**5.2 The spectrum of** L*<sup>ν</sup>*

**Proposition 5.2.** *If k* 6¼ *m, then*

$$
\lambda\_k = \sqrt{J\_1 + J\_2 + J\_3}.
$$

where

$$\begin{aligned} J\_1 &= \left(\frac{(1+k-m)\_m}{m!(k-m+1)}\right)^2 \sum\_{n=0}^{\infty} A\_n \frac{\Gamma(2n+2k-2m+6-1)\Gamma(4\nu-2m-1)}{\Gamma(2n+2k-4m+4\nu+6)}, \\ J\_2 &= \left(\frac{\alpha\_k^{\nu,m}}{2\nu-m-1}\right)^2 \sum\_{n=0}^{\infty} A\_n \frac{\Gamma(4\nu-2m-1)\Gamma(2n+2)}{\Gamma(2n+4\nu-2m+1)}.\end{aligned}$$

and

$$J\_3 = \frac{(1+k-m)\_m a\_k^{\nu,m}}{m!(k-m+1)(2\nu-m-1)} \left( \sum\_{n=0}^{\infty} A\_n \frac{\Gamma(k-m+2)\Gamma(4\nu-2m-1)}{\Gamma(4\nu-k-3m)} \right),$$

*If k* ¼ *m then*

$$
\lambda\_k^2 = \frac{\alpha\_k^{\nu, m} (2(2\nu - m) - 1)}{8(\pi(2\nu - m + 1))} \sum\_{n=0}^{\infty} \frac{B\_n}{n + 2\nu - m}. \tag{39}
$$

where

$$B\_n = \sum\_{n=0}^{\infty} \frac{\Gamma(-m+1)\Gamma(2\nu-m)\Gamma(2(\nu-m)+1)}{n!\Gamma(2(\nu-m)\Gamma(2\nu-m+2))}.$$

*Proof.* If *k* 6¼ *m*. We have

$$\left(\mathcal{L}\_{\nu}\left(\boldsymbol{\phi}\_{k}^{\nu,m}\right)\right)(\boldsymbol{z}) = \frac{\pi \boldsymbol{\gamma}\_{k}^{\nu,m} (I\_{3} + I\_{4})}{2(k - m)} \boldsymbol{\varepsilon}^{(k - m)\nu}.$$

$$\text{We set } H = \left(L^{2}(\mathbb{D}), \left(1 - |\boldsymbol{\xi}|^{2}\right)^{2\nu - 2} d\mu(\boldsymbol{\xi})\right), I\_{3} = I\_{3}(\rho), \text{ and } I\_{4} = I\_{4}(\rho) \text{ we have}$$

$$\begin{split} \lambda\_{k}^{2} &= \left<\mathcal{L}\_{\nu}\left(\boldsymbol{\theta}\_{k}^{\nu,m}\right), \mathcal{L}\_{\nu}\left(\boldsymbol{\theta}\_{k}^{\nu,m}\right)\right>\_{H} \\ &= \frac{\pi^{2} \boldsymbol{\gamma}\_{k}^{\nu,m}}{(k - m)} \int\_{0}^{1} \left(I\_{3}(\rho) + I\_{4}(\rho)\right)^{2} \rho d\rho. \end{split}$$

Calculus of Ð <sup>1</sup> <sup>0</sup>ð Þ *I*3ð Þ*ρ* 2 *ρdρ*.

$$I\_3(\rho) = \frac{(1+k-m)\_m}{m!(k-m+1)} \rho^{k-m+2} (1-\rho^2)^{2\nu-m-1} \,\_2F\_1 \left( \begin{matrix} -m+1, 2(\nu-m)+k \\ 2+k-m \end{matrix} \; | \rho^2 \right).$$

Since

$${}\_{2}F\_{1}\left(\begin{array}{c} -m+1, \, 2(\nu-m)+k\\2+k-m \end{array}\Big|\rho^{2}\right)=\sum\_{n=0}^{\infty}\frac{(-m+1)\_{n}(2(\nu-m)+k)\_{n}}{(2+k-m)\_{n}}\frac{\rho^{2n}}{n!},$$

then

$$\left(I\_3(\rho)\right)^2 = \left(\frac{(1+k-m)\_m}{m!(k-m+1)}\right)^2 \sum\_{n=0}^{\infty} A\_n \rho^{2n} \left(1-\rho^2\right)^{4\nu-2m-2}.$$

*The Singular Values of the Logarithmic Potential Transform on Bound States Spaces DOI: http://dx.doi.org/10.5772/intechopen.107090*

where

$$A\_n = \frac{1}{n!} \sum\_{i=0}^n \frac{(-m+1)\_i (-m+1)\_{n-i} (2(\nu-m)+k)\_i (2(\nu-m)+k)\_{n-i}}{(2(\nu-m))\_i (2(\nu-m))\_{n-i}}.$$

Thus

$$J\_1 = \int\_0^1 (I\_3(\rho))^2 \rho d\rho = \left(\frac{(1+k-m)\_m}{m!(k-m+1)}\right)^2 \sum\_{n=0}^\infty A\_n \int\_0^1 \rho^{2n+2k-2m+6-1} (1-\rho^2)^{4\nu-2m-1-1} d\rho.$$

Use the fact that

$$\int\_0^1 t^{a-1} (1-t)^{\beta-1} dt = \frac{\Gamma(a)\Gamma(\beta)}{\Gamma(a+\beta)},$$

implies

$$\int\_0^1 \left( I\_3(\rho) \right)^2 \rho d\rho = \left( \frac{(1+k-m)\_m}{m!(k-m+1)} \right)^2 \sum\_{n=0}^\infty A\_n \frac{\Gamma(2n+2k-2m+6-1)\Gamma(4\nu-2m-1)}{\Gamma(2n+2k-4m+4\nu+6)}.\tag{40}$$

Calculus of Ð <sup>1</sup> <sup>0</sup>ð Þ *I*4ð Þ*ρ* 2 *ρdρ*. In the same,

$$J\_2 = \int\_0^1 (I\_4(\rho))^2 \rho d\rho = \left(\frac{a\_k^{\nu, m}}{2\nu - m - 1}\right)^2 \sum\_{n=0}^\infty A\_n \frac{\Gamma(4\nu - 2m - 1)\Gamma(2n + 2)}{\Gamma(2n + 4\nu - 2m + 1)}.\tag{41}$$

$$\begin{aligned} \text{Calculus of } 2 \int\_0^1 (I\_3(\rho))(I\_4(\rho)) \rho d\rho. \\ J\_3 = 2 \int\_0^1 (I\_3(\rho))(I\_4(\rho)) \rho d\rho = \frac{(1+k-m)\_m a\_k^{\nu, m}}{m!(k-m+1)(2\nu-m-1)} \sum\_{n=0}^\infty A\_n \frac{\Gamma(k-m+2)\Gamma(4\nu-2m-1)}{\Gamma(4\nu-k-3m)}. \end{aligned} \tag{42}$$

$$\begin{array}{l} \text{If } k = m. \\ \text{Since} \end{array}$$

$$
\begin{pmatrix}
\,\_3F\_2\left(\begin{matrix}-m+1, & 2\nu-m, & 2(\nu-m)+1\\2(\nu-m), & 2\nu-m+2\end{matrix}\bigg|1-\rho^2\right)\end{pmatrix} = \sum\_{n=0}^{m} \frac{\Gamma(-m+1)\Gamma(2\nu-m)\Gamma(2(\nu-m)+1)}{n!\Gamma(2(\nu-m)\Gamma(2\nu-m+2))} \left(1-\rho^2\right)^n.
\tag{43}
$$

$$
\lambda\_k^2 = \frac{a\_k^{\nu,m}(2(2\nu-m)-1)}{8(\pi(2\nu-m+1))} \sum\_{n=0}^{\infty} B\_n \int\_0^1 \left(1-\rho^2\right)^{n+2\nu-m-1} d\rho = \frac{a\_k^{\nu,m}(2(2\nu-m)-1)}{8(\pi(2\nu-m+1))} \sum\_{n=0}^{\infty} \frac{B\_n}{n+2\nu-m},\tag{44}
$$

where

$$B\_n = \sum\_{\nu=0}^{\infty} \frac{\Gamma(-m+1)\Gamma(2\nu-m)\Gamma(2(\nu-m)+1)}{n!\Gamma(2(\nu-m)\Gamma(2\nu-m+2))}.$$
