from $0$ to $\pi$. This yields (after changing to the variable $x=\cos\theta$) \begin{align*} &\mathrel{\phantom{=}}{} \sum_{m=0}^\infty A_{m,n}\int_{-1}^1 \,\mathrm dx\ P_m^n(x)P_l^n(x)\\ &=\sum_{m=0}^\infty B_{m,n}\alpha\int_{-1}^1 \,\mathrm dx\,\sqrt{1-x^2}P_m^n(x)P_l^n(x) + \sum_{m=0}^\infty B_{m,n}\beta\int_{-1}^1 \,\mathrm dx\,x P_m^n(x)P_l^n(x). \end{align*} From this paper I have found expressions for the integrals $$ \int_{-1}^1 \,\mathrm dx\,P_m^n(x)P_l^n(x), \qquad \int_{-1}^1 \,\mathrm dx\,xP_m^n(x)P_l^n(x), $$ but I have not found an expression for $$ \int_{-1}^1 \,\mathrm dx\,\sqrt{1-x^2}P_m^n(x)P_l^n(x) $$ and I do not have much experience dealing with associated Legendre polynomials.


$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}\def\threej#1#2#3#4#5#6{\begin{pmatrix}#1&#3&#5\\#2&#4&#6\end{pmatrix}}$Note that $P_1^1(x) = -\sqrt{1 - x^2}$, thus$$ \int_{-1}^1 \sqrt{1 - x^2} P_m^n(x) P_l^n(x) \,\d x = -\int_{-1}^1 P_1^1(x) P_m^n(x) P_l^n(x) \,\d x. $$ Referring to the results by Dong and Renato,\begin{align*} &\peq I(l_1, m_1; l_2, m_2; l_3, m_3) = \int_{-1}^1 P_{l_1}^{m_1}(x) P_{l_2}^{m_2}(x) P_{l_3}^{m_3}(x) \,\d x\\ &= \sqrt{\frac{(l_1 + m_1)! (l_2 + m_2)! (l_3 + m_3)!}{(l_1 - m_1)! (l_2 - m_2)! (l_3 - m_3)!}} \sum_{l_{12}} \sum_{l_{123}} G_{12} G_{123} \sqrt{\frac{(l_{123} - m_{123})!}{(l_{123} + m_{123})!}} · I(l_{123}, m_{123}), \end{align*} where:

  1. $m_{12} = m_1 + m_2$, $m_{123} = m_1 + m_2 + m_3$,
  2. $|l_1 - l_2| \leqslant l_{12} \leqslant l_1 + l_2$, $l_{12} \geqslant m_{12}$, and $l_{12} + l_1 + l_2$ is even,
  3. $|l_{12} - l_3| \leqslant l_{123} \leqslant l_{12} + l_3$, $l_{123} \geqslant m_{123}$, and $l_{123} + l_3 + l_{123}$ is even,
  4. \begin{align*} G_{12} &= (-1)^{m_{12}} (2l_{12} + 1) \threej{l_1}{0}{l_2}{0}{l_{12}}{0} \threej{l_1}{m_1}{l_2}{m_2}{l_{12}}{-m_{12}},\\ G_{123} &= (-1)^{m_{123}} (2l_{123} + 1) \threej{l_{12}}{0}{l_3}{0}{l_{123}}{0} \threej{l_{12}}{m_{12}}{l_3}{m_3}{l_{123}}{-m_{123}}, \end{align*}
  5. \begin{align*} &\peq I(l_{123}, m_{123}) = \int_{-1}^1 P_{l_{123}}^{m_{123}}(x) \,\d x\\ &= ((-1)^{m_{123}} + (-1)^{l_{123}}) 2^{m_{123} - 2} m_{123} · \frac{Γ\left( \dfrac{l_{123}}{2} \right) Γ\left( \dfrac{l_{123} + m_{123} + 1}{2} \right)}{\left( \dfrac{l_{123} - m_{123}}{2} \right)! · Γ\left( \dfrac{l_{123} + 2}{2} \right)}. \end{align*}

Now plugging in $(l_1, m_1; l_2, m_2; l_3, m_3) = (1, 1; m, n; l, n)$ yields the result.

  • $\begingroup$ Excellent!..... $\endgroup$ – Yuri Negometyanov Jun 10 '18 at 7:06

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