# Orthogonal polynomials for the weight $x^{\nu}\exp(-x-t/x)$

S. Yakubovich
Department of Mathematics, Fac. Sciences of University of Porto,
Rua do Campo Alegre, 687; 4169-007 Porto (Portugal)

###### Abstract.

Orthogonal polynomials for the weight $x^{\nu}\exp(-x-t/x),\
x,t>0,\nu\in\mathbb{R}$ are investigated. Differential-difference equations,
recurrence relations, explicit representations, generating functions and
Rodrigues-type formula are obtained.

###### Key words and phrases:

Classical orthogonal polynomials, Laguerre polynomials, modified Bessel
function, generalized hypergeometric function

###### 2000 Mathematics Subject Classification:

33C10, 42C05, 44A15

E-mail<EMAIL_ADDRESS>
The work was partially supported by CMUP, which is financed by national funds
through FCT(Portugal), under the project with reference UIDB/00144/2020.

##  1\. Introduction and preliminary results

Let $\nu\in\mathbb{R},\ t>0$ be parameters and consider the sequence of
orthogonal polynomials $\\{P_{n}^{\nu}(x,t)\\}_{n\geq 0}$ of degree $n$,
satisfying the orthogonality condition

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{m}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu}dx=\delta_{n,m},$

where $\delta_{n,m},\ n,m\in\mathbb{N}_{0}$ is the Kronecker symbol. Admitting
the limit case $t=0$ and $\nu>-1$, it gives normalized Laguerre polynomials

$None$
$P_{n}^{\nu}(x,0)\equiv{\tilde{L}}_{n}^{\nu}(x)=\left({n!\over\Gamma(n+\nu+1)}\right)^{1/2}L_{n}^{\nu}(x),$

where $\\{L_{n}^{\nu}(x)\\}_{n\geq 0}$ are classical Laguerre polynomials [2],
Vol. II and $\Gamma(z)$ is the Euler gamma function [2], Vol. I. It satisfies
the three term recurrence relation

$None$
$xL_{n}^{\nu}(x)=-(n+1)L_{n+1}^{\nu}(x)+(2n+\nu+1)L_{n}^{\nu}(x)-(n+\nu)L_{n-1}^{\nu}(x),\
n\in\mathbb{N}.$

As it follows from the theory of orthogonal polynomials, the three term
recurrence relation for the sequence $\\{P_{n}^{\nu}(x,t)\\}_{n\geq 0}$ can be
written in the form

$None$
$xP_{n}^{\nu}(x,t)=A_{n+1}(t)P_{n+1}^{\nu}(x,t)+B_{n}(t)P_{n}^{\nu}(x,t)+A_{n}(t)P_{n-1}^{\nu}(x,t),$

where $P_{-1}^{\nu}(x,t)\equiv 0,\
P_{n}^{\nu}(x,t)=a_{n}(t)x^{n}+b_{n}(t)x^{n-1}+\dots,\quad a_{n}(t)\neq 0$ and

$None$ $A_{n}(t)={a_{n-1}(t)\over a_{n}(t)},\quad B_{n}(t)={b_{n}(t)\over
a_{n}(t)}-{b_{n+1}(t)\over a_{n+1}(t)}.$

As a consequence of (1.4) the Christoffel-Darboux formula takes place

$None$
$\sum_{k=0}^{n}P_{k}^{\nu}(x,t)P_{k}^{\nu}(y,t)=A_{n+1}(t)\frac{P_{n+1}^{\nu}(x,t)P_{n}^{\nu}(y,t)-P_{n}^{\nu}(x,t)P_{n+1}^{\nu}(y,t)}{x-y}.$

Using the value of the integral [2], Vol. II

$None$
$\int_{0}^{\infty}e^{-t/x-x}x^{\nu-1}dx=2t^{\nu/2}K_{\nu}\left(2\sqrt{t}\right)\equiv\rho_{\nu}(t),\
t>0,$

where $K_{\nu}(z)$ is the modified Bessel function or Macdonald function [6],
we can find the moments of the weight $x^{\nu}\exp(-x-t/x)$. Namely, we find

$None$ $\int_{0}^{\infty}e^{-t/x-x}x^{\nu+n}dx=\rho_{\nu+n+1}(t),\quad
n\in\mathbb{N}_{0}.$

The asymptotic behavior of the modified Bessel function at infinity and near
the origin [2], Vol. II gives the corresponding values for the function
$\rho_{\nu},\ \nu\in\mathbb{R}$. Precisely, we have

$None$ $\rho_{\nu}(t)=O\left(t^{(\nu-|\nu|)/2}\right),\ t\to 0,\ \nu\neq
0,\quad\rho_{0}(t)=O(\log t),\ t\to 0,$

$None$ $\rho_{\nu}(t)=O\left(t^{\nu/2-1/4}e^{-2\sqrt{t}}\right),\
t\to+\infty.$

Moreover, it can be represented in terms of Laguerre polynomials (cf. [4],
Vol. II, Entry 2.19.4.13 )

$None$ ${(-1)^{n}t^{n}\over n!}\
\rho_{\nu}(t)=\int_{0}^{\infty}x^{\nu+n-1}e^{-x-t/x}L_{n}^{\nu}(x)dx,\quad
n\in\mathbb{N}_{0}.$

Further, it has a relationship with the Riemann-Liouville fractional integral
[6]

$None$
$\left(I_{-}^{\alpha}f\right)(t)={1\over\Gamma(\alpha)}\int_{t}^{\infty}(x-t)^{\alpha-1}f(x)dx,\quad{\rm
Re}\alpha>0,$

namely, we get the formula

$None$ $\rho_{\nu}(t)=\left(I_{-}^{\nu}\rho_{0}\right)(t),\ \nu>0.$

Hence the index law for fractional integrals immediately implies

$None$
$\rho_{\nu+\mu}(t)=\left(I_{-}^{\nu}\rho_{\mu}\right)(t)=\left(I_{-}^{\mu}\rho_{\nu}\right)(t).$

The corresponding definition of the fractional derivative presumes the
relation $D^{\mu}_{-}=-DI_{-}^{1-\mu}$. Hence for the ordinary $n$-th
derivative of $\rho_{\nu}$ we find

$None$ $D^{n}\rho_{\nu}(t)=(-1)^{n}\rho_{\nu-n}(t),\quad n\in\mathbb{N}_{0}.$

Recalling (1.7) and integrating by parts, it is not difficult to establish the
following recurrence relation for $\rho_{\nu}$

$None$
$\rho_{\nu+1}(t)=\nu\rho_{\nu}(t)+t\rho_{\nu-1}(t),\quad\nu\in\mathbb{R}.$

In the operator form it can be written as follows

$None$ $\rho_{\nu+1}(t)=\left(\nu-tD\right)\rho_{\nu}(t).$

Finally in this section we observe that up to a normalization factor the
orthogonality (1.1) is equivalent to the following $n$ equalities

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)\ e^{-x-t/x}x^{\nu+m}dx=0,\quad
m=0,1,\dots,n-1.$

Besides, taking into account (1.4), (1.5), we get the identities

$None$ $\int_{0}^{\infty}[P_{n}^{\nu}(x,t)]^{2}\
e^{-x-t/x}x^{\nu+1}dx=B_{n}(t),$

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)\ e^{-x-t/x}x^{\nu+n}dx={1\over
a_{n}(t)},$

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu+n+1}dx=-{b_{n+1}(t)\over a_{n+1}(t)a_{n}(t)},$

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu+1}dx=A_{n}(t),$

and with (1.8) it yields

$P_{0}^{\nu}(x,t)=[\rho_{\nu+1}(t)]^{-1/2},\quad
P_{1}^{\nu}(x,t)=\left[\rho_{\nu+1}(t)(\rho_{\nu+3}(t)\rho_{\nu+1}(t)-\rho^{2}_{\nu+2}(t))\right]^{-1/2}$

$None$ $\times\left[-\rho_{\nu+1}(t)x+\rho_{\nu+2}(t)\right],$

where $\rho_{\nu+3}(t)\rho_{\nu+1}(t)-\rho^{2}_{\nu+2}(t)>0$ for all $t>0$.
The latter fact follows immediately from the Ismail integral representation of
the quotient of functions $\rho_{\nu},\rho_{\nu+1}$ [3]

$None$
${\rho_{\nu}(x)\over\rho_{\nu+1}(x)}={1\over\pi^{2}}\int_{0}^{\infty}{y^{-1}dy\over(x+y)\left[J_{\nu+1}^{2}(2\sqrt{y})+Y^{2}_{\nu+1}(2\sqrt{y})\right]}$

and the Nicholson’s integral [2], Vol. II

$None$
$J_{\nu}^{2}(2\sqrt{x})+Y^{2}_{\nu}(2\sqrt{x})={8\over\pi^{2}}\int_{0}^{\infty}K_{0}\left(4\sqrt{x}\
\sinh(t)\right)\cosh(2\nu t)dt.$

##  2\. Differential and differential-difference equations

We begin this section with the following auxiliary lemma.

Lemma 1. Let $\nu\in\mathbb{R},\ t>0,\ n\in\mathbb{N}$. Then it has the
identities

$None$ $\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu-1}dx={1\over t}\left[B_{n}(t)-\nu-1-2n\right],$

$None$ $\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu-2}dx={1\over t}\left[1-{\nu\over
t}\left[B_{n}(t)-\nu-1-2n\right]\right],$

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu-1}dx={1\over t}\left[A_{n}(t)+{b_{n}(t)\over
a_{n}(t)A_{n}(t)}\right],$

$None$ $\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu-2}dx=-{1\over t}\left[{\nu\over
t}\left[A_{n}(t)+{b_{n}(t)\over a_{n}(t)A_{n}(t)}\right]+{n\over
A_{n}(t)}\right].$

###### Proof.

In fact, employing (1.18), (1.19), (1.20) and integration by parts we derive

$\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu-1}dx={1\over
t}\left[\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu+1}dx\right.$

$\left.-(\nu+1)\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu}dx-2\int_{0}^{\infty}P_{n}^{\nu}(x,t){\partial\over\partial
x}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu+1}dx\right]$

$={1\over t}\left[B_{n}(t)-\nu-1-2n\right],$

which proves (2.1). Then, analogously, recalling (1.1) and using (2.1), we get

$\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu-2}dx={1\over
t}\left[\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu}dx\right.$

$\left.-\nu\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu-1}dx-2\int_{0}^{\infty}P_{n}^{\nu}(x,t){\partial\over\partial
x}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu}dx\right]$

$={1\over t}\left[1-{\nu\over t}\left[B_{n}(t)-\nu-1-2n\right]\right].$

Concerning (2.3), we deduce, invoking (1.1), (1.5), (1.20), (1.21), (1.22),

$\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu-1}dx={1\over
t}\left[\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu+1}dx\right.$

$\left.-(\nu+1)\int_{0}^{\infty}P_{n}^{\nu}(x,t)\
P_{n-1}^{\nu}(x,t)e^{-x-t/x}x^{\nu}dx-\int_{0}^{\infty}P_{n}^{\nu}(x,t){\partial\over\partial
x}[P_{n-1}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu+1}dx\right.$

$\left.-\int_{0}^{\infty}P_{n-1}^{\nu}(x,t){\partial\over\partial
x}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu+1}dx\right]={1\over
t}\left[A_{n}(t)+{b_{n}(t)\over a_{n}(t)A_{n}(t)}\right].$

Finally, a similar reasoning gives via (2.3)

$\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu-2}dx={1\over
t}\left[\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{n-1}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu}dx\right.$

$\left.-\nu\int_{0}^{\infty}P_{n}^{\nu}(x,t)\
P_{n-1}^{\nu}(x,t)e^{-x-t/x}x^{\nu-1}dx-\int_{0}^{\infty}P_{n}^{\nu}(x,t){\partial\over\partial
x}[P_{n-1}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu}dx\right.$

$\left.-\int_{0}^{\infty}P_{n-1}^{\nu}(x,t){\partial\over\partial
x}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu}dx\right]=-{1\over t}\left[{\nu\over
t}\left[A_{n}(t)+{b_{n}(t)\over a_{n}(t)A_{n}(t)}\right]+{n\over
A_{n}(t)}\right],$

and we establish (2.4), completing the proof of Lemma 1.

∎

Now we are ready to prove the following theorem.

Theorem 1. Let $n\in\mathbb{N}$. Orthogonal polynomials $P_{n}^{\nu}(x,t)$
satisfy the first order linear differential-difference equation

$None$ $x^{2}{\partial\over\partial x}[P_{n}^{\nu}(x,t)]=\left[nx-
A_{n}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}\right]P_{n}^{\nu}(x,t)+A_{n}(t)\left[x+B_{n}(t)-\nu-1-2n\right]P_{n-1}^{\nu}(x,t).$

###### Proof.

Since ${\partial\over\partial x}[P_{n}^{\nu}(x,t)]$ is a polynomial of degree
$n-1$, we write it in the form

$None$ ${\partial\over\partial
x}[P_{n}^{\nu}(x,t)]=\sum_{k=0}^{n-1}c_{n,k}(t)P_{k}^{\nu}(x,t),$

where, owing to the orthogonality,

$None$ $c_{n,k}(t)=\int_{0}^{\infty}{\partial\over\partial
x}[P_{n}^{\nu}(x,t)]P_{k}^{\nu}(x,t)\ e^{-x-t/x}x^{\nu}dx.$

Then, integrating by parts and using the orthogonality relation, we obtain

$None$ $c_{n,k}(t)=-t\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{k}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu-2}dx-\nu\int_{0}^{\infty}P_{n}^{\nu}(x,t)P_{k}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu-1}dx.$

Moreover, we observe that

$\int_{0}^{\infty}P_{n}^{\nu}(y,t)\sum_{k=0}^{n-1}P_{k}^{\nu}(y,t)P_{k}^{\nu}(x,t)\
e^{-y-t/y}y^{\nu}x^{-1}dy=0,$

$\int_{0}^{\infty}P_{n}^{\nu}(y,t)\sum_{k=0}^{n-1}P_{k}^{\nu}(y,t)P_{k}^{\nu}(x,t)\
e^{-y-t/y}y^{\nu}x^{-2}dy=0.$

Therefore from (2.6), (2.8) and Christoffel-Darboux formula (1.6) we derive

${\partial\over\partial
x}[P_{n}^{\nu}(x,t)]=-t\sum_{k=0}^{n-1}P_{k}^{\nu}(x,t)\int_{0}^{\infty}P_{n}^{\nu}(y,t)P_{k}^{\nu}(y,t)\
e^{-y-t/y}y^{\nu}\left[{1\over y^{2}}-{1\over x^{2}}\right]dy$

$-\nu\sum_{k=0}^{n-1}P_{k}^{\nu}(x,t)\int_{0}^{\infty}P_{n}^{\nu}(y,t)P_{k}^{\nu}(y,t)\
e^{-y-t/y}y^{\nu}\left[{1\over y}-{1\over x}\right]dy$

$=-tA_{n}(t)P_{n}^{\nu}(x,t)\int_{0}^{\infty}P_{n}^{\nu}(y,t)P_{n-1}^{\nu}(y,t)\
e^{-y-t/y}y^{\nu}\left[{1\over xy^{2}}+{1\over yx^{2}}\right]dy$

$+tA_{n}(t)P_{n-1}^{\nu}(x,t)\int_{0}^{\infty}\left[P_{n}^{\nu}(y,t)\right]^{2}\
e^{-y-t/y}y^{\nu}\left[{1\over xy^{2}}+{1\over yx^{2}}\right]dy$

$-{\nu\over
x}A_{n}(t)P_{n}^{\nu}(x,t)\int_{0}^{\infty}P_{n}^{\nu}(y,t)P_{n-1}^{\nu}(y,t)\
e^{-y-t/y}y^{\nu-1}dy$

$None$ $+{\nu\over
x}A_{n}(t)P_{n-1}^{\nu}(x,t)\int_{0}^{\infty}\left[P_{n}^{\nu}(y,t)\right]^{2}\
e^{-y-t/y}y^{\nu-1}dy.$

Hence by virtue of Lemma 1 equalities (2.9) become

${1\over A_{n}(t)}{\partial\over\partial x}[P_{n}^{\nu}(x,t)]={1\over
x}P_{n}^{\nu}(x,t)\left[\left({\nu\over t}-{1\over
x}\right)\left[A_{n}(t)+{b_{n}(t)\over a_{n}(t)A_{n}(t)}\right]+{n\over
A_{n}(t)}\right]$

$+{1\over x}P_{n-1}^{\nu}(x,t)\left[1-\left({\nu\over t}-{1\over
x}\right)\left[B_{n}(t)-\nu-1-2n\right]\right]$

$-{\nu\over xt}P_{n}^{\nu}(x,t)\left[A_{n}(t)+{b_{n}(t)\over
a_{n}(t)A_{n}(t)}\right]+{\nu\over
xt}P_{n-1}^{\nu}(x,t)\left[B_{n}(t)-\nu-1-2n\right].$

Hence after simplification we arrive at the differential-difference equation
(2.5). Theorem 1 is proved. ∎

Corollary 1. Denoting by $a_{n,0}(t)$ the free term of the polynomial
$P_{n}^{\nu}(x,t)$, it has the value

$None$ $a_{n,0}(t)={1\over
a_{n}(t)\rho_{\nu+1}(t)}\prod_{k=1}^{n}\frac{B_{k}(t)-\nu-1-2k}{A^{2}_{k}(t)+{b_{k}(t)\over
a_{k}(t)}}.$

###### Proof.

In fact, letting $x=0$ in (2.5), we find the recurrence relation

$None$ $\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]a_{n,0}(t)=\left[B_{n}(t)-\nu-1-2n\right]A_{n}(t)a_{n-1,0}(t).$

Hence formula (2.10) comes immediately, solving recurrence (2.11) with the use
of (1.5) and (1.23).

∎

Remark 1. In the limit case $t=0$ we use (1.2), (1.5), (1.8) to have
$\rho_{\nu+1}(0)=\Gamma(\nu+1)$,

$None$ $a_{n}(0)={(-1)^{n}\over[n!\Gamma(n+\nu+1)]^{1/2}},\quad
b_{n}(0)=(-1)^{n+1}\left({n(n+\nu)\over(n-1)!\Gamma(n+\nu)}\right)^{1/2},$

$B_{n}(0)=2n+\nu+1,\quad A_{n}(0)=-(n(n+\nu))^{1/2},\quad
a_{n,0}(0)={1\over\Gamma(\nu+1)}\left({\Gamma(n+\nu+1)\over n!}\right)^{1/2},$

$None$ $\lim_{t\to 0}\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]\bigg{[}B_{n}(t)-\nu-1-2n\bigg{]}^{-1}=-n.$

Moreover, taking the limit of the indeterminate form under the product sign in
(2.10) when $t\to 0$, it implies the identity

$None$
$\prod_{k=1}^{n}\frac{B^{\prime}_{k}(0)}{(-1)^{k}(k!\Gamma(k+\nu+1))^{1/2}[b^{\prime}_{k}(0)+k(k+\nu)a^{\prime}_{k}(0)]-2(k(k+\nu))^{1/2}A^{\prime}_{k}(0)}={(-1)^{n}\over
n!}.$

Now, assuming that polynomial coefficients in (1.4), (1.5) $a_{n,k}(t)\in
C^{1}(\mathbb{R}_{+}),\ a_{n,n}(t)\equiv a_{n}(t),\ a_{n,n-1}(t)=b_{n}(t)$ as
functions of $t$, we differentiate equality (1.1) by $t$ to obtain

$0={\partial\over\partial
t}\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\
e^{-x-t/x}x^{\nu}dx=2\int_{0}^{\infty}P_{n}^{\nu}(x,t){\partial\over\partial
t}\left[P_{n}^{\nu}(x,t)\right]\ e^{-x-t/x}x^{\nu}dx$

$-\int_{0}^{\infty}\left[P_{n}^{\nu}(x,t)\right]^{2}\ e^{-x-t/x}x^{\nu-1}dx,$

where the differentiation under the integral sign can be easily motivated by
virtue of the absolute and uniform convergence. Then, appealing to (1.20) and
(2.1), it implies the equality

$None$ ${a^{\prime}_{n}(t)\over a_{n}(t)}={B_{n}(t)-\nu-1-2n\over 2t}.$

This first order differential equation with respect to the leading term
$a_{n}(t)$ can be uniquely solved under initial condition (2.12), and we get

$None$ $a_{n}(t)={(-1)^{n}\over\left(n!\Gamma(n+\nu+1)\right)^{1/2}}\
\exp\left({1\over 2}\int_{0}^{t}{B_{n}(y)-\nu-1-2n\over y}dy\right),$

where the integral under the exponential function exists since, evidently,
(see (2.13), (2.15))

$None$ $\lim_{t\to 0}{B_{n}(t)-\nu-1-2n\over
t}=B_{n}^{\prime}(0)=2{a^{\prime}_{n}(0)\over a_{n}(0)}.$

Moreover, employing (1.5), we find

$None$ $A_{n+1}(t)=-((n+1)(n+1+\nu))^{1/2}\ \exp\left({1\over
2}\int_{0}^{t}{B_{n}(y)-B_{n+1}(y)+2\over y}dy\right).$

Hence after differentiation we obtain

$None$ $A^{\prime}_{n+1}(0)={1\over
2}((n+1)(n+1+\nu))^{1/2}\left[B^{\prime}_{n+1}(0)-B^{\prime}_{n}(0)\right].$

On the other hand, via (1.15), (1.23) we have

$None$ $B_{0}(t)={\rho_{\nu+2}(t)\over\rho_{\nu+1}(t)},\quad
B^{\prime}_{0}(t)={\rho_{\nu+2}(t)\rho_{\nu}(t)\over\rho^{2}_{\nu+1}(t)}-1,\quad
B^{\prime}_{0}(0)={1\over\nu},\ \nu>0.$

Hence it gives the equality

$None$
$B^{\prime}_{n+1}(0)={1\over\nu}+2\sum_{k=0}^{n}{A^{\prime}_{k+1}(0)\over((k+1)(k+1+\nu))^{1/2}}.$

The following theorem shows the first order partial differential-difference
equation whose solutions are polynomials $P_{n}^{\nu}(x,t)$.

Theorem 2. Orthogonal polynomials $P_{n}^{\nu}(x,t)$ satisfy the first order
partial differential-difference equation

$None$ $\left(t{\partial\over\partial t}+x{\partial\over\partial
x}\right)P_{n}^{\nu}(x,t)=\left(t{a_{n}^{\prime}(t)\over
a_{n}(t)}+n\right)P_{n}^{\nu}(x,t)+A_{n}(t)P_{n-1}^{\nu}(x,t).$

###### Proof.

Indeed, differentiating both sides of (1.18) with respect to $t$, we have

$None$ $\int_{0}^{\infty}{\partial\over\partial t}[P_{n}^{\nu}(x,t)]\
e^{-x-t/x}x^{\nu+m}dx-\int_{0}^{\infty}P_{n}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu+m-1}dx=0,\quad m=0,1,\dots,n-1.$

Meanwhile, the second integral on the left-hand side of (2.23) can be treated
via integration by parts which gives

$\int_{0}^{\infty}P_{n}^{\nu}(x,t)\ e^{-x-t/x}x^{\nu+m-1}dx={1\over
t}\int_{0}^{\infty}P_{n}^{\nu}(x,t)\ e^{-x-t/x}x^{\nu+m+1}dx$

$-{1\over t}\int_{0}^{\infty}{\partial\over\partial x}[P_{n}^{\nu}(x,t)]\
e^{-x-t/x}x^{\nu+m+1}dx-{\nu+m+1\over t}\int_{0}^{\infty}P_{n}^{\nu}(x,t)\
e^{-x-t/x}x^{\nu+m}dx,$

and, combining with (2.23), (1.4) and (1.18), we obtain

$None$ $\int_{0}^{\infty}\left[\left(t{\partial\over\partial
t}+x{\partial\over\partial
x}\right)P_{n}^{\nu}(x,t)-A_{n}(t)P_{n-1}^{\nu}(x,t)\right]\
e^{-x-t/x}x^{\nu+m}dx=0,\quad m=0,1,\dots,n-1.$

Hence by unicity we therefore have

$None$ $\left(t{\partial\over\partial t}+x{\partial\over\partial
x}\right)P_{n}^{\nu}(x,t)-A_{n}(t)P_{n-1}^{\nu}(x,t)=c_{n}(t)P_{n}^{\nu}(x,t).$

The function $c_{n}(t)$ is defined, equating coefficients of $x^{n}$ on both
sides in (2.25) to find

$c_{n}(t)=t{a^{\prime}_{n}(t)\over a_{n}(t)}+n.$

This completes the proof of Theorem 2.

∎

Corollary 2. Equation $(2.22)$ can be written in the form

$\left(t{\partial\over\partial t}+x{\partial\over\partial
x}\right)P_{n}^{\nu}(x,t)={1\over
2}\left[B_{n}(t)-\nu-1\right]P_{n}^{\nu}(x,t)+A_{n}(t)P_{n-1}^{\nu}(x,t).$

###### Proof.

The proof is immediate with the use of (2.15).

∎

Corollary 3. Let $t>0$. The following equalities take place

$None$ ${d\over dt}\left[{b_{n}(t)\over a_{n}(t)}\right]={1\over
t}\left[A_{n}^{2}(t)+{b_{n}(t)\over a_{n}(t)}\right],$

$None$ $B_{n}^{\prime}(t)={1\over
t}\left[A_{n}^{2}(t)-A_{n+1}^{2}(t)+B_{n}(t)\right],$

$None$ $a_{n,0}^{\prime}(t)={B_{n}(t)-\nu-1\over 2t}a_{n,0}(t)+{A_{n}(t)\over
t}a_{n-1,0}(t).$

###### Proof.

The proof is immediate, equating coefficients of $x^{n-1}$ and free terms on
both sides of (2.22), and the use of (1.5), (2.15).

∎

Further, differentiating (1.19) by $t$ and employing (1.1), (1.21), we deduce

$B_{n}^{\prime}(t)=2\int_{0}^{\infty}P_{n}^{\nu}(x,t){\partial\over\partial
t}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu+1}dx-1$

$=-2{a_{n}^{\prime}(t)b_{n+1}(t)\over
a_{n+1}(t)a_{n}(t)}+2{b_{n}^{\prime}(t)\over a_{n}(t)}-1,$

i.e. (see (1.5))

$None$ $B_{n}^{\prime}(t)=2B_{n}(t){a_{n}^{\prime}(t)\over a_{n}(t)}+2{d\over
dt}\left[{b_{n}(t)\over a_{n}(t)}\right]-1.$

Corollary 4. Let $\nu>0,\ n\in\mathbb{N}_{0}$. It has the values

$None$ $A_{n}^{\prime}(0)=0,\quad B_{n}^{\prime}(0)={1\over\nu},\quad
a_{n}^{\prime}(0)={(-1)^{n}\over 2\nu[n!\Gamma(n+\nu+1)]^{1/2}},$

$None$ $b_{n}^{\prime}(0)=(-1)^{n+1}{n(n+\nu+2)\over
2\nu[n!\Gamma(n+\nu+1)]^{1/2}},\quad a_{n,0}^{\prime}(0)={1\over
2\nu\Gamma(\nu+1)}\left({\Gamma(n+\nu+1)\over n!}\right)^{1/2}.$

###### Proof.

In fact, taking a limit in (2.26) when $t\to 0$, we find

$\lim_{t\to 0}{d\over
dt}\left[A_{n}^{2}(t)\right]=2A_{n}(0)A^{\prime}_{n}(0)=0.$

Hence via (2.13) $A_{n}^{\prime}(0)=0$ and the sum in (2.21) is zero. Thus
$B_{n}^{\prime}(0)={1\over\nu}$. The value for $a_{n}^{\prime}(0)$ comes from
(2.12), (2.17) and $b_{n}^{\prime}(0)$ is obtained from (2.29). Finally,
$a_{n,0}^{\prime}(0)$ is a consequence of (2.28) when $t\to 0$.

∎

Now, assuming that polynomial coefficients are twice continuously
differentiable functions of $t$, we solve a simple Cauchy problem for the
first order differential equation (2.26) to find

$None$ ${b_{n}(t)\over a_{n}(t)}=2t\int_{0}^{t}{A_{n}(y)A_{n}^{\prime}(y)\over
y}dy-A^{2}_{n}(t)-{nt\over\nu}.$

Moreover, via (1.5), (2.27) and integration by parts we derive

$B_{n}(t)=t\int_{0}^{t}{\left(A^{2}_{n}(y)-A^{2}_{n+1}(y)\right)^{\prime}\over
y}dy+A^{2}_{n+1}(t)-A^{2}_{n}(t)+{t\over\nu}$

$None$
$=2n+\nu+1+t\int_{0}^{t}\left[A^{2}_{n}(y)-A^{2}_{n+1}(y)+2n+\nu+1\right]{dy\over
y^{2}}+{t\over\nu},$

$None$
$B^{\prime}_{n}(t)=\int_{0}^{t}{\left(A^{2}_{n}(y)-A^{2}_{n+1}(y)\right)^{\prime}\over
y}dy+{1\over\nu}.$

Corollary 5. Coefficients $A_{n}$ satisfy the following second kind nonlinear
differential-difference equation

$None$
$A_{n}^{\prime\prime}(t)A_{n}(t)-\left(A_{n}^{\prime}(t)\right)^{2}-{A^{2}_{n}(t)\over
2t^{2}}\bigg{[}A_{n-1}^{2}(t)-2A^{2}_{n}(t)+A_{n+1}^{2}(t)-2\bigg{]}=0.$

###### Proof.

Indeed, from (1.5) and (2.18) we get

$None$ ${A_{n}^{\prime}(t)\over A_{n}(t)}={B_{n-1}(t)-B_{n}(t)+2\over
2t},\quad t>0.$

Then by virtue of (2.27)

$2t{d\over dt}\left[t{A_{n}^{\prime}(t)\over A_{n}(t)}\right]=t{d\over
dt}\left[B_{n-1}(t)-B_{n}(t)\right]=B_{n-1}(t)-B_{n}(t)+A_{n-1}^{2}(t)-2A^{2}_{n}(t)+A_{n+1}^{2}(t)$

and the result follows via (2.36) and simple differentiation.

∎

Corollary 6. The free term $a_{n,0}(t)$ can be determined by the formula

$None$ $a_{n,0}(t)=(-1)^{n}(1+\nu)_{n}\
a_{n}(t)\exp\left(\int_{0}^{t}\left[n+{A_{n}^{2}(y)+{b_{n}(y)\over
a_{n}(y)}\over B_{n}(y)-\nu-1-2n}\right]{dy\over y}\right),$

where $(z)_{n}$ is the Pochhammer symbol.

###### Proof.

In fact, invoking (2.11), (2.15), (2.28) and since $B_{n}(t)-\nu-1-2n\neq 0,\
t>0$ via (2.1), we write the differential equation

${a_{n,0}^{\prime}(t)\over a_{n,0}(t)}={a_{n}^{\prime}(t)\over
a_{n}(t)}+{1\over t}\left[n+{A_{n}^{2}(t)+{b_{n}(t)\over a_{n}(t)}\over
B_{n}(t)-\nu-1-2n}\right],$

which can be uniquely solved by formula (2.37), owing to inicial conditions
(2.12) and where the integral converges under condition (2.13).

∎

Theorem 3. Orthogonal polynomials $P_{n}^{\nu}(x,t)$ obey the second order
differential equation

$x^{4}\bigg{[}x+B_{n}(t)-\nu-1-2n\bigg{]}{\partial^{2}\over\partial
x^{2}}[P_{n}^{\nu}(x,t)]-x^{2}\bigg{[}x^{2}+\left[x+B_{n}(t)-\nu-1-2n\right]\bigg{.}$

$\times\left[(2n-3)x-A_{n}^{2}(t)-A_{n-1}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}-{b_{n-1}(t)\over a_{n-1}(t)}\right.$

$\bigg{.}\bigg{.}+\left[x+B_{n-1}(t)-\nu+1-2n\right][x-B_{n-1}(t)]\bigg{]}\bigg{]}{\partial\over\partial
x}[P_{n}^{\nu}(x,t)]$

$+\bigg{[}\left[x+B_{n}(t)-\nu-1-2n\right]\bigg{[}A^{2}_{n}(t)\left[x+B_{n}(t)-\nu-1-2n\right]\left[x+B_{n-1}(t)-\nu+1-2n\right]\bigg{.}\bigg{.}$

$\bigg{.}+\left[nx-A_{n}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}\right]\bigg{[}(n-1)x-A_{n-1}^{2}(t)-{b_{n-1}(t)\over
a_{n-1}(t)}+\left[x+B_{n-1}(t)-\nu+1-2n\right][x-B_{n-1}(t)]\bigg{]}\bigg{]}$

$None$
$\bigg{.}\bigg{.}-x^{2}\bigg{[}n\left[B_{n}(t)-\nu-2n-1\right]+A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\bigg{]}\bigg{]}P_{n}^{\nu}(x,t)=0.$

###### Proof.

Differentiating both sides of (2.5) with respect to $x$, we have

$x^{2}{\partial^{2}\over\partial
x^{2}}[P_{n}^{\nu}(x,t)]=nP_{n}^{\nu}(x,t)+\left[(n-2)x-A_{n}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}\right]{\partial\over\partial x}[P_{n}^{\nu}(x,t)]$

$+A_{n}(t)P_{n-1}^{\nu}(x,t)+A_{n}(t)\left[x+B_{n}(t)-\nu-1-2n\right]{\partial\over\partial
x}[P_{n-1}^{\nu}(x,t)].$

But, owing to (1.4) and (2.5), the latter equality becomes

$x^{4}{\partial^{2}\over\partial
x^{2}}[P_{n}^{\nu}(x,t)]=x^{2}\left[(n-2)x-A_{n}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}\right]{\partial\over\partial x}[P_{n}^{\nu}(x,t)]$

$+\left[nx^{2}-A^{2}_{n}(t)\left[x+B_{n}(t)-\nu-1-2n\right]\left[x+B_{n-1}(t)-\nu+1-2n\right]\right]P_{n}^{\nu}(x,t)$

$+\left[{x^{2}\over
x+B_{n}(t)-\nu-1-2n}+(n-1)x-A_{n-1}^{2}(t)-{b_{n-1}(t)\over
a_{n-1}(t)}\right.$

$\bigg{.}+\left[x+B_{n-1}(t)-\nu+1-2n\right][x-B_{n-1}(t)]\bigg{]}A_{n}(t)\left[x+B_{n}(t)-\nu-1-2n\right]P_{n-1}^{\nu}(x,t).$

Finally, recalling (2.5) to express $P_{n-1}^{\nu}(x,t)$, we end up with the
equation (2.37), completing the proof of Theorem 3.

∎

##  3\. Explicit representations. Recurrence relations for coefficients

In this section we will deduce recurrence relations for coefficients of
orthogonal polynomials $P_{n}^{\nu}(x,t)$ and their explicit representations.
In fact, we have

Theorem 4. Let $n\in\mathbb{N}_{0},\ t>0$. Then the following identities hold

$None$
$A_{n+1}^{2}(t)+B_{n}^{2}(t)+A_{n}^{2}(t)-(2n+\nu+2)B_{n}(t)+2{b_{n}(t)\over
a_{n}(t)}-t=0,$

$None$
$A^{2}_{n}(t)\left[B_{n}(t)-\nu-1-2n\right]\left[B_{n-1}(t)-\nu+1-2n\right]-\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]^{2}+t\bigg{[}A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\bigg{]}=0,$

$None$
$t\bigg{[}B_{n-1}(t)+B_{n}(t)\bigg{]}^{\prime}+\bigg{[}B_{n-1}(t)+B_{n}(t)\bigg{]}\bigg{[}B_{n-1}(t)-B_{n}(t)+1\bigg{]}-(2n+\nu)\bigg{[}B_{n-1}(t)-B_{n}(t)\bigg{]}=0.$

###### Proof.

In order to prove (3.1), we appeal to the three term recurrence relation (1.4)
and the orthogonality condition (1.1). Hence, integrating by parts, we derive

$A_{n+1}^{2}(t)+B_{n}^{2}(t)+A_{n}^{2}(t)=\int_{0}^{\infty}[xP_{n}^{\nu}(x,t)]^{2}\
e^{-x-t/x}x^{\nu}dx=(\nu+2)\int_{0}^{\infty}[P_{n}^{\nu}(x,t)]^{2}\
e^{-x-t/x}x^{\nu+1}dx$

$+t\int_{0}^{\infty}[P_{n}^{\nu}(x,t)]^{2}\
e^{-x-t/x}x^{\nu}dx+2\int_{0}^{\infty}P_{n}^{\nu}(x,t){d\over
dx}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu+2}dx$

$=(\nu+2)B_{n}(t)+t+2nB_{n}(t)+2A_{n}(t)\int_{0}^{\infty}xP_{n-1}^{\nu}(x,t){d\over
dx}[P_{n}^{\nu}(x,t)]\ e^{-x-t/x}x^{\nu}dx$

$t+(2n+\nu+2)B_{n}(t)-2n\ A_{n}(t){b_{n}(t)\over a_{n-1}(t)}+2(n-1)A_{n}(t)\
{b_{n}(t)\over a_{n-1}(t)}$

$=t+(2n+\nu+2)B_{n}(t)-2{b_{n}(t)\over a_{n}(t)}.$

This gives (3.1). On the other hand, writing (3.1) for the index $n-1$ in the
form

$None$ $A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}-t=B_{n-1}(t)\left[2n+\nu-1-B_{n-1}(t)\right]-A_{n-1}^{2}(t)-{b_{n-1}(t)\over
a_{n-1}(t)},$

we let $x=0$ in (2.38) to find the equality

$A^{2}_{n}(t)\left[B_{n}(t)-\nu-1-2n\right]\left[B_{n-1}(t)-\nu+1-2n\right]\bigg{.}\bigg{.}$

$None$ $\bigg{.}+\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]\bigg{[}A_{n-1}^{2}(t)+{b_{n-1}(t)\over
a_{n-1}(t)}+\left[B_{n-1}(t)-2n-\nu+1\right]B_{n-1}(t)\bigg{]}=0.$

Hence a simple comparison leads to (3.2). Finally, writing (3.1) for $n-1$ and
subtracting one equality from another, we use (2.27) to establish (3.3).

∎

Corollary 7. Differential equation $(2.38)$ reduces to the equality

$x^{2}\left(x+B_{n}(t)-\nu-1-2n\right){\partial^{2}\over\partial
x^{2}}[P_{n}^{\nu}(x,t)]$

$-\bigg{[}x^{3}+\bigg{[}B_{n}(t)-2(\nu+n+1)\bigg{]}x^{2}-\bigg{[}t+\left(B_{n}(t)-\nu-1-2n\right)(\nu+2)\bigg{]}x-t\left(B_{n}(t)-\nu-1-2n\right)\bigg{]}{\partial\over\partial
x}[P_{n}^{\nu}(x,t)]$

$+\bigg{[}nx^{2}-\left({b_{n}(t)\over
a_{n}(t)}-n\left(B_{n}(t)-2\nu-3n-1\right)\right)x+\left(B_{n}(t)-\nu-2n-1\right)\bigg{[}A^{2}_{n}(t)-n(n+\nu+1)\bigg{]}\bigg{.}$

$None$ $\bigg{.}+\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]\left(2n+\nu-B_{n}(t)\right)\bigg{]}P_{n}^{\nu}(x,t)=0.$

###### Proof.

In fact, appealing to (3.4), (3.5), we get from (2.38)

$x^{3}\bigg{[}x+B_{n}(t)-\nu-1-2n\bigg{]}{\partial^{2}\over\partial
x^{2}}[P_{n}^{\nu}(x,t)]-x\bigg{[}x^{2}+\left[x+B_{n}(t)-\nu-1-2n\right]\bigg{.}$

$\times\left[(2n-3)x-A_{n}^{2}(t)-A_{n-1}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}-{b_{n-1}(t)\over a_{n-1}(t)}\right.$

$\bigg{.}\bigg{.}+\left[x+B_{n-1}(t)-\nu+1-2n\right][x-B_{n-1}(t)]\bigg{]}\bigg{]}{\partial\over\partial
x}[P_{n}^{\nu}(x,t)]$

$+\bigg{[}\left[x+B_{n}(t)-\nu-1-2n\right]\bigg{[}A^{2}_{n}(t)\left[B_{n}(t)+B_{n-1}(t)+x-2\nu-4n\right]+n(x^{2}-(n+\nu)x-t)\bigg{]}$

$+\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]\bigg{[}\left[B_{n}(t)-\nu-1-2n\right](2n+\nu-x)+x(2n+\nu-1-x)\bigg{]}$

$None$ $\bigg{.}-nx\left[B_{n}(t)-\nu-2n-1\right]\bigg{]}P_{n}^{\nu}(x,t)=0.$

Hence, letting $x=0$ in (3.7), we find the identity

$None$
$A^{2}_{n}(t)\left[B_{n}(t)+B_{n-1}(t)-2\nu-4n\right]+(2n+\nu)\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]-nt=0.$

Consequently, (3.7) becomes

$x^{3}\bigg{[}x+B_{n}(t)-\nu-1-2n\bigg{]}{\partial^{2}\over\partial
x^{2}}[P_{n}^{\nu}(x,t)]-x\bigg{[}x^{2}+\left[x+B_{n}(t)-\nu-1-2n\right]\bigg{.}$

$\times\left[(2n-3)x-A_{n}^{2}(t)-A_{n-1}^{2}(t)-{b_{n}(t)\over
a_{n}(t)}-{b_{n-1}(t)\over a_{n-1}(t)}\right.$

$\bigg{.}\bigg{.}+\left[x+B_{n-1}(t)-\nu+1-2n\right][x-B_{n-1}(t)]\bigg{]}\bigg{]}{\partial\over\partial
x}[P_{n}^{\nu}(x,t)]$

$+x\bigg{[}x\left(n(x-n-\nu)-{b_{n}(t)\over
a_{n}(t)}\right)+\left(B_{n}(t)-\nu-2n-1\right)\left(A^{2}_{n}(t)+n(x-n-\nu-1)\right)\bigg{.}$

$\bigg{.}+\left[A_{n}^{2}(t)+{b_{n}(t)\over a_{n}(t)}\right]\left(2n+\nu-
B_{n}(t)\right)\bigg{]}P_{n}^{\nu}(x,t)=0.$

Dividing by $x$ and appealing to (3.4), we arrive at (3.6).

∎

Remark 2. For the limit case $t=0$ we employ (1.5), (2.12), (2.13) to reduce
(3.6) to the classical differential equation for Laguerre polynomials.

The following theorem gives the integro-differential-difference equation for
orthogonal polynomials $P_{n}^{\nu}(x,t)$. Precisely, it has

Theorem 5. Orthogonal polynomials $P_{n}^{\nu}(x,t)$ satisfy the integral-
difference equation of the form

$P_{n}^{\nu}(x,t)={P_{n-1}^{\nu}(x,t)\over
a_{n-1,0}(t)}-a_{n}(t)\exp\left({b_{n}(t)\over x\
a_{n}(t)}\right)\int_{0}^{t}\exp\left(-{b_{n}(y)\over x\
a_{n}(y)}\right){d\over dy}\left[{P_{n-1}^{\nu}(x,y)\over a_{n-1,0}(y)\
a_{n}(y)}\right]dy$

$None$ $+(-1)^{n}(n-1)!\ a_{n}(t)\exp\left({1\over x}\left[{b_{n}(t)\over
a_{n}(t)}+n(n+\nu)\right]\right)\left[n\
L_{n}^{\nu}(x)-\Gamma(1+\nu)\left({n!(n+\nu)\over\Gamma(n+\nu)}\right)^{1/2}L_{n-1}^{\nu}(x)\right].$

###### Proof.

Recalling the differential-difference equations (2.5), (2.28), the recurrence
relation (2.11) and Corollary 2, we rewrite equation (2.22) in the form

${\partial\over\partial t}\left[{P_{n}^{\nu}(x,t)\over
a_{n,0}(t)}\right]+{1\over t}\left[n+{A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\over B_{n}(t)-\nu-1-2n}-{1\over x}\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]\right]{P_{n}^{\nu}(x,t)\over a_{n,0}(t)}$

$None$ $=-{1\over xt}\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\right]{P_{n-1}^{\nu}(x,t)\over a_{n-1,0}(t)},\quad x\neq 0.$

Then by virtue of (2.26), (2.37) it implies

${\partial\over\partial t}\left[{P_{n}^{\nu}(x,t)\over
a_{n,0}(t)}\right]+\left[{a^{\prime}_{n,0}(t)\over
a_{n,0}(t)}-{a^{\prime}_{n}(t)\over a_{n}(t)}-{1\over x}{d\over
dt}\left[{b_{n}(t)\over a_{n}(t)}\right]\right]{P_{n}^{\nu}(x,t)\over
a_{n,0}(t)}=-{1\over x}{d\over dt}\left[{b_{n}(t)\over
a_{n}(t)}\right]{P_{n-1}^{\nu}(x,t)\over a_{n-1,0}(t)}.$

Solving this first order differential equation in terms of $P_{n}^{\nu}$ under
initial condition (1.2) and taking into account (2.12), we find

$P_{n}^{\nu}(x,t)=-{a_{n}(t)\over x}\exp\left({b_{n}(t)\over x\
a_{n}(t)}\right)\int_{0}^{t}\exp\left(-{b_{n}(y)\over x\
a_{n}(y)}\right){d\over dy}\left[{b_{n}(y)\over
a_{n}(y)}\right]{P_{n-1}^{\nu}(x,y)\over a_{n-1,0}(y)\ a_{n}(y)}dy$

$None$ $+(-1)^{n}n!\ a_{n}(t)\exp\left({1\over x}\left[{b_{n}(t)\over
a_{n}(t)}+n(n+\nu)\right]\right)L_{n}^{\nu}(x).$

Finally, integrating by parts in (3.11), we end up with (3.9). Theorem 5 is
proved.

∎

Concerning recurrence relations for the coefficients of the polynomials
$P_{n}^{\nu}$, we have the following result.

Theorem 6. For the orthogonal polynomial
$P_{n}^{\nu}(x,t)=\sum_{k=0}^{n}a_{n,k}(t)x^{k},\ a_{n,n}(t)\equiv a_{n}(t),\
a_{n,n-1}(t)=b_{n}(t)$ its coefficients fullfil the differential-recurrence
relations

$None$
$a_{n,k}(t)=a_{n,0}(t)\sum_{m=k}^{n}{a_{m}(t)\left[a_{m}(t)a^{\prime}_{m,k-1}(t)-a_{m,k-1}(t)a_{m}^{\prime}(t)\right]\over\left[a_{m}(t)b_{m}^{\prime}(t)-b_{m}(t)a_{m}^{\prime}(t)\right]a_{m,0}(t)},\quad
k=1,\dots,n.$

###### Proof.

By virtue of (2.26) and taking into account that $A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}\neq 0,\ t>0$ (see (2.1), (3.2)), we write equality (3.10) in the
form

${a^{2}_{n}(t)\over\left[a_{n}(t)b_{n}^{\prime}(t)-b_{n}(t)a_{n}^{\prime}(t)\right]a_{n,0}(t)}\left[{\partial\over\partial
t}\left[P_{n}^{\nu}(x,t)\right]-{a_{n}^{\prime}(t)\over
a_{n}(t)}P_{n}^{\nu}(x,t)\right]={1\over x}\left[{P_{n}^{\nu}(x,t)\over
a_{n,0}(t)}-{P_{n-1}^{\nu}(x,t)\over a_{n-1,0}(t)}\right].$

Hence it yields

$\sum_{m=1}^{n}{a_{m}(t)\over\left[a_{m}(t)b_{m}^{\prime}(t)-b_{m}(t)a_{m}^{\prime}(t)\right]a_{m,0}(t)}\sum_{k=1}^{m}\left[a_{m}(t)a^{\prime}_{m,k-1}(t)-a_{m,k-1}(t)a_{m}^{\prime}(t)\right]x^{k}={P_{n}^{\nu}(x,t)\over
a_{n,0}(t)}-1.$

Therefore changing the order of summation on the left-hand side of the latter
equality, we arrive at the representation

$P_{n}^{\nu}(x,t)=a_{n,0}(t)+\sum_{k=1}^{n}x^{k}\sum_{m=k}^{n}{a_{m}(t)\left[a_{m}(t)a^{\prime}_{m,k-1}(t)-a_{m,k-1}(t)a_{m}^{\prime}(t)\right]a_{n,0}(t)\over\left[a_{m}(t)b_{m}^{\prime}(t)-b_{m}(t)a_{m}^{\prime}(t)\right]a_{m,0}(t)},$

which leads to (3.12).

∎

The three term recurrence relation (1.4) can be written in another form.
Indeed, it has

Theorem 7. Orthogonal polynomials $P_{n}^{\nu}$ satisfy the following
recurrence relation

$\bigg{[}A_{n+1}^{2}(t)+{b_{n+1}(t)\over a_{n+1}(t)}+{x\over
2}\left[B_{n}(t)-\nu-1-2n\right]\bigg{]}\bigg{[}A_{n+1}(t)P_{n+1}^{\nu}(x,t)+A_{n}(t)P_{n-1}^{\nu}(x,t)\bigg{]}$

$+\bigg{[}xtB^{\prime}_{n}(t)+A^{2}_{n}(t)\left[B_{n-1}(t)-\nu+1-2n\right]-A^{2}_{n+1}(t)\left[B_{n+1}(t)-\nu-3-2n\right]\bigg{.}$

$None$ $\bigg{.}-\left(x-B_{n}(t)\right)\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}+{x\over
2}\left[B_{n}(t)-\nu-1-2n\right]\right]\bigg{]}P_{n}^{\nu}(x,t)=0.$

###### Proof.

Differentiating the three term recurrence relation (1.4), we have

$x{\partial\over\partial
t}[P_{n}^{\nu}(x,t)]=A^{\prime}_{n+1}(t)P_{n+1}^{\nu}(x,t)+B^{\prime}_{n}(t)P_{n}^{\nu}(x,t)+A^{\prime}_{n}(t)P_{n-1}^{\nu}(x,t)$

$+A_{n+1}(t){\partial\over\partial
t}[P_{n+1}^{\nu}(x,t)]+B_{n}(t){\partial\over\partial
t}[P_{n}^{\nu}(x,t)]+A_{n}(t){\partial\over\partial t}[P_{n-1}^{\nu}(x,t)].$

Then, using (1.4), (2.5) and Corollary 2, after straightforward
simplifications the latter equality becomes

$\bigg{[}xtA^{\prime}_{n+1}(t)+A_{n+1}(t)\left[A_{n+1}^{2}(t)+{b_{n+1}(t)\over
a_{n+1}(t)}+{x\over
2}\left[B_{n+1}(t)-\nu-3-2n\right]\right]\bigg{]}P_{n+1}^{\nu}(x,t)$

$+\bigg{[}xtB^{\prime}_{n}(t)+A^{2}_{n}(t)\left[B_{n-1}(t)-\nu+1-2n\right]-A^{2}_{n+1}(t)\left[B_{n+1}(t)-\nu-3-2n\right]\bigg{.}$

$\bigg{.}-\left(x-B_{n}(t)\right)\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}+{x\over
2}\left[B_{n}(t)-\nu-1-2n\right]\right]\bigg{]}P_{n}^{\nu}(x,t)$

$\bigg{.}+\bigg{[}xtA^{\prime}_{n}(t)+A_{n}(t)\left[A_{n-1}^{2}(t)+{b_{n-1}(t)\over
a_{n-1}(t)}+\left[B_{n-1}(t)-{x\over
2}\right]\left[B_{n-1}(t)-\nu+1-2n\right]\right]\bigg{.}$

$\bigg{.}-\left(B_{n}(t)-x\right)A_{n}(t)\left[B_{n}(t)-\nu-1-2n\right]\bigg{]}P_{n-1}^{\nu}(x,t)=0.$

It can be rewritten, appealing to (2.36), and we obtain

$\bigg{[}A_{n+1}^{2}(t)+{b_{n+1}(t)\over a_{n+1}(t)}+{x\over
2}\left[B_{n}(t)-\nu-1-2n\right]\bigg{]}A_{n+1}(t)P_{n+1}^{\nu}(x,t)$

$+\bigg{[}xB_{n}(t)+A^{2}_{n}(t)\left[x+B_{n-1}(t)-\nu+1-2n\right]-A^{2}_{n+1}(t)\left[x+B_{n+1}(t)-\nu-3-2n\right]\bigg{.}$

$\bigg{.}-\left(x-B_{n}(t)\right)\left[A_{n}^{2}(t)+{b_{n}(t)\over
a_{n}(t)}+{x\over
2}\left[B_{n}(t)-\nu-1-2n\right]\right]\bigg{]}P_{n}^{\nu}(x,t)$

$+\bigg{[}A_{n-1}^{2}(t)+{b_{n-1}(t)\over
a_{n-1}(t)}+B_{n-1}(t)\left[B_{n-1}(t)-\nu+1-2n\right]\bigg{.}$

$None$ $\bigg{.}-\left(B_{n}(t)-{x\over
2}\right)\left[B_{n}(t)-\nu-1-2n\right]\bigg{]}A_{n}(t)P_{n-1}^{\nu}(x,t)=0.$

Hence, taking into account (2.27), (3.4), equality (3.14) leads to the final
form (3.15).

∎

Corollary 8. The following identity takes place

$None$
$tB_{n}(t)B^{\prime}_{n}(t)+A^{2}_{n}(t)\left[B_{n-1}(t)-\nu+1-2n\right]-A^{2}_{n+1}(t)\left[B_{n+1}(t)-\nu-3-2n\right]=0.$

###### Proof.

The proof is immediate, letting $x=0$ in (3.14) and recalling (2.27), (3.4).

∎

Corollary 9. Coefficients $B_{n}$ obey the following integral-recurrence
relation

$B_{n}(t)+B_{n-1}(t)-2n-\nu+1=\exp\left(\int_{0}^{t}{B_{n-1}(y)-B_{n}(y)+2\over
y}dy\right)$

$None$
$\times\left(2\int_{0}^{t}\exp\left(-\int_{0}^{y}{B_{n-1}(u)-B_{n}(u)+2\over
u}du\right)\left[B_{n-1}(y)-2n-\nu+1\right]{dy\over y}+2n+\nu+1\right).$

###### Proof.

In fact, writing (3.3) in the form

${d\over dt}\bigg{[}B_{n}(t)+B_{n-1}(t)-2n-\nu+1\bigg{]}+{1\over
t}\left[B_{n-1}(t)-B_{n}(t)+2\right]\left[B_{n-1}(t)+B_{n}(t)-2n-\nu+1\right]$

$-{2\over t}\left[B_{n-1}(t)-2n-\nu+1\right]=0,$

we solve a simple Cauchy problem for the first order differential equation to
obtain (3.17).

∎

In the sequel, let us consider polynomial coefficients (3.12) $a_{n,k}(t)$ as
functions of $\nu$ as well, i.e. $a_{n,k}\equiv a_{n,k}^{\nu}$. Then,
returning to the formula (1.8) for the moments, we represent orthogonal
polynomials $P_{n}^{\nu}$ in terms of the Hankel determinant

$None$
$P^{\nu}_{n}(x,t)={1\over[G^{\nu}_{n-1}(t)G^{\nu}_{n}(t)]^{1/2}}\begin{vmatrix}\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+2}(t)&\dots&\dots&\dots&\rho_{\nu+n+2}(t)\\\
\dots&\dots&\dots&\dots&\dots\\\ \vdots&\ddots&\ddots&\ddots&\vdots\\\
\rho_{\nu+n}(t)&\dots&\dots&\dots&\rho_{\nu+2n}(t)\\\ 1&x&\dots&\dots&x^{n}\\\
\end{vmatrix},$

where

$None$
$G^{\nu}_{n}(t)=\begin{vmatrix}\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+2}(t)&\dots&\dots&\dots&\rho_{\nu+n+2}(t)\\\
\dots&\dots&\dots&\dots&\dots\\\ \vdots&\ddots&\ddots&\ddots&\vdots\\\
\rho_{\nu+n+1}(t)&\dots&\dots&\dots&\rho_{\nu+2n+1}(t)\\\ \end{vmatrix}.$

Hence, employing the Laplace theorem to the last row, we find from (3.18),
(3.19)

$G^{\nu}_{n}(t)=(-1)^{n}\rho_{\nu+n+1}(t)G^{\nu+1}_{n-1}(t)+\rho_{\nu+2n+1}(t)G^{\nu}_{n-1}(t)+(-1)^{n+1}\sum_{k=2}^{n}(-1)^{k}\rho_{\nu+n+k}(t)G^{\nu}_{n,k}(t)$

$None$
$=[G^{\nu}_{n-1}(t)G^{\nu}_{n}(t)]^{1/2}\sum_{k=0}^{n}\rho_{\nu+n+k+1}(t)a^{\nu}_{n,k}(t),$

where

$None$
$G^{\nu}_{n,k}(t)=\begin{vmatrix}\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&\dots&\rho_{\nu+k-1}(t)&\rho_{\nu+k+1}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\dots&\rho_{\nu+k}(t)&\rho_{\nu+k+2}(t)&\dots&\rho_{\nu+n+2}(t)\\\
\dots&\dots&\dots&\dots&\dots&\dots&\dots\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n}(t)&\dots&\dots&\rho_{\nu+n+k-2}(t)&\rho_{\nu+n+k}(t)&\dots&\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

In particular, we easily find from (3.18), (3.20) the expression for the free
coefficient $a_{n,0}^{\nu}(t)$ in terms of the determinant by the formula

$None$
$a_{n,0}^{\nu}(t)={(-1)^{n}G^{\nu}_{n,1}(t)\over\left[G^{\nu}_{n-1}(t)G^{\nu}_{n}(t)\right]^{1/2}}.$

Moreover, since the leading term $a_{n}^{\nu}$ (cf. [2], Vol. II) has the
representation

$None$ $\left|a_{n}^{\nu}(t)\right|=\bigg{[}{G^{\nu}_{n-1}(t)\over
G^{\nu}_{n}(t)}\bigg{]}^{1/2},$

we find from (3.22), (3.23) the equality

$None$ ${G^{\nu+1}_{n-1}(t)\over G^{\nu}_{n}(t)}=(-1)^{n}a_{n}^{\nu}(t)\
a_{n,0}^{\nu}(t).$

Therefore it yields

$(-1)^{n}a_{n}^{\nu}(t)\ a_{n,0}^{\nu}(t)={G^{\nu+1}_{n-1}(t)\over
G^{\nu+1}_{n}(t)}{G^{\nu+1}_{n}(t)\over G^{\nu}_{n}(t)}$

$=(-1)^{n+1}\left[a_{n}^{\nu+1}(t)\right]^{2}\
a_{n+1,0}^{\nu}(t)a_{n+1}^{\nu}(t){G^{\nu}_{n+1}(t)\over G^{\nu}_{n}(t)}$

$=(-1)^{n+1}\left[{a_{n}^{\nu+1}(t)\over a_{n+1}^{\nu}(t)}\right]^{2}\
a_{n+1,0}^{\nu}(t)a_{n+1}^{\nu}(t).$

Hence, recalling (2.11), we derive the identities

$\left[{a_{n}^{\nu+1}(t)\over
a_{n+1}^{\nu}(t)}\right]^{2}=-A_{n+1}^{\nu}(t){a_{n,0}^{\nu}(t)\over
a_{n+1,0}^{\nu}(t)}=\frac{\left[A^{\nu}_{n+1}(t)\right]^{2}+{b^{\nu}_{n+1}(t)\over
a^{\nu}_{n+1}(t)}}{2n+\nu+3-B^{\nu}_{n+1}(t)},$

$None$
$\left[a_{n}^{\nu+1}(t)\right]^{2}=\frac{\left[a^{\nu}_{n}(t)\right]^{2}+a^{\nu}_{n+1}(t)b^{\nu}_{n+1}(t)}{2n+\nu+3-B^{\nu}_{n+1}(t)},$

$None$
$\prod_{k=0}^{n}\left[a_{k}^{\nu+1}(t)\right]^{2}={(-1)^{n+1}a_{n+1}^{\nu}(t)\over
a_{n+1,0}^{\nu}(t)}\prod_{k=0}^{n}\left[a_{k}^{\nu}(t)\right]^{2}.$

Moreover, (3.20), (3.23) imply

$None$ $\sum_{k=0}^{n}\rho_{\nu+n+k+1}(t)a^{\nu}_{n,k}(t)={1\over
a_{n}^{\nu}(t)},$

and this agrees with (1.20).

Corollary 10. Coefficients $B^{\nu}_{n}(t),A^{\nu}_{n+1}(t)$ $(1.5)$ of the
three term recurrence relation $(1.4)$ are equal, correspondingly,

$None$ $B^{\nu}_{n}(t)=2n+\nu+1+t\bigg{[}{{G^{\nu}}^{\prime}_{n-1}(t)\over
G^{\nu}_{n-1}(t)}-{{G^{\nu}}^{\prime}_{n}(t)\over G^{\nu}_{n}(t)}\bigg{]},$

$None$
$A_{n+1}^{\nu}(t)={\left[G^{\nu}_{n-1}(t)G^{\nu}_{n+1}(t)\right]^{1/2}\over
G^{\nu}_{n}(t)}.$

###### Proof.

Indeed, appealing to (2.16) and (3.23), we have

$None$ ${G^{\nu}_{n-1}(t)\over G^{\nu}_{n}(t)}={1\over n!\Gamma(n+\nu+1)}\
\exp\left(\int_{0}^{t}{B_{n}(y)-\nu-1-2n\over y}dy\right).$

Therefore,

$G^{\nu}_{n}(t)=n!\Gamma(n+\nu+1)\
G^{\nu}_{n-1}(t)\exp\left(\int_{0}^{t}{\nu+1+2n-B_{n}(y)\over y}dy\right),$

and, solving this recurrence relation owing to (1.5) and (3.19), we easily
derive

$G^{\nu}_{n}(t)=\prod_{k=0}^{n}k!\Gamma(k+\nu+1)\exp\left(\int_{0}^{t}\bigg{[}{b_{n+1}^{\nu}(y)\over
a_{n+1}^{\nu}(y)}+(n+1)(n+\nu+1)\bigg{]}{dy\over y}\right).$

Consequently,

$None$ ${G^{\nu}}^{\prime}_{n}(t)\equiv{dG^{\nu}_{n}\over
dt}={G^{\nu}_{n}(t)\over t}\bigg{[}{b_{n+1}^{\nu}(t)\over
a_{n+1}^{\nu}(t)}+(n+1)(n+\nu+1)\bigg{]}.$

Hence equality (3.28) follows immediately from (1.5). Formula (3.29) is a
direct consequence of (1.5), (3.23) or (2.18), (3.30). ∎

Corollary 11. The following equality holds

$None$ $t^{2}{d\over dt}\bigg{[}{{G^{\nu}}^{\prime}_{n}\over
G^{\nu}_{n}}\bigg{]}=A_{n+1}^{2}-(n+1)(n+1+\nu).$

###### Proof.

The proof is immediate, involving differentiation in (3.31) and employing
(2.26).

∎

On the other hand, the key identity (1.16) for the moments and properties of
the determinants allow to treat (3.19) when $n=2,3,\dots,$ as follows

$tG^{\nu}_{n}(t)=\begin{vmatrix}t\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
t\rho_{\nu+2}(t)&\dots&\dots&\dots&\rho_{\nu+n+2}(t)\\\
\dots&\dots&\dots&\dots&\dots\\\ \vdots&\ddots&\ddots&\ddots&\vdots\\\
t\rho_{\nu+n+1}(t)&\dots&\dots&\dots&\rho_{\nu+2n+1}(t)\\\ \end{vmatrix}$

$=-\begin{vmatrix}(\nu+2)\rho_{\nu+2}(t)&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
(\nu+3)\rho_{\nu+3}(t)&\rho_{\nu+3}(t)&\dots&\dots&\rho_{\nu+n+2}(t)\\\
\dots&\dots&\dots&\dots&\dots\\\ \vdots&\vdots&\ddots&\ddots&\vdots\\\
(\nu+n+2)\rho_{\nu+n+2}(t)&\rho_{\nu+n+2}(t)&\dots&\dots&\rho_{\nu+2n+1}(t)\\\
\end{vmatrix}$

$=-\begin{vmatrix}0&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+3}(t)&\rho_{\nu+3}(t)&\dots&\dots&\rho_{\nu+n+2}(t)\\\
2\rho_{\nu+4}(t)&\rho_{\nu+4}(t)&\dots&\dots&\dots\\\
\vdots&\vdots&\ddots&\ddots&\vdots\\\
n\rho_{\nu+n+2}(t)&\rho_{\nu+n+2}(t)&\dots&\dots&\rho_{\nu+2n+1}(t)\\\
\end{vmatrix}.$

Continuing this process, we find via Laplace’s theorem and (1.16)

$None$
$(-1)^{n+1}t^{n-1}G^{\nu}_{n}(t)=\begin{vmatrix}0&0&\dots&0&\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)&\rho_{\nu+n+1}(t)&\rho_{\nu+n+2}(t)\\\
2\rho_{\nu+4}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n+2}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)\\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\\
n\rho_{\nu+n+2}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+2n}(t)&\rho_{\nu+2n}(t)&\rho_{\nu+2n+1}(t)\\\
\end{vmatrix}.$

On the other hand, using continuously (1.16) from the last column in (3.19),
we arrive in the same manner at the equality

$None$
$G^{\nu}_{n}(t)=\begin{vmatrix}\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&0&0&\dots&0\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&2\rho_{\nu+4}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+1}(t)&\rho_{\nu+n+2}(t)&n\rho_{\nu+n+2}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

Let us consider the following double sequence of determinants

$None$
$H_{i,j}^{\nu,n}(t)=\begin{vmatrix}\rho_{\nu+i}(t)&\rho_{\nu+j}(t)&0&0&\dots&0\\\
\rho_{\nu+i+1}(t)&\rho_{\nu+j+1}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+i+2}(t)&\rho_{\nu+j+2}(t)&2\rho_{\nu+4}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+i}(t)&\rho_{\nu+n+j}(t)&n\rho_{\nu+n+2}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+2n}(t)\\\
\end{vmatrix},$

where $(i,j)\in\mathbb{Z}^{2}$. Hence it is not difficult to observe from
(3.33), (3.34), (3.35) that $H_{i,j}^{\nu,n}(t)=-H_{j,i}^{\nu,n}(t)$ and

$None$ $H_{j,j}^{\nu,n}(t)=0,\ j\in\mathbb{Z},\quad
H_{j,j+1}^{\nu,n}(t)=(-1)^{j+1}t^{j-1}G^{\nu}_{n}(t),\quad j=1,\dots,n.$

Generally, we find via (1.16)

$H_{i,j}^{\nu,n}(t)=(\nu+i-1)H_{i-1,j}^{\nu,n}(t)+tH_{i-2,j}^{\nu,n}(t)$

$+\begin{vmatrix}0&\rho_{\nu+j}(t)&0&0&\dots&0\\\
\rho_{\nu+i}(t)&\rho_{\nu+j+1}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
2\rho_{\nu+i+1}(t)&\rho_{\nu+j+2}(t)&2\rho_{\nu+4}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
n\rho_{\nu+n+i-1}(t)&\rho_{\nu+n+j}(t)&n\rho_{\nu+n+2}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}$

$=(\nu+i-1)H_{i-1,j}^{\nu,n}(t)+tH_{i-2,j}^{\nu,n}(t)$

$-n!\rho_{\nu+j}(t)\begin{vmatrix}\rho_{\nu+i}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+i+1}(t)&\rho_{\nu+4}(t)&\rho_{\nu+5}(t)&\dots&\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+i-1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)&\dots&\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

Hence,

$H_{i,j}^{\nu,n}(t)=(\nu+i-1)H_{i-1,j}^{\nu,n}(t)+tH_{i-2,j}^{\nu,n}(t)$

$None$
$-n!\rho_{\nu+j}(t)\begin{vmatrix}\rho_{\nu+i}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+i+1}(t)&\rho_{\nu+4}(t)&\rho_{\nu+5}(t)&\dots&\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+i-1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)&\dots&\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

Analogously, it has

$H_{i,j}^{\nu,n}(t)=(\nu+j-1)H_{i,j-1}^{\nu,n}(t)+tH_{i,j-2}^{\nu,n}(t)$

$None$
$+n!\rho_{\nu+i}(t)\begin{vmatrix}\rho_{\nu+j}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+j+1}(t)&\rho_{\nu+4}(t)&\rho_{\nu+5}(t)&\dots&\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+j-1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)&\dots&\rho_{\nu+2n}(t)\\\
\end{vmatrix},$

Thus we find

$(\nu+i-1)H_{i-1,j}^{\nu,n}(t)+tH_{i-2,j}^{\nu,n}(t)$

$-n!\rho_{\nu+j}(t)\begin{vmatrix}\rho_{\nu+i}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+i+1}(t)&\rho_{\nu+4}(t)&\rho_{\nu+5}(t)&\dots&\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+i-1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)&\dots&\rho_{\nu+2n}(t)\\\
\end{vmatrix}$

$=(\nu+j-1)H_{i,j-1}^{\nu,n}(t)+tH_{i,j-2}^{\nu,n}(t)$

$None$
$+n!\rho_{\nu+i}(t)\begin{vmatrix}\rho_{\nu+j}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+j+1}(t)&\rho_{\nu+4}(t)&\rho_{\nu+5}(t)&\dots&\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n+j-1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)&\dots&\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

As an immediate consequence, when $(i,j)\in[3,\dots,n+1]\times[3,\dots,n+1]$
the following recurrence relation holds

$None$
$(\nu+i-1)H_{i-1,j}^{\nu,n}(t)+(\nu+j-1)H_{j-1,i}^{\nu,n}(t)+t\bigg{[}H_{i-2,j}^{\nu,n}(t)+H_{j-2,i}^{\nu,n}(t)\bigg{]}=0.$

Recalling (3.36), (3.37), it yields

$None$
$H_{n+2,n+1}^{\nu,n}(t)=(-1)^{n+1}\bigg{[}t^{n-1}G^{\nu}_{n}(t)-n!\rho_{\nu+n+1}(t)G^{\nu+2}_{n-1}(t)\bigg{]}.$

On the other hand, differentiating determinant (3.19) by virtue of (1.15) and
applying the same process as above to the derivative, we deduce the equalities

$t{dG^{\nu}_{n}\over
dt}=-\begin{vmatrix}t\rho_{\nu}(t)&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
t\rho_{\nu+1}(t)&\dots&\dots&\dots&\rho_{\nu+n+2}(t)\\\
\dots&\dots&\dots&\dots&\dots\\\ \vdots&\ddots&\ddots&\ddots&\vdots\\\
t\rho_{\nu+n}(t)&\dots&\dots&\dots&\rho_{\nu+2n+1}(t)\\\ \end{vmatrix}$

$=(\nu+1)G_{n}^{\nu}(t)+\begin{vmatrix}0&\rho_{\nu+2}(t)&\dots&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\dots&\dots&\rho_{\nu+n+2}(t)\\\
2\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\dots&\dots\\\
\vdots&\ddots&\ddots&\ddots&\vdots\\\
n\rho_{\nu+n+1}(t)&\rho_{\nu+n+2}(t)&\dots&\dots&\rho_{\nu+2n+1}(t)\\\
\end{vmatrix}$

$=(\nu+1)G_{n}^{\nu}(t)+\begin{vmatrix}0&\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&0&\dots&0&0\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)\\\
2\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\rho_{\nu+5}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n}(t)&2\rho_{\nu+n+1}(t)\\\
\vdots&\ddots&\ddots&\ddots&\vdots&\vdots&\vdots\\\
n\rho_{\nu+n+1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+3}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+n-1}(t)&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}$

$=(\nu+1)G_{n}^{\nu}(t)+\begin{vmatrix}0&\rho_{\nu+2}(t)&0&0&\dots&0&0\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)\\\
2\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&2\rho_{\nu+4}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n}(t)&2\rho_{\nu+n+1}(t)\\\
\vdots&\ddots&\ddots&\ddots&\vdots&\vdots&\vdots\\\
n\rho_{\nu+n+1}(t)&\rho_{\nu+n+2}(t)&n\rho_{\nu+n+2}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+n-1}(t)&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}$

$+t\begin{vmatrix}0&\rho_{\nu+2}(t)&\rho_{\nu+1}(t)&0&\dots&0&0\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\rho_{\nu+2}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)\\\
2\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\rho_{\nu+3}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n}(t)&2\rho_{\nu+n+1}(t)\\\
\vdots&\ddots&\ddots&\ddots&\vdots&\vdots&\vdots\\\
n\rho_{\nu+n+1}(t)&\rho_{\nu+n+2}(t)&\rho_{\nu+n+1}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+n-1}(t)&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

Consequently, it gives by virtue of (3.19), (1.16)

$t{dG^{\nu}_{n}\over
dt}=(\nu+1)G_{n}^{\nu}(t)-n!\rho_{\nu+2}(t)G_{n-1}^{\nu+1}(t)$

$None$ $+t^{2}\begin{vmatrix}\rho_{\nu}(t)&\rho_{\nu+1}(t)&0&0&\dots&0&0\\\
\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&\rho_{\nu+2}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&2\rho_{\nu+3}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n}(t)&2\rho_{\nu+n+1}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)&n\rho_{\nu+n+1}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+n-1}(t)&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

But from (3.35) we observe

${d\over
dt}\bigg{[}H_{0,1}^{\nu,n}(t)\bigg{]}=-\begin{vmatrix}\rho_{\nu-1}(t)&\rho_{\nu+1}(t)&0&0&\dots&0\\\
\rho_{\nu}(t)&\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+1}(t)&\rho_{\nu+3}(t)&2\rho_{\nu+4}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n-1}(t)&\rho_{\nu+n+1}(t)&n\rho_{\nu+n+2}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}$

$-\begin{vmatrix}\rho_{\nu}(t)&\rho_{\nu+1}(t)&0&0&\dots&0\\\
\rho_{\nu+1}(t)&\rho_{\nu+2}(t)&\rho_{\nu+2}(t)&\rho_{\nu+4}(t)&\dots&\rho_{\nu+n+1}(t)\\\
\rho_{\nu+2}(t)&\rho_{\nu+3}(t)&2\rho_{\nu+3}(t)&2\rho_{\nu+5}(t)&\dots&2\rho_{\nu+n+2}(t)\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\rho_{\nu+n}(t)&\rho_{\nu+n+1}(t)&n\rho_{\nu+n+1}(t)&n\rho_{\nu+n+3}(t)&\dots&n\rho_{\nu+2n}(t)\\\
\end{vmatrix}.$

Hence we derive from (3.42) the identity

$None$ $t{dG^{\nu}_{n}\over
dt}-(\nu+1)G_{n}^{\nu}(t)+n!\rho_{\nu+2}(t)G_{n-1}^{\nu+1}(t)+t^{2}\bigg{[}{d\over
dt}\bigg{[}H_{0,1}^{\nu,n}(t)\bigg{]}+H_{-1,1}^{\nu,n}(t)\bigg{]}=0.$

Remark 3. By using these features an interesting open question is to obtain a
differential-difference recurrence relation for determinants (3.19).

##  4\. Rodrigues-type formula. Generating function

Let us expand the function $e^{-t/x}P^{\nu}_{n}(x,t)$ in terms of the Laguerre
polynomials $L_{n}^{\nu}(x)$. It gives

$None$
$e^{-t/x}P^{\nu}_{n}(x,t)=\sum_{k=0}^{\infty}d_{n,k}^{\nu}(t)L_{k}^{\nu}(x),$

where

$None$
$d_{n,k}^{\nu}(t)={k!\over\Gamma(k+\nu+1)}\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu}dx.$

But from (1.18) we have $d_{n,k}^{\nu}(t)=0,\ k=0,\dots,n-1$. The uniform
estimate with respect to $k$ for coefficients $d_{n,k}^{\nu}(t)$ is given by
the following lemma.

Lemma 2. Let $t>0\ \nu>-1$. Coefficients $d_{n,k}^{\nu}(t),\
n,k\in\mathbb{N}_{0}$ satisfy the upper bound of the form

$None$ $\left|d_{n,k}^{\nu}(t)\right|\leq{k!\
h_{n}^{\nu}(t)\over\Gamma(k+\nu+1)},$

where

$None$
$h_{n}^{\nu}(t)=2^{\nu-1/2}\int_{0}^{t}Q_{n}^{\nu}(t-y)\rho^{1/2}_{2\nu+1}(2y){dy\over\sqrt{y}}$

and

$None$ $Q_{n}^{\nu}(x)=\sum_{m=0}^{n}\left|a_{n,m}^{\nu}(t)\right|{x^{m}\over
m!}.$

###### Proof.

Writing polynomial $P^{\nu}_{n}$ in the explicit form, equality (4.2) becomes

$None$
$d_{n,k}^{\nu}(t)={k!\over\Gamma(k+\nu+1)}\sum_{m=0}^{n}a_{n,m}^{\nu}(t)\int_{0}^{\infty}e^{-x-t/x}L_{k}^{\nu}(x)x^{\nu+m}dx.$

Recalling (1.11), the latter integral can be represented as follows

$\int_{0}^{\infty}e^{-x-t/x}L_{k}^{\nu}(x)x^{\nu+m}dx=(-1)^{k+m+1}{d^{k-m-1}\over
dt^{k-m-1}}\int_{0}^{\infty}x^{\nu+k-1}e^{-x-t/x}L_{k}^{\nu}(x)dx$

$None$ $={(-1)^{m+1}\over k!}{d^{k-m-1}\over
dt^{k-m-1}}\bigg{[}t^{k}\rho_{\nu}(t)\bigg{]}={(-1)^{m+1}\over k!\
m!}\int_{0}^{t}(t-y)^{m}{d^{k}\over
dy^{k}}\bigg{[}y^{k}\rho_{\nu}(y)\bigg{]}dy,$

where we mean (cf. (1.12))

${d^{-q}f\over
dt^{-q}}\equiv\left(I_{+}^{q}f\right)(t)={1\over(q-1)!}\int_{0}^{t}(t-y)^{q-1}f(y)dy,\quad
q\in\mathbb{N}_{0}.$

Taking into account (4.6), coefficients $d_{n,k}^{\nu}(t)$ can be written in
the operator form

$None$
$d_{n,k}^{\nu}(t)=-{k!\over\Gamma(k+\nu+1)}P_{n}^{\nu}\bigg{(}-I_{+},t\bigg{)}\bigg{\\{}{d^{k}\over
dt^{k}}\bigg{[}t^{k}\rho_{\nu}(t)\bigg{]}\bigg{\\}}.$

Meanwhile, via (1.11) and the Rodrigues formula for Laguerre polynomials it
has

${d^{k}\over
dy^{k}}\bigg{[}y^{k}\rho_{\nu}(y)\bigg{]}=k!\int_{0}^{\infty}x^{\nu-1}e^{-x-y/x}L_{k}^{\nu}(x)dx=\int_{0}^{\infty}x^{-1}e^{-y/x}{d^{k}\over
dx^{k}}\left[e^{-x}x^{\nu+k}\right]dx.$

Integrating by parts with the use of Entry 1.1.3.2 on p. 4 in [1], we get

${d^{k}\over
dy^{k}}\bigg{[}y^{k}\rho_{\nu}(y)\bigg{]}=(-1)^{k}\int_{0}^{\infty}{d^{k}\over
dx^{k}}\left[x^{-1}e^{-y/x}\right]e^{-x}x^{\nu+k}dx$

$=k!\int_{0}^{\infty}e^{-x-y/x}x^{\nu-1}L_{k}\left({y\over x}\right)dx,$

where $L_{k}$ are Laguerre polynomials of the index zero. Then owing to (1.7),
Schwarz’s inequality and orthogonality of Laguerre polynomials, we find the
estimate

$\left|{d^{k}\over dy^{k}}\bigg{[}y^{k}\rho_{\nu}(y)\bigg{]}\right|\leq
k!\left(\int_{0}^{\infty}e^{-yx-2/x}x^{-2(\nu+1)}dx\right)^{1/2}\left(\int_{0}^{\infty}e^{-yx}\left[L_{k}\left(yx\right)\right]^{2}dx\right)^{1/2}$

$={2^{\nu-1/2}k!\over\sqrt{y}}\rho^{1/2}_{2\nu+1}(2y).$

Therefore, returning to (4.6), (4.7), we derive

$\left|d_{n,k}^{\nu}(t)\right|\leq{2^{\nu-1/2}k!\over\Gamma(k+\nu+1)}\int_{0}^{t}\sum_{m=0}^{n}\left|a_{n,m}^{\nu}(t)\right|{(t-y)^{m}\over
m!}\rho^{1/2}_{2\nu+1}(2y){dy\over\sqrt{y}},$

which implies (4.3) and completes the proof of Lemma 2.

∎

Corollary 12. Under the condition $\nu>3/2$ the Laguerre series $(4.1)$
converges absolutely and uniformly on closed intervals of $\mathbb{R}_{+}$.

###### Proof.

In fact, Stirling’s asymptotic formula for gamma-function [2], Vol. I yields

${k!\over\Gamma(k+\nu+1)}=O\left(k^{-\nu}\right),\quad k\to\infty.$

Since Laguerre polynomials $L_{k}^{\nu}(x)$ behave as
$O\left(k^{\nu/2-1/4}\right),\ k\to\infty$ uniformly on closed intervals
$x\in[\alpha,\beta]$ of $\mathbb{R}_{+}$, the absolute and uniform convergence
of the series (4.1) is guaranteed under the assumption $\nu>3/2$.

∎

Further, writing (4.1) as follows

$e^{-t/x}P^{\nu}_{n}(x,t)=\sum_{k=n}^{\infty}d_{n,k}^{\nu}(t)L_{k}^{\nu}(x)=x^{-\nu}e^{x}\sum_{k=n}^{\infty}{d_{n,k}^{\nu}(t)\over
k!}{d^{k}\over dx^{k}}\left[x^{k+\nu}e^{-x}\right]$

$=x^{-\nu}e^{x}\sum_{k=0}^{\infty}{d_{n,k+n}^{\nu}(t)\over(k+n)!}{d^{k+n}\over
dx^{k+n}}\left[x^{k+n+\nu}e^{-x}\right]$

$=x^{-\nu}e^{x}\sum_{k=0}^{\infty}{d_{n,k+n}^{\nu}(t)\
k!\over(k+n)!}{d^{n}\over
dx^{n}}\bigg{[}e^{-x}x^{\nu+n}L_{k}^{n+\nu}(x)\bigg{]},$

the problem arises to interchange the order of differentiation and summation.
It can be done by virtue of the absolute and uniform convergence of the series

$\sum_{k=0}^{\infty}{d_{n,k+n}^{\nu}(t)\over(k+n)!}{d^{k+j}\over
dx^{k+j}}\left[x^{k+n+\nu}e^{-x}\right]$

$None$
$=x^{-\nu-n+j}e^{x}\sum_{k=0}^{\infty}{d_{n,k+n}^{\nu}(t)(k+j)!\over(k+n)!}L_{k+j}^{n+\nu-j}(x),\quad
j=1,2,\dots,n,\ n\in\mathbb{N}$

on closed intervals of $\mathbb{R}_{+}$. Indeed, this is an immediate
consequence of the bound (4.3) and Corollary 12. Thus we arrive at the
following Rodrigues-type formula for polynomials $P^{\nu}_{n}$

$None$ $P^{\nu}_{n}(x,t)=x^{-\nu}e^{x+t/x}{d^{n}\over
dx^{n}}\bigg{[}e^{-x}x^{\nu+n}\sum_{k=0}^{\infty}{d_{n,k+n}^{\nu}(t)\
k!\over(k+n)!}\ L_{k}^{n+\nu}(x)\bigg{]}.$

Recurrence relations for coefficients $d_{n,k}^{\nu}(t)$ are given by

Theorem 8. Coefficients $d_{n,k}^{\nu}(t)$ obey recurrence relations of the
form

$\left(2k+\nu+1-(k+1)B_{n}^{\nu}(t)\right)d_{n,k}^{\nu}(t)-\left(k+1\right)A_{n+1}^{\nu}(t)d_{n+1,k}^{\nu}(t)-(k+\nu+1)d_{n,k+1}^{\nu}(t)$

$None$ $-\left(k+1\right)A_{n}^{\nu}(t)d_{n-1,k}^{\nu}(t)-k\
d_{n,k-1}^{\nu}(t)=0,$

$(2k+n+1+\nu)A_{n+1}^{\nu}(t)d_{n+1,k}^{\nu}(t)-(k+1+\nu)A_{n+1}^{\nu}(t)d_{n+1,k+1}^{\nu}(t)$

$+\bigg{[}(1+n)B_{n}^{\nu}(t)-(k+\nu+2)(2k+\nu+1)-t+\left[A_{n}^{\nu}(t)\right]^{2}-A_{n}^{\nu}(t)\left[B^{\nu}_{n}(t)-\nu-1-2n\right]\bigg{]}d_{n,k}^{\nu}(t)$

$+(k+\nu+1)(k+\nu+2-B^{\nu}_{n}(t))d_{n,k+1}^{\nu}(t)-(k+\nu+1)A_{n}^{\nu}(t)d_{n-1,k+1}^{\nu}(t)$

$+\bigg{[}\left(2k+n+\nu+1+B^{\nu}_{n-1}(t)\right)A_{n}^{\nu}(t)-\bigg{[}[A^{\nu}_{n}(t)]^{2}+{b^{\nu}_{n}(t)\over
a^{\nu}_{n}(t)}\bigg{]}\bigg{]}d_{n-1,k}^{\nu}(t)$

$None$
$+k(k+\nu+2)d_{n,k-1}^{\nu}(t)+A_{n}^{\nu}(t)A_{n-1}^{\nu}(t)d_{n-2,k}^{\nu}(t)=0.$

###### Proof.

Indeed, relation (4.11) follows immediately from the three recurrence
relations (1.3), (1.4) for Laguerre polynomials and polynomials $P_{n}^{\nu}$,
respectively. In order to prove (4.12), we integrate by parts in (4.2) to
obtain

$d_{n,k}^{\nu}(t)={k!\over
t\Gamma(k+\nu+1)}\bigg{[}\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu+2}dx\bigg{.}$

$-(\nu+2)\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu+1}dx+\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k-1}^{\nu+1}(x)x^{\nu+2}dx$

$None$ $\bigg{.}-\int_{0}^{\infty}e^{-x-t/x}{\partial\over\partial
x}\left[P^{\nu}_{n}(x,t)\right]L_{k}^{\nu}(x)x^{\nu+2}dx\bigg{]}.$

But using the known relation for Laguerre polynomials [5]

$xL_{k-1}^{\nu+1}(x)=(k+\nu)L_{k-1}^{\nu}(x)-kL_{k}^{\nu}(x),$

equality (4.13) becomes

$d_{n,k}^{\nu}(t)={k!\over
t\Gamma(k+\nu+1)}\bigg{[}\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu+2}dx\bigg{.}$

$-(\nu+2+k)\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu+1}dx+(k+\nu)\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k-1}^{\nu}(x)x^{\nu+1}dx$

$None$ $\bigg{.}-\int_{0}^{\infty}e^{-x-t/x}{\partial\over\partial
x}\left[P^{\nu}_{n}(x,t)\right]L_{k}^{\nu}(x)x^{\nu+2}dx\bigg{]}.$

Then, employing again recurrence relations (1.3), (1.4) and differential-
difference equation (2.5), we derive

${k!\over
t\Gamma(k+\nu+1)}\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu+2}dx$

$={k!\over
t\Gamma(k+\nu+1)}\int_{0}^{\infty}e^{-x-t/x}\left[A_{n+1}^{\nu}(t)P^{\nu}_{n+1}(x,t)+B_{n}^{\nu}(t)P^{\nu}_{n}(x,t)+A_{n}^{\nu}(t)P^{\nu}_{n-1}(x,t)\right]$

$\times\left[(2k+1+\nu)L_{k}^{\nu}(x)-(k+1)L_{k+1}^{\nu}(x)-(k+\nu)L_{k-1}^{\nu}(x)\right]x^{\nu}dx$

$={1\over
t}\bigg{[}(2k+1+\nu)\bigg{[}A_{n+1}^{\nu}(t)d_{n+1,k}^{\nu}(t)+B_{n}^{\nu}(t)d_{n,k}^{\nu}(t)+A_{n}^{\nu}(t)d_{n-1,k}^{\nu}(t)\bigg{]}\bigg{.}$

$-(k+1+\nu)\bigg{[}A_{n+1}^{\nu}(t)d_{n+1,k+1}^{\nu}(t)+B_{n}^{\nu}(t)d_{n,k+1}^{\nu}(t)+A_{n}^{\nu}(t)d_{n-1,k+1}^{\nu}(t)\bigg{]}$

$\bigg{.}-k\bigg{[}A_{n+1}^{\nu}(t)d_{n+1,k-1}^{\nu}(t)+B_{n}^{\nu}(t)d_{n,k-1}^{\nu}(t)+A_{n}^{\nu}(t)d_{n-1,k-1}^{\nu}(t)\bigg{]}\bigg{]},$

${k!(k+\nu+2)\over
t\Gamma(k+\nu+1)}\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k}^{\nu}(x)x^{\nu+1}dx$

$={k+\nu+2\over
t}\bigg{[}(2k+1+\nu)d_{n,k}^{\nu}(t)-(k+1+\nu)d_{n,k+1}^{\nu}(t)-kd_{n,k-1}^{\nu}(t)\bigg{]},$

${k!\over
t\Gamma(k+\nu)}\int_{0}^{\infty}e^{-x-t/x}P^{\nu}_{n}(x,t)L_{k-1}^{\nu}(x)x^{\nu+1}dx$

$={k\over
t}\bigg{[}A_{n+1}^{\nu}(t)d_{n+1,k-1}^{\nu}(t)+B_{n}^{\nu}(t)d_{n,k-1}^{\nu}(t)+A_{n}^{\nu}(t)d_{n-1,k-1}^{\nu}(t)\bigg{]},$

${k!\over t\Gamma(k+\nu+1)}\int_{0}^{\infty}e^{-x-t/x}{\partial\over\partial
x}\left[P^{\nu}_{n}(x,t)\right]L_{k}^{\nu}(x)x^{\nu+2}dx$

$={k!\over
t\Gamma(k+\nu+1)}\int_{0}^{\infty}e^{-x-t/x}L_{k}^{\nu}(x)x^{\nu}\bigg{[}\left[nx-\left[A^{\nu}_{n}(t)\right]^{2}-{b^{\nu}_{n}(t)\over
a^{\nu}_{n}(t)}\right]P_{n}^{\nu}(x,t)\bigg{.}$

$\bigg{.}+A^{\nu}_{n}(t)\left[x+B^{\nu}_{n}(t)-\nu-1-2n\right]P_{n-1}^{\nu}(x,t)\bigg{]}dx$

$={A^{\nu}_{n}(t)\over
t}\left[B^{\nu}_{n}(t)-\nu-1-2n\right]d_{n,k}^{\nu}(t)-{1\over
t}\bigg{[}\left[A^{\nu}_{n}(t)\right]^{2}+{b^{\nu}_{n}(t)\over
a^{\nu}_{n}(t)}\bigg{]}d_{n-1,k}^{\nu}(t)$

$+{n\over
t}\bigg{[}A_{n+1}^{\nu}(t)d_{n+1,k}^{\nu}(t)+B_{n}^{\nu}(t)d_{n,k}^{\nu}(t)+A_{n}^{\nu}(t)d_{n-1,k}^{\nu}(t)\bigg{]}$

$+{A^{\nu}_{n}(t)\over
t}\bigg{[}A_{n}^{\nu}(t)d_{n,k}^{\nu}(t)+B_{n-1}^{\nu}(t)d_{n-1,k}^{\nu}(t)+A_{n-1}^{\nu}(t)d_{n-2,k}^{\nu}(t)\bigg{]}.$

Hence, substituting these values into (4.13), we get after simplification the
desired relation (4.12).

∎

Finally, defining as usual the generating function $G(x,w,t)$ in terms of the
series

$None$ $G(x,w,t)=\sum_{n=0}^{\infty}P^{\nu}_{n}(x,t){w^{n}\over n!},\quad
x>0,\ w\in\mathbb{C},$

it can be written due to (4.10) in the form

$None$ $G(x,w,t)=x^{-\nu}e^{x+t/x}\sum_{n=0}^{\infty}{w^{n}\over
n!}{d^{n}\over
dx^{n}}\bigg{[}e^{-x}x^{\nu+n}\sum_{k=0}^{\infty}{d_{n,k+n}^{\nu}(t)\
k!\over(k+n)!}\ L_{k}^{n+\nu}(x)\bigg{]}.$

The convergence of the series (4.15) is guaranteed at least in
$L_{2}\left(\mathbb{R}_{+};e^{-x-t/x}x^{\nu}dx\right)$. Indeed, by virtue of
the Minkowski inequality it has

$\bigg{(}\int_{0}^{\infty}e^{-x-t/x}\bigg{|}\sum_{n=N}^{\infty}P^{\nu}_{n}(x,t){w^{n}\over
n!}\bigg{|}^{2}x^{\nu}dx\bigg{)}^{1/2}\leq\sum_{n=N}^{\infty}{|w|^{n}\over
n!}\bigg{(}\int_{0}^{\infty}e^{-x-t/x}\left[P^{\nu}_{n}(x,t)\right]^{2}x^{\nu}dx\bigg{)}^{1/2}$

$=\sum_{n=N}^{\infty}{|w|^{n}\over n!}\to 0,\quad N\to\infty.$

## References

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