**Proposition 6.1.**

$$
\lambda\_k \sim C \sqrt{k^{m-4\nu+1}}, \text{ as } k \to \infty,
$$

where *C* is a constant. *Proof.* If *k*> *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

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

The limit of *λ<sup>k</sup>* as *k* ! ∞. We use the formula

$$\frac{\Gamma(k+a)}{k=b} \sim k^{a-b}$$

we have

$$J\_1 \sim \left(\frac{k^{-1-m}}{m!}\right)^2 \sum\_{n=0}^{\infty} A\_n \Gamma(4\nu - 2m - 1)(2k)^{2m - 4\nu - 1} \sim k^{-4\nu - 1} 2^{2m - 4\nu - 1} \Gamma(4\nu - 2m - 1) \sum\_{n=0}^{\infty} \frac{A\_n}{m!}.\tag{45}$$

$$J\_2 = \mathcal{O}\_{k \sim \infty}(\mathbf{1})\tag{46}$$

In the same

$$J\_3 \sim k^{m-4\nu+1} \frac{a\_k^{\nu,m} \Gamma(4\nu - 2m - 1)}{m!(2\nu - m - 1)} \sum\_{n=0}^{\infty} A\_n. \tag{47}$$

Therefore

$$
\lambda\_k \sim C \sqrt{k^{m-4\nu+1}},
$$

where *C* is a constant.

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