As the previous

$$\begin{split} \int\_{\rho^{2}}^{1} (1-t)^{2\nu-m-2} {}\_{2}F\_{1} \left( \begin{matrix} -m, 2(\nu-m)+k \\ 1+k-m \end{matrix} \bigg| t \right) dt &= {}\_{k}^{\nu,m} \Big| \int\_{\rho^{2}}^{1} (1-t)^{2\nu-m-2} {}\_{2}F\_{1} \left( \begin{matrix} -m+1, 2\nu-m \\ 1-t \end{matrix} \bigg| t \right) dt \\ &= (-1)^{m} {}\_{k}^{\nu,m} \Big| \int\_{\rho^{2}-1}^{0} t^{2\nu-m-2} {}\_{2}F\_{1} \left( \begin{matrix} -m+1, 2\nu-m \\ 2(\nu-m) \end{matrix} \bigg| t \right) dt \\ &= \frac{{}a\_{k}^{\nu,m}}{2\nu-m-1} \Big( 1-\rho^{2} \Big)^{2\nu-m-1} {}\_{3}F\_{2} \left( \begin{matrix} -m+1, 2\nu-m, 2\nu-m-1 \\ 2(\nu-m), 2\nu-m \end{matrix} \bigg| 1-\rho^{2} \right) \end{split}$$

also

$${}\_{3}F\_{2}\left( \begin{array}{c} -m+1, 2\nu-m, 2\nu-m-1\\ 2(\nu-m), 2\nu-m \end{array} \bigg| 1-\rho^{2} \right) = \,\_{2}F\_{1}\left( \begin{array}{c} -m+1, 2\nu-m-1\\ 2(\nu-m) \end{array} \bigg| 1-\rho^{2} \right)$$

Now if *k*< *m*. We have

$$\phi\_k^{\nu,m}(\mathbf{z}) = (-1)^k \sqrt{\frac{2(\nu-m)-1}{\pi} \frac{k! \Gamma(2(\nu-m)+m)}{m! \Gamma(2(\nu-m)+k)}} \left(\mathbf{1} - |\mathbf{z}|^2\right)^{-m} \overline{\mathbf{z}}^{m-k} P\_k^{(m-k, \frac{2(\nu-m)-1}{2})} \left(\mathbf{1} - 2|\mathbf{z}|^2\right).$$

By the formula

$$\frac{\Gamma(m+1)}{\Gamma(m-s+1)}P\_{m}^{(-s,-a)}(u) = \frac{\Gamma(m+a+1)}{\Gamma(m-s+a+1)} \left(\frac{u-1}{2}\right)^s P\_{m-s}^{(s,a)}(u), 1 \le s \le m,\tag{38}$$

and put *s* ¼ *m* � *k* and *α* ¼ 2ð Þ� *ν* � *m* 1, we have

$$P\_k^{(m-k,\ 2(\nu-m)-1)}\left(\mathbf{1}-2|z|^2\right) = \frac{m!\Gamma(k+a+1)}{k!\Gamma(m+a+1)}P\_m^{(k-m,\ 2(\nu-m)-1)}\left(\mathbf{1}-2|z|^2\right),$$

substituting in the expression of *ϕ<sup>ν</sup>*,*<sup>m</sup> <sup>k</sup>* ð Þ*z* , we get

$$
\phi\_k^{\nu,m}(x) = (-1)^m \sqrt{\frac{2(\nu - m) - 1}{\pi} \frac{m! \Gamma(2(\nu - m) + k)}{k! \Gamma(2(\nu - m) + m)}} \left(1 - |x|^2\right)^{-m} x^{k - m} P\_m^{(k - m, \frac{\nu}{2}(\nu - m) - 1)} \left(1 - 2|x|^2\right),
$$

it is the same formula for *k*> *m*, which proves the same formula of L*<sup>ν</sup> ϕ<sup>ν</sup>*,*<sup>m</sup> k* � �ð Þ*<sup>z</sup>* if *k*> *m*.

**Remark 5.1.** *By the previous formula in [9], we have*

$${}\_{2}F\_{1}\left(\begin{array}{c} -m+1,2(\nu-m)+k\\2(\nu-m) \end{array}\Big|\rho^{2}\right) = \frac{k!\Gamma(2+k-m)}{\Gamma(1-2(\nu-m))}{}\_{2}F\_{1}\left(\begin{array}{c} -m+1,2(\nu-m)+k\\2(\nu-m) \end{array}\Big|\mathbf{1}-\rho^{2}\right).$$
