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Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I

Alexander Apelblat

^1,†

and

Juan Luis González-Santander

^2,*,†

Department of Chemical Engineering, Ben Gurion University of the Negev, Beer Sheva 84105, Israel

Department of Mathematics, Universidad de Oviedo, 33007 Oviedo, Spain

Author to whom correspondence should be addressed.

^†

These authors contributed equally to this work.

Axioms 2023, 12(4), 381; https://doi.org/10.3390/axioms12040381

Submission received: 24 February 2023 / Revised: 7 April 2023 / Accepted: 14 April 2023 / Published: 16 April 2023

(This article belongs to the Special Issue Dedicated to Professor Hari Mohan Srivastava on the Occasion of His 82nd Birthday)

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Abstract

In this paper, first derivatives of the Whittaker function

M_{κ, μ} (x)

are calculated with respect to the parameters. Using the confluent hypergeometric function, these derivarives can be expressed as infinite sums of quotients of the digamma and gamma functions. Moreover, from the integral representation of

M_{κ, μ} (x)

it is possible to obtain these parameter derivatives in terms of finite and infinite integrals with integrands containing elementary functions (products of algebraic, exponential, and logarithmic functions). These infinite sums and integrals can be expressed in closed form for particular values of the parameters. For this purpose, we have obtained the parameter derivative of the incomplete gamma function in closed form. As an application, reduction formulas for parameter derivatives of the confluent hypergeometric function are derived, along with finite and infinite integrals containing products of algebraic, exponential, logarithmic, and Bessel functions. Finally, reduction formulas for the Whittaker functions

M_{κ, μ} (x)

and integral Whittaker functions

{Mi}_{κ, μ} (x)

and

{mi}_{κ, μ} (x)

are calculated.

Keywords:

derivatives with respect to parameters; Whittaker functions; integral Whittaker functions; incomplete gamma functions; sums of infinite series of psi and gamma; finite and infinite logarithmic integrals and Bessel functions

MSC:

33B15; 33B20; 33C10; 33C15; 33C20; 33C50; 33E20

1. Introduction

Introduced in 1903 by Whittaker [1], the

M_{κ, μ} (z)

and

W_{κ, μ} (z)

functions are defined as follows:

\begin{matrix} M_{κ, μ} (z) & = & z^{μ + 1 / 2} e^{- z / 2}_{1} F_{1} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| z), \\ 2 μ & \neq & - 1, - 2, \dots \end{matrix}

(1)

and

\begin{matrix} W_{κ, μ} (z) & = & \frac{Γ (- 2 μ)}{Γ (\frac{1}{2} - μ - κ)} M_{κ, μ} (z) + \frac{Γ (2 μ)}{Γ (\frac{1}{2} + μ - κ)} M_{κ, - μ} (z), \\ 2 μ & \neq & \pm 1, \pm 2, \dots \end{matrix}

(2)

respectively, where

Γ (x)

denotes the gamma function and

z \in C \ (- \infty, 0]

. These functions, called Whittaker functions, are closely associated with the following confluent hypergeometric function (Kummer function):

_{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| z) = \frac{Γ (b)}{Γ (a)} \sum_{n = 0}^{\infty} \frac{Γ (a + n)}{Γ (b + n)} \frac{z^{n}}{n!},

(3)

where

_{p} F_{q} (\begin{matrix} a_{1}, \dots, a_{p} \\ b_{1}, \dots, b_{q} \end{matrix}| z)

denotes the generalized hypergeometric function.

For particular values of the parameters

κ

and

μ

, the Whittaker functions can be reduced to a variety of elementary and special functions. Whittaker [1] discussed the connection between the functions defined in (1) and (2) and many other special functions, such as the modified Bessel function, the incomplete gamma functions, the parabolic cylinder function, the error functions, the logarithmic and the cosine integrals, and the generalized Hermite and Laguerre polynomials. Monographs and treatises dealing with special functions [2,3,4,5,6,7,8,9,10] present properties of the Whittaker functions with more or less extension.

The Whittaker functions are frequently applied in various areas of mathematical physics (see for example [11,12,13]), such as the well-known solution of the Schrödinger equation for the harmonic oscillator [14].

M_{κ, μ} (x)

and

W_{κ, μ} (x)

are usually treated as functions of variable x with fixed values of the parameters

κ

and

μ

. However, there are other investigations which consider

κ

and

μ

as variables. For instance, Laurenzi [15] discussed methods to calculate derivatives of

M_{κ, 1 / 2} (x)

and

W_{κ, 1 / 2} (x)

with respect to

κ

when this parameter is an integer. Using the Mellin transform, Buschman [16] showed that the derivatives of the Whittaker functions with respect to the parameters for certain particular values of these parameters can be expressed in finite sums of Whittaker functions. López and Sesma [17] considered the behaviour of

M_{κ, μ} (x)

as a function of

κ

. They derived a convergent expansion in ascending powers of

κ

and an asymptotic expansion in descending powers of

κ

. Using series of Bessel functions and Buchholz polynomials, Abad and Sesma [18] presented an algorithm for the calculation of the nth derivative of the Whittaker functions with respect to the

κ

parameter. Becker [19] investigated certain integrals with respect to the

μ

parameter. Ancarini and Gasaneo [20] presented a general case of differentiation of generalized hypergeometric functions with respect to the parameters in terms of infinite series containing the digamma function. In addition, Sofostasios and Brychkov [21] considered derivatives of hypergeometric functions and classical polynomials with respect to the parameters.

The primary focus of this research is a systematic investigation of the first derivatives of

M_{κ, μ} (x)

with respect to the parameters. We primarily base our findings on two distinct methods. The first pertains to the series representation of

M_{κ, μ} (x)

, whereas the second pertains to the integral representations of

M_{κ, μ} (x)

. Regarding the first approach, direct differentiation of (1) with respect to the parameters leads to infinite sums of quotients of digamma and gamma functions. It is possible to calculate such sums in closed form for particular values of the parameters. The parameter differentiation of the integral representations of

M_{κ, μ} (x)

leads to finite and infinite integrals of elementary functions, such as products of algebraic, exponential, and logarithmic functions. These integrals are similar to those investigated by Kölbig [22] and Geddes et al. [23]. As in the case of the first approach, it is possible to calculate such integrals in closed form for some particular values of the parameters.

In the Appendices, we calculate the first derivative of the incomplete gamma functions

γ (ν, x)

and

Γ (ν, x)

with respect to the parameter

ν

. These results are used when we calculate several of the integrals found in the second approach mentioned above. In addition, we calculate new reduction formulas of the integral Whittaker functions which we recently introduced in [24]. These are defined in a similar way as other integral functions in the mathematical literature:

\begin{matrix} {Mi}_{κ, μ} (x) & = & \int_{0}^{x} \frac{M_{κ, μ} (t)}{t} d t, \end{matrix}

(4)

\begin{matrix} {mi}_{κ, μ} (x) & = & \int_{x}^{\infty} \frac{M_{κ, μ} (t)}{t} d t . \end{matrix}

(5)

Finally, we include a list of reduction formulas for the Whittaker function

M_{κ, μ} (x)

in the Appendices.

2. Parameter Differentiation of $M_{κ, μ}$ via Kummer Function $_{1} F_{1}$

As mentioned above, the Whittaker function

M_{κ, μ} (x)

is closely related to the confluent hypergeometric function

_{1} F_{1} (a; b; x)

. Likewise, the parameter derivatives of

M_{κ, μ} (x)

are related to the parameter derivatives of

_{1} F_{1} (a; b; x)

. Below, we introduce the following notation set by Ancarini and Gasaneo [20].

Definition 1.

Define the parameter derivatives of the confluent hypergeometric function as

G^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) = \frac{\partial}{\partial a} [_{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| x)],

(6)

and

H^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) = \frac{\partial}{\partial b} [_{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| x)] .

(7)

According to (3), we have

\begin{matrix} G^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) & = & \frac{Γ (b)}{Γ (a)} \sum_{n = 0}^{\infty} \frac{Γ (a + n)}{Γ (b + n)} [ψ (a + n) - ψ (a)] \frac{x^{n}}{n!}, \\ H^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) & = & - \frac{Γ (b)}{Γ (a)} \sum_{n = 0}^{\infty} \frac{Γ (a + n)}{Γ (b + n)} [ψ (b + n) - ψ (b)] \frac{x^{n}}{n!} . \end{matrix}

Additionally, according to [25], we have

G^{(1)} (\begin{matrix} a \\ b \end{matrix}| z) = \frac{z}{b} \sum_{m = 0}^{\infty} \frac{{(a)}_{m} z^{m}}{{(b + 1)}_{m} {(2)}_{m}}_{2} F_{2} (\begin{matrix} 1, a + m + 1 \\ m + 2, b + m + 1 \end{matrix}| z),

(8)

and

H^{(1)} (\begin{matrix} a \\ b \end{matrix}| z) = - \frac{a z}{b^{2}} \sum_{m = 0}^{\infty} \frac{{(a + 1)}_{m} {(b)}_{m} z^{m}}{{[{(b + 1)}_{m}]}^{2} {(2)}_{m}}_{2} F_{2} (\begin{matrix} 1, a + m + 1 \\ m + 2, b + m + 1 \end{matrix}| z) .

(9)

Because one of the integral representations of the confluent hypergeometric function is ([6], Section 6.5.1)

\begin{matrix} _{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| x) = \frac{Γ (b)}{Γ (a) Γ (b - a)} \int_{0}^{1} e^{x t} t^{a - 1} {(1 - t)}^{b - a - 1} d t \\ Re b > Re a > 0, \end{matrix}

(10)

by direct differentiation of (10) with respect to parameters a and b we obtain

\begin{matrix} G^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) & = & [ψ (b) - ψ (a)]_{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| x) \\ + \frac{Γ (b)}{Γ (a) Γ (b - a)} \int_{0}^{1} e^{x t} t^{a - 1} {(1 - t)}^{b - a - 1} ln (\frac{t}{1 - t}) d t, \end{matrix}

and

\begin{matrix} H^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) & = & - [ψ (b) - ψ (b - a)]_{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| x) \\ + \frac{Γ (b)}{Γ (a) Γ (b - a)} \int_{0}^{1} e^{x t} t^{a - 1} {(1 - t)}^{b - a - 1} ln (1 - t) d t . \end{matrix}

Because our main focus is the systematic investigation of the parameter derivatives of

M_{κ, μ} (x)

, we present these parameter derivatives as Theorems throughout the paper and the corresponding results for

G^{(1)} (a; b; x)

and

H^{(1)} (a; b; x)

as Corollaries. Additionally, note that all the results regarding

G^{(1)} (a; b; x)

can be transformed according to the next Theorem.

Theorem 1.

The following transformation holds true:

G^{(1)} (\begin{matrix} a \\ b \end{matrix}| x) = - e^{x} G^{(1)} (\begin{matrix} b - a \\ b \end{matrix}| - x) .

Proof.

Differentiate Kummer’s transformation formula ([8], Equation 13.2.39) with respect to a:

_{1} F_{1} (\begin{matrix} a \\ b \end{matrix}| x) = e^{x}_{1} F_{1} (\begin{matrix} b - a \\ b \end{matrix}| - x)

to obtain the desired result. □

2.1. Derivative with Respect to the First Parameter $\partial M_{κ, μ} (x) / \partial κ$

Using (1) and (3), the first derivative of

M_{κ, μ} (x)

with respect to the first parameter

κ

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial κ} & = & ψ (\frac{1}{2} + μ - κ) M_{κ, μ} (x) \\ - \frac{Γ (1 + 2 μ)}{Γ (\frac{1}{2} + μ - κ)} x^{μ + 1 / 2} e^{- x / 2} S_{1} (κ, μ, x), \end{matrix}

(11)

where

ψ (x)

denotes the digamma function and

S_{1} (κ, μ, x) = \sum_{n = 0}^{\infty} \frac{Γ (\frac{1}{2} + μ - κ + n)}{Γ (1 + 2 μ + n)} ψ (\frac{1}{2} + μ - κ + n) \frac{x^{n}}{n!} .

(12)

Theorem 2.

For

2 μ \notin Z^{-}

and for

x \in R

x \neq 0

, the following parameter derivative formula of

M_{κ, μ} (x)

holds true:

{\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = - μ - 1 / 2} = - \frac{x^{μ + 3 / 2}}{2 μ + 1} e^{x / 2}_{2} F_{2} (\begin{matrix} 1, 1 \\ 2 (μ + 1), 2 \end{matrix}| - x) .

(13)

Proof.

For

κ = - μ - 1 / 2

, Equation (11) becomes

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = - μ - 1 / 2} \\ = & x^{μ + 1 / 2} e^{- x / 2} [ψ (1 + 2 μ) \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} - \sum_{n = 0}^{\infty} ψ (2 μ + 1 + n) \frac{x^{n}}{n!}] . \end{matrix}

Apply ([26], Equation 6.2.1(60))

\sum_{k = 0}^{\infty} \frac{t^{k}}{k!} ψ (k + a) = e^{t} [ψ (a) + \frac{t}{a}_{2} F_{2} (\begin{matrix} 1, 1 \\ a + 1, 2 \end{matrix}| - t)]

(14)

to obtain (13), completing the proof. □

Corollary 1.

For

a \in R

a \neq 0

, and for

x \in R

, the following reduction formula holds true:

G^{(1)} (\begin{matrix} a \\ a \end{matrix}| x) = \frac{x e^{x}}{a}_{2} F_{2} (\begin{matrix} 1, 1 \\ a + 1, 2 \end{matrix}| - x) .

(15)

Proof.

Direct differentiation of (1) yields

\frac{\partial M_{κ, μ} (x)}{\partial κ} = - x^{μ + 1 / 2} e^{- x / 2} G^{(1)} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x),

(16)

thus, by comparing (16) with

κ = - μ - \frac{1}{2}

to (13), we arrive at (15), as we wanted to prove. □

Corollary 2.

For

a \in R

a \neq 0

and for

x \in R

, the following sum holds true:

\sum_{m = 0}^{\infty} \frac{γ (m + 1, x)}{(m + a) m!} = \frac{x}{a}_{2} F_{2} (\begin{matrix} 1, 1 \\ a + 1, 2 \end{matrix}| - x),

where

γ (ν, z)

denotes the lower incomplete gamma function (A1).

Proof.

According to (8) and the reduction formula ([9], Equation 7.11.1(15))

_{1} F_{1} (\begin{matrix} 1 \\ b \end{matrix}| z) = (b - 1) z^{1 - b} e^{z} γ (b - 1, z),

we have

\begin{matrix} G^{(1)} (\begin{matrix} a \\ a \end{matrix}| x) & = & \frac{x}{a} \sum_{m = 0}^{\infty} \frac{{(a)}_{m} x^{m}}{{(a + 1)}_{m} (m + 1)!}_{1} F_{1} (\begin{matrix} 1 \\ m + 2 \end{matrix}| x) \\ = & e^{x} \sum_{m = 0}^{\infty} \frac{γ (m + 1, x)}{(m + a) m!} . \end{matrix}

(17)

Comparing (15) to (17) completes the proof. □

Table 1 presents explicit expressions for particular values of (13) and

x \in R

, obtained with the help of the MATHEMATICA program. Note that the

Shi (x)

and

Chi (x)

functions are defined in (61) and (62), respectively.

Next, we present other reduction formula of

\partial M_{κ, μ} (x) / \partial κ

from the result found in [15] for

x \in R

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = n, μ = 1 / 2} \\ = & [ln |x| - ψ (n + 1) - Ei (x)] M_{n, 1 / 2} (x) + \sum_{ℓ = 0}^{n - 1} (a_{ℓ} + b_{ℓ} e^{x}) M_{ℓ, 1 / 2} (x), \end{matrix}

(18)

where

Ei (x)

denotes the exponential integral, and for

n, ℓ = 1, 2, \dots

a_{ℓ} = \frac{1}{n} (\frac{n + ℓ}{n - ℓ})

(19)

and

b_{ℓ} = \{\begin{matrix} \frac{1}{n} \sum_{k = 0}^{n - ℓ - 1} \frac{{(ℓ)}_{k} 2^{k}}{{(ℓ + n)}_{k}}, & ℓ = 1, 2, \dots \\ 0, & ℓ = 0 . \end{matrix}

(20)

In order to calculate the finite sum provided in (20), we derive the following Lemma.

Lemma 1.

The following finite sum holds true

\forall n, ℓ = 1, 2, \dots

S (n, ℓ) = \sum_{k = 0}^{n - ℓ - 1} \frac{{(ℓ)}_{k} 2^{k}}{{(ℓ + n)}_{k}} = Re [_{2} F_{1} (\begin{matrix} 1, ℓ \\ ℓ + n \end{matrix}| 2)] .

(21)

Proof.

Split the sum in two as

S (n, ℓ) = \underset{S_{1} (n, ℓ)}{\underset{︸}{\sum_{k = 0}^{\infty} \frac{{(ℓ)}_{k} {(1)}_{k} 2^{k}}{k! {(ℓ + n)}_{k}}}} - \underset{S_{2} (n, ℓ)}{\underset{︸}{\sum_{k = n - ℓ}^{\infty} \frac{{(ℓ)}_{k} {(1)}_{k} 2^{k}}{k! {(ℓ + n)}_{k}}}},

where

S_{1} (n, ℓ) =_{2} F_{1} (\begin{matrix} 1, ℓ \\ ℓ + n \end{matrix}| 2),

and

\begin{matrix} S_{2} (n, ℓ) & = & 2^{n - ℓ} \sum_{s = 0}^{\infty} \frac{{(ℓ)}_{s + n - ℓ} {(1)}_{s} 2^{s}}{s! {(ℓ + n)}_{s + n - ℓ}} \\ = & 2^{n - ℓ} \frac{{(ℓ)}_{n}}{{(n)}_{n}} \sum_{s = 0}^{\infty} \frac{{(n)}_{s} {(1)}_{s} 2^{s}}{s! {(2 n)}_{s}} \\ = & 2^{n - ℓ} \frac{{(ℓ)}_{n}}{{(n)}_{n}}_{2} F_{1} (\begin{matrix} 1, n \\ 2 n \end{matrix}| 2) . \end{matrix}

Take

a = 1

b = n

, and

z = 2

in the quadratic transformation ([8], Equation 15.18.3)

\begin{matrix} _{2} F_{1} (\begin{matrix} a, b \\ 2 b \end{matrix}| z) \\ = & {(1 - z)}^{- a / 2}_{2} F_{1} (\begin{matrix} \frac{a}{2}, b - \frac{a}{2} \\ b + \frac{1}{2} \end{matrix}| \frac{z^{2}}{4 (z - 1)}), \end{matrix}

to obtain

_{2} F_{1} (\begin{matrix} 1, n \\ 2 n \end{matrix}| 2) = i_{2} F_{1} (\begin{matrix} \frac{1}{2}, n - \frac{1}{2} \\ n + \frac{1}{2} \end{matrix}| 1) .

Now, apply Gauss’s summation theorem ([8], Equation 15.4.20)

\begin{matrix} _{2} F_{1} (\begin{matrix} a, b \\ c \end{matrix}| 1) & = & \frac{Γ (c) Γ (c - a - b)}{Γ (c - a) Γ (c - b)}, \\ Re (c - a - b) & > & 0, \end{matrix}

and the formula ([7], Equation 43:4:3)

Γ (n + \frac{1}{2}) = \frac{(2 n - 1)!!}{2^{n}} \sqrt{π},

to arrive at

_{2} F_{1} (\begin{matrix} 1, n \\ 2 n \end{matrix}| 2) = i π \frac{(2 n - 1)!!}{2^{n} (n - 1)!} .

Therefore,

S_{2} (n, ℓ)

is a pure imaginary number. Because

S (n, ℓ)

is a real number, we conclude that

S (n, ℓ) = Re [S_{1} (n, ℓ)]

, as we wanted to prove. □

Theorem 3.

The following reduction formula holds true for

n = 1, 2, \dots

and

x \in R

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = n, μ = 1 / 2} \\ = & \frac{2}{n} sinh (\frac{x}{2}) + \frac{x e^{- x / 2}}{n} \\ \{[ln |x| + γ - H_{n} - Ei (x)] L_{n - 1}^{(1)} (x) \\ + \sum_{ℓ = 1}^{n - 1} (\frac{n + ℓ}{n - ℓ} - e^{x} Re [_{2} F_{1} (\begin{matrix} 1, ℓ \\ ℓ + n \end{matrix}| 2)]) \frac{L_{ℓ - 1}^{(1)} (x)}{ℓ}\}, \end{matrix}

(22)

where

L_{n}^{(α)} (x)

denotes the Laguerre polynomials (A14) and

H_{n} = \sum_{k = 1}^{n} \frac{1}{k}

the n-th harmonic number.

Proof.

From (21) and (20), we can see that

b_{ℓ} = Re [_{2} F_{1} (\begin{matrix} 1, ℓ \\ ℓ + n \end{matrix}| 2)], ℓ = 1, 2, \dots

(23)

Additionally, according to ([8], Equation 13.18.1),

M_{0, 1 / 2} (x) = 2 sinh (\frac{x}{2}) .

(24)

By performing the transformations

κ \to κ + 1

κ \to 0,

and

n \to n - 1

in (A13), we obtain

\forall n = 1, 2, \dots

M_{n, 1 / 2} (x) = \frac{x e^{- x / 2}}{n} L_{n - 1}^{(1)} (x) .

(25)

Finally, we have the following for

n = 1, 2, \dots

([27], Equation 1.3.7):

ψ (n + 1) = - γ + H_{n} .

(26)

Now, insert (19) and (20)–(26) in (18) to arrive at (22), as we wanted to prove. □

Corollary 3.

The following reduction formula holds true for

n = 1, 2, \dots

and

x \in R

\begin{matrix} G^{(1)} (\begin{matrix} 1 - n \\ 2 \end{matrix}| x) \\ = & \frac{1}{n} \{\frac{1 - e^{x}}{x} - [ln |x| + γ - H_{n} - Ei (x)] L_{n - 1}^{(1)} (x) \\ - \sum_{ℓ = 1}^{n - 1} (\frac{n + ℓ}{n - ℓ} - e^{x} Re [_{2} F_{1} (\begin{matrix} 1, ℓ \\ ℓ + n \end{matrix}| 2)]) \frac{L_{ℓ - 1}^{(1)} (x)}{ℓ}\} . \end{matrix}

Proof.

Consider (16) and (22) to arrive at the desired result. □

In Table 2, we collect particular cases of (22) for

x \in R

obtained with the help of the MATHEMATICA program.

2.2. Derivative with Respect to the Second Parameter $\partial M_{κ, μ} (x) / \partial μ$

Using (1) and (3), the first derivative of

M_{κ, μ} (x)

with respect to the parameter

μ

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial μ} \\ = & [ln x + 2 ψ (1 + 2 μ) - ψ (\frac{1}{2} + μ - κ)] M_{κ, μ} (x) \\ + x^{μ + 1 / 2} e^{- x / 2} \frac{Γ (1 + 2 μ)}{Γ (\frac{1}{2} + μ - κ)} [S_{1} (κ, μ, x) - S_{2} (κ, μ, x)], \end{matrix}

(27)

where

S_{1} (κ, μ, x)

is provided in (12) and the series

S_{2} (κ, μ, x)

S_{2} (κ, μ, x) = 2 \sum_{n = 0}^{\infty} \frac{Γ (\frac{1}{2} + μ - κ + n)}{Γ (1 + 2 μ + n)} ψ (1 + 2 μ + n) \frac{x^{n}}{n!} .

(28)

Theorem 4.

For

μ \neq - 1 / 2

and

x \in R

, the following parameter derivative formula of

M_{κ, μ} (x)

holds true:

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial μ}|}_{κ = - μ - 1 / 2} \\ = & x^{μ + 1 / 2} e^{x / 2} [ln x - \frac{x}{1 + 2 μ}_{2} F_{2} (\begin{matrix} 1, 1 \\ 2 (μ + 1), 2 \end{matrix}| - x)] . \end{matrix}

(29)

Proof.

For

κ = - μ - 1 / 2

, we have

S_{2} (κ, μ, x) = 2 S_{1} (κ, μ, x)

; therefore, (27) becomes

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial μ}|}_{κ = - μ - 1 / 2} \\ = & [ln x + ψ (1 + 2 μ)] M_{- μ - 1 / 2, μ} (x) - x^{μ + 1 / 2} e^{- x / 2} S_{1} (- μ - \frac{1}{2}, μ, x), \end{matrix}

where

S_{1} (- μ - \frac{1}{2}, μ, x) = \sum_{n = 0}^{\infty} ψ (1 + 2 μ + n) \frac{x^{n}}{n!} .

Thus, using (14),

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial μ}|}_{κ = - μ - 1 / 2} \\ = & [ln x + ψ (1 + 2 μ)] M_{- μ - 1 / 2, μ} (x) \\ - x^{μ + 1 / 2} e^{x / 2} [ψ (1 + 2 μ) + \frac{x}{1 + 2 μ}_{2} F_{2} (\begin{matrix} 1, 1 \\ 2 μ + 2, 2 \end{matrix}| - x)] . \end{matrix}

(30)

Because, according to (1) and (3),

M_{- μ - 1 / 2, μ} (x) = x^{μ + 1 / 2} e^{x / 2},

(30) now takes the simple form provided in (29), as we wanted to prove. □

Corollary 4.

For

a \in R

a \neq 0

, and

x \in R

, the following reduction formula holds true:

H^{(1)} (\begin{matrix} a \\ a \end{matrix}| x) = - \frac{x e^{x}}{a}_{2} F_{2} (\begin{matrix} 1, 1 \\ a + 1, 2 \end{matrix}| - x) .

(31)

Proof.

Direct differentiation of (1) yields

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial μ} & = & ln x M_{κ, μ} (x) + x^{μ + 1 / 2} e^{- x / 2} \\ [G^{(1)} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x) + 2 H^{(1)} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x)], \end{matrix}

(32)

thus, comparing (32) with

κ = - μ - \frac{1}{2}

to (29) and taking into account (15), we arrive at (31), as we wanted to prove. □

Using (29), the derivative of

M_{κ, μ} (x)

with respect

μ

can be calculated for particular values of

κ

and

μ

with

x \in R

; as obtained with the help of MATHEMATICA, these are presented in Table 3.

Note that for

μ = - 1 / 2

, we obtain an indeterminate expression in (29). For this case, we present the following result.

Theorem 5.

The following parameter derivative formula of

M_{κ, μ} (x)

holds true for

x \in R

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial μ}|}_{κ = 0} \\ = & 4^{μ} \sqrt{x} Γ (1 + μ) \{I_{μ} (\frac{x}{2}) [ln 4 + ψ (1 + μ)] + \frac{\partial I_{μ} (x / 2)}{\partial μ}\}, \end{matrix}

(33)

where

I_{ν} (x)

denotes the modified Bessel function.

Proof.

Differentiating with respect to

μ

the expression ([8], Equation 13.18.8)

M_{0, μ} (x) = 4^{μ} Γ (1 + μ) \sqrt{x} I_{μ} (\frac{x}{2}),

(34)

we obtain (33), as we wanted to prove. □

The order derivative of the modified Bessel function

I_{μ} (x)

is provided in terms of the Meijer-G function and the generalized hypergeometric function

\forall Re x > 0, μ \geq 0

[28]:

\begin{matrix} \frac{\partial I_{μ} (x)}{\partial μ} & = & - \frac{μ I_{μ} (x)}{2 \sqrt{π}} G_{2, 4}^{3, 1} (x^{2} |\begin{matrix} \frac{1}{2}, 1 \\ 0, 0, μ, - μ \end{matrix}) \\ - \frac{K_{μ} (x)}{Γ^{2} (μ + 1)} {(\frac{x}{2})}^{2 μ}_{2} F_{3} (\begin{matrix} μ, μ + \frac{1}{2} \\ μ + 1, μ + 1, 2 μ + 1 \end{matrix}| x^{2}), \end{matrix}

(35)

where

K_{ν} (x)

is the modified Bessel function of the second kind, or in terms of generalized hypergeometric functions, only

\forall Re x > 0, μ > 0

μ \notin Z

[29]:

\begin{matrix} \frac{\partial I_{μ} (x)}{\partial μ} \\ = & I_{μ} (x) [\frac{x^{2}}{4 (1 - μ^{2})}_{3} F_{4} (\begin{matrix} 1, 1, \frac{3}{2} \\ 2, 2, 2 - μ, 2 + μ \end{matrix}| x^{2}) + ln (\frac{x}{2}) - ψ (μ) - \frac{1}{2 μ}] \\ - I_{- μ} (x) \frac{π csc (π μ)}{2 Γ^{2} (μ + 1)} {(\frac{x}{2})}^{2 μ}_{2} F_{3} (\begin{matrix} μ, μ + \frac{1}{2} \\ μ + 1, μ + 1, 2 μ + 1 \end{matrix}| x^{2}) . \end{matrix}

(36)

There are different expressions for the order derivatives of the Bessel functions [30,31]. This subject is summarized in [32], where more general results are presented in terms of convolution integrals, while order derivatives of Bessel functions are found for particular values of the order.

Using (33), (35), and (36), derivatives of

M_{κ, μ} (x)

with respect to

μ

can be calculated for

x \in R

; these are presented in Table 4 as obtained with the help of MATHEMATICA.

3. Parameter Differentiation of $M_{κ, μ}$ via Integral Representations

3.1. Derivative with Respect to the First Parameter $\partial M_{κ, μ} (x) / \partial κ$

Integral representations of

M_{κ, μ} (x)

can be obtained via integral representations of confluent hypergeometric functions ([6], Section 7.4.1); thus,

\begin{matrix} M_{κ, μ} (x) \end{matrix}

\begin{matrix} = & \frac{x^{μ + 1 / 2} e^{- x / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} d t \end{matrix}

(37)

\begin{matrix} = & \frac{x^{μ + 1 / 2} e^{x / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} \int_{0}^{1} e^{- x t} t^{μ + κ - 1 / 2} {(1 - t)}^{μ - κ - 1 / 2} d t \\ Re (μ \pm κ + \frac{1}{2}) > 0, \end{matrix}

(38)

where

B (a, b) = \frac{Γ (a) Γ (b)}{Γ (a + b)}

(39)

denotes the beta function. In order to calculate the first derivative of

M_{κ, μ} (x)

with respect to parameter

κ

, we introduce the following finite logarithmic integrals.

Definition 2.

\begin{matrix} I_{1} (κ, μ; x) & = & \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln (\frac{1 - t}{t}) d t, \end{matrix}

(40)

\begin{matrix} I_{2} (κ, μ; x) & = & \int_{0}^{1} e^{- x t} t^{μ + κ - 1 / 2} {(1 - t)}^{μ - κ - 1 / 2} ln (\frac{t}{1 - t}) d t . \end{matrix}

(41)

Differentiation of (37) and (38) with respect to parameter

κ

yields, respectively,

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial κ} & = & [ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2})] M_{κ, μ} (x) \\ + \frac{x^{μ + 1 / 2} e^{- x / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} I_{1} (κ, μ; x) \end{matrix}

(42)

\begin{matrix} = & [ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2})] M_{κ, μ} (x) \\ + \frac{x^{μ + 1 / 2} e^{x / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} I_{2} (κ, μ; x), \end{matrix}

(43)

Note that from (42) and (43) we have

I_{2} (κ, μ; x) = e^{- x} I_{1} (κ, μ; x) .

(44)

Likewise, we can depart from other integral respresentations of

M_{κ, μ} (x)

([6], Section 7.4.1) (note that there are several typos in this reference regarding these integral representations) to obtain

\begin{matrix} M_{κ, μ} (x) & = & \frac{2^{- 2 μ} x^{μ + 1 / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} \\ \int_{- 1}^{1} e^{x t / 2} {(1 + t)}^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} d t \end{matrix}

(45)

\begin{matrix} = & \frac{2^{- 2 μ} x^{μ + 1 / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} \\ \int_{- 1}^{1} e^{- x t / 2} {(1 + t)}^{μ + κ - 1 / 2} {(1 - t)}^{μ - κ - 1 / 2} d t \\ Re (μ \pm κ + \frac{1}{2}) > 0, \end{matrix}

(46)

and consequently, we have

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial κ} & = & [ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2})] M_{κ, μ} (x) \\ + \frac{2^{- 2 μ} x^{μ + 1 / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} I_{3} (κ, μ; x) \end{matrix}

(47)

\begin{matrix} = & [ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2})] M_{κ, μ} (x) \\ + \frac{2^{- 2 μ} x^{μ + 1 / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} I_{4} (κ, μ; x), \end{matrix}

(48)

where we have defined the following logarithmic integrals.

Definition 3.

\begin{matrix} I_{3} (κ, μ; x) & = & \int_{- 1}^{1} e^{x t / 2} {(1 + t)}^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln (\frac{1 - t}{1 + t}) d t, \end{matrix}

(49)

\begin{matrix} I_{4} (κ, μ; x) & = & \int_{- 1}^{1} e^{- x t / 2} {(1 + t)}^{μ + κ - 1 / 2} {(1 - t)}^{μ - κ - 1 / 2} ln (\frac{1 + t}{1 - t}) d t . \end{matrix}

(50)

Note that from (47) and (48), we have

I_{3} (κ, μ; x) = I_{4} (κ, μ; x) = 2^{2 μ} e^{- x / 2} I_{1} (κ, μ; x) .

(51)

Because

I_{2} (κ, μ; x)

I_{3} (κ, μ; x)

, and

I_{4} (κ, μ; x)

are reduced to the calculation of

I_{1} (κ, μ; x)

, we next calculate the latter integral.

Theorem 6.

The following integral holds true for

x \in R

\begin{matrix} I_{1} (κ, μ; x) \\ = & B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2}) \\ \{[ψ (\frac{1}{2} + μ + κ) - ψ (\frac{1}{2} + μ - κ)]_{1} F_{1} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x) \\ - G^{(1)} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x)\} . \end{matrix}

(52)

Proof.

Comparing (42) to (16) and taking into account (1), we arrive at (52), as we wanted to prove. □

Corollary 5.

For

κ = 0

, Equation (52) is reduced to

I_{1} (0, μ; x) = - B (μ + \frac{1}{2}, μ + \frac{1}{2}) G^{(1)} (\begin{matrix} \frac{1}{2} + μ \\ 1 + 2 μ \end{matrix}| x) .

(53)

Theorem 7.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following integral holds true for

x \in R

I_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) = e^{x} F (- ℓ, m - ℓ, - x) - F (ℓ, m, x),

(54)

where

\begin{matrix} F (s, k, z) \\ = & \sum_{n = 0}^{k} {(- 1)}^{n} (\binom{k}{n}) \frac{d^{n + k - s}}{d z^{n + k - s}} [\frac{ln z - Chi (z) - Shi (z) + γ}{z}], \end{matrix}

(55)

and the functions

Shi (z)

and

Chi (z)

denote the hyperbolic sine and cosine integrals.

Proof.

From the definition of

I_{1} (κ, μ; x)

provided in (40), we have

\begin{matrix} I_{1} (κ, μ; x) & = & \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln (1 - t) d t \\ - \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln t d t . \end{matrix}

We can change the variables

τ = 1 - t

in the first integral above to arrive at

I_{1} (κ, μ; x) = e^{x} I_{1} (- κ, μ; - x) - I_{1} (κ, μ; x),

(56)

where we have set

I_{1} (κ, μ; x) = \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln t d t .

(57)

Taking into account the binomial theorem and the integral (A9) calculated in Appendix A, i.e.,

\int_{0}^{1} e^{x t} t^{m} ln t d t = \frac{- 1}{{(m + 1)}^{2}}_{2} F_{2} (\begin{matrix} m + 1, m + 1 \\ m + 2, m + 2 \end{matrix}| x),

we can calculate

\begin{matrix} I_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) \\ = & \int_{0}^{1} e^{x t} t^{m - ℓ} {(1 - t)}^{m} ln t d t \\ = & \sum_{n = 0}^{m} (\binom{m}{n}) {(- 1)}^{n} \int_{0}^{1} e^{x t} t^{m + n - ℓ} ln t d t \\ = & \sum_{n = 0}^{m} (\binom{m}{n}) \frac{{(- 1)}^{n + 1}}{{(n + m - ℓ + 1)}^{2}}_{2} F_{2} (\begin{matrix} n + m - ℓ + 1, n + m - ℓ + 1 \\ n + m - ℓ + 2, n + m - ℓ + 2 \end{matrix}| x) . \end{matrix}

(58)

Now, we can apply the differentiation formula ([8], Equation 16.3.1)

\frac{d^{n}}{d z^{n}}_{p} F_{q} (\begin{matrix} a_{1}, \dots, a_{p} \\ b_{1}, \dots, b_{q} \end{matrix}| z) = \frac{{(a_{1})}_{n} \dots {(a_{p})}_{n}}{{(b_{1})}_{n} \dots {(b_{q})}_{n}}_{p} F_{q} (\begin{matrix} a_{1} + n, \dots, a_{p} + n \\ b_{1} + n, \dots, b_{q} + n \end{matrix}| z),

to obtain

I_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) = \sum_{n = 0}^{m} (\binom{m}{n}) {(- 1)}^{n + 1} \frac{d^{n + m - ℓ}}{d x^{n + m - ℓ}}_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, 2 \end{matrix}| x) .

(59)

According to ([9], Equation 7.12.2(67)), we have

_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, 2 \end{matrix}| x) = \frac{Ei (x) - ln (- x) - γ}{x},

(60)

In order to obtain similar expressions to those obtained in Table 1, we can derive an alternative form of (60). Indeed, from the definition of the hyperbolic sine and cosine integrals ([8], Equations 6.2.15–6.2.16),

\forall z \in C

\begin{matrix} Shi (z) & = & \int_{0}^{z} \frac{sinh t}{t} d t \end{matrix}

(61)

\begin{matrix} Chi (z) & = & γ + ln z + \int_{0}^{z} \frac{cosh t - 1}{t} d t, \end{matrix}

(62)

it is easy to prove that

\begin{matrix} Shi (- z) & = & - Shi (z), \end{matrix}

(63)

\begin{matrix} Chi (- z) & = & Chi (z) - ln z + ln (- z) . \end{matrix}

(64)

Additionally, from the definition of a complementary exponential integral ([8], Equation 6.2.3)

Ein (z) = \int_{0}^{z} \frac{1 - e^{- t}}{t} d t

and the property

\forall x > 0

([8], Equation 6.2.7)

Ei (- x) = - Ein (x) + ln x + γ,

it is easy to prove that

Ei (- x) = Chi (x) - Shi (x),

thus, taking into account (63) and (64), we have

Ei (x) = Chi (x) - ln x + ln (- x) + Shi (x) .

(65)

We can insert (65) in (60) to obtain

_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, 2 \end{matrix}| x) = \frac{Chi (x) - ln x + Shi (x) - γ}{x} .

(66)

Finally, by substituting (66) in (59) while taking into account (55), we arrive at

\begin{matrix} I_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) \\ = & \sum_{n = 0}^{m} (\binom{m}{n}) {(- 1)}^{n + 1} \frac{d^{n + m - ℓ}}{d x^{n + m - ℓ}} [\frac{Chi (x) - ln x + Shi (x) - γ}{x}] \\ = & F (ℓ, m, x) . \end{matrix}

Similarly, we can calculate

I_{1} (- \frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; - x) = F (- ℓ, m - ℓ, - x) .

(67)

Finally, according to (56), we arrive at (54), as we wanted to prove. □

Table 5 shows the integral

I_{1} (κ, μ; x)

for

x \in R

and particular values of the parameters

κ

and/or

μ

obtained from (52) and (54) with the aid of MATHEMATICA program.

Theorem 8.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following reduction formula holds true for

x \in R

\begin{matrix} M_{ℓ / 2, m + (1 - ℓ) / 2} (x) \\ = & (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) {(- 1)}^{m - ℓ} x^{ℓ / 2 - m} \\ [e^{x / 2} P (- ℓ, m - ℓ, - x) - e^{- x / 2} P (ℓ, m, x)], \end{matrix}

(68)

where we have set the polynomials:

P (s, k, z) = \sum_{n = 0}^{k} (\binom{k}{n}) (2 k - s - n)! z^{n} .

(69)

Proof.

According to the definition of

M_{κ, μ} (x)

(1), we have

M_{ℓ / 2, m + (1 - ℓ) / 2} (x) = x^{m + 1 - ℓ / 2} e^{- x / 2}_{1} F_{1} (\begin{matrix} m + 1 - ℓ \\ 2 (m + 1) - ℓ \end{matrix}| x) .

(70)

Applying the property ([7], Equation 18:5:1)

{(- x)}_{n} = {(- 1)}^{n} {(x - n + 1)}_{n}

and the reduction formula ([9], Equation 7.11.1(12))

\begin{matrix} _{1} F_{1} (\begin{matrix} n \\ m \end{matrix}| z) = \frac{(m - 2)! {(1 - m)}_{n}}{(n - 1)!} z^{1 - m} \\ \{\sum_{k = 0}^{m - n - 1} \frac{{(1 + n - m)}_{k}}{k! {(2 - m)}_{k}} z^{k} - e^{z} \sum_{k = 0}^{n - 1} \frac{{(1 - n)}_{k}}{k! {(2 - m)}_{k}} {(- z)}^{k}\}, \end{matrix}

where

n, m = 1, 2, \dots

and

m > n

, after some algebra we arrive at

\begin{matrix} _{1} F_{1} (\begin{matrix} m + 1 - ℓ \\ 2 (m + 1) - ℓ \end{matrix}| x) \\ = & (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) {(- 1)}^{m + 1 - ℓ} x^{ℓ - 2 m} \\ \{\sum_{k = 0}^{m} (\binom{m}{k}) (2 m - ℓ - k)! x^{k} - e^{x} \sum_{k = 0}^{m - ℓ} (\binom{m - ℓ}{k}) (2 m - ℓ - k)! {(- x)}^{k}\} . \end{matrix}

(71)

We can now insert (71) in (70) to obtain (68), as we wanted to prove. □

In addition to (68), other reduction formulas for the Whittaker function

M_{κ, μ} (x)

are presented in Appendix C. A large list of reduction formulas for

M_{κ, μ} (x)

is available in [24] and in other monographs dealing with the special functions [2,3,4,5,6,7,8,9,10,26].

Theorem 9.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following reduction formula holds true for

x \in R

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = ℓ / 2, μ = m + (1 - ℓ) / 2} \\ = & (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) x^{ℓ / 2 - m} e^{- x / 2} \\ \{{(- 1)}^{m - ℓ} (H_{m - ℓ} - H_{m}) [e^{x} P (- ℓ, m - ℓ, - x) - P (ℓ, m, x)] \\ + x^{2 m + 1 - ℓ} [e^{x} F (- ℓ, m - ℓ, - x) - F (ℓ, m, x)]\} . \end{matrix}

(72)

Proof.

According to (42), we have

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = ℓ / 2, μ = m + (1 - ℓ) / 2} \\ = & [ψ (m - ℓ + 1) - ψ (m + 1)] M_{ℓ / 2, m + (1 - ℓ) / 2} (x) \\ + \frac{x^{m + 1 + ℓ / 2} e^{- x / 2}}{B (m + 1, m - ℓ + 1)} I_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) . \end{matrix}

Now, we can apply (39) and the property (26) to obtain

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial κ}|}_{κ = ℓ / 2, μ = m + (1 - ℓ) / 2} \\ = & (H_{m - ℓ} - H_{m}) M_{ℓ / 2, m + (1 - ℓ) / 2} (x) \\ + (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) x^{m + 1 - ℓ / 2} e^{- x / 2} I_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) . \end{matrix}

Finally, by applying the results provided in (54) and (68), we arrive at (72), as we wanted to prove. □

Corollary 6.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following reduction formula holds true for

x \in R

\begin{matrix} G^{(1)} (\begin{matrix} m + 1 - ℓ \\ 2 (m + 1) - ℓ \end{matrix}| x) \\ = & (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) \\ \{{(- 1)}^{m - ℓ} x^{ℓ - 2 m - 1} (H_{m - ℓ} - H_{m}) [P (ℓ, m, x) - e^{x} P (- ℓ, m - ℓ, - x)] \\ + F (ℓ, m, x) - e^{x} F (- ℓ, m - ℓ, - x)\} . \end{matrix}

(73)

Proof.

Set (16) for

κ = \frac{ℓ}{2}

and

μ = m + \frac{1 - ℓ}{2}

and compare the result to (72). □

Table 6 shows the first derivative of

M_{κ, μ} (x)

with respect to the

κ

parameter for particular values of

κ

and

μ

and for

x \in R

, which are calculated from (72) and are not contained in Table 1.

3.2. Application to the Calculation of Infinite Integrals

Additional integral representations of the Whittaker function

M_{κ, μ} (x)

in terms of Bessel functions ([6], Section 6.5.1) are known:

\begin{matrix} M_{κ, μ} (x) \end{matrix}

\begin{matrix} = & \frac{Γ (1 + 2 μ) x^{1 / 2} e^{- x / 2}}{Γ (μ - κ + \frac{1}{2})} \int_{0}^{\infty} e^{- t} t^{- κ - 1 / 2} I_{2 μ} (2 \sqrt{x t}) d t \end{matrix}

(74)

\begin{matrix} = & \frac{Γ (1 + 2 μ) x^{1 / 2} e^{x / 2}}{Γ (μ + κ + \frac{1}{2})} \int_{0}^{\infty} e^{- t} t^{κ - 1 / 2} J_{2 μ} (2 \sqrt{x t}) d t \\ Re (- \frac{1}{2} - μ + κ) > 0 . \end{matrix}

(75)

Let us next introduce the following infinite logarithmic integrals.

Definition 4.

\begin{matrix} H_{1} (κ, μ; x) & = & \int_{0}^{\infty} e^{- t} t^{- κ - 1 / 2} I_{2 μ} (2 \sqrt{x t}) ln t d t, \end{matrix}

(76)

\begin{matrix} H_{2} (κ, μ; x) & = & \int_{0}^{\infty} e^{- t} t^{κ - 1 / 2} J_{2 μ} (2 \sqrt{x t}) ln t d t . \end{matrix}

(77)

Differentiation of (74) and (75) with respect to the

κ

parameter respectively yields

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial κ} \end{matrix}

\begin{matrix} = & ψ (μ - κ + \frac{1}{2}) M_{κ, μ} (x) - \frac{Γ (1 + 2 μ) x^{1 / 2} e^{- x / 2}}{Γ (μ - κ + \frac{1}{2})} H_{1} (κ, μ; x) \end{matrix}

(78)

\begin{matrix} = & - ψ (μ + κ + \frac{1}{2}) M_{κ, μ} (x) + \frac{Γ (1 + 2 μ) x^{1 / 2} e^{x / 2}}{Γ (μ + κ + \frac{1}{2})} H_{2} (κ, μ; x) . \end{matrix}

(79)

Note that from (42) and (78) we have

\begin{matrix} H_{1} (κ, μ; x) \\ = & \frac{Γ (μ - κ + \frac{1}{2}) ψ (μ + κ + \frac{1}{2})}{Γ (1 + 2 μ) \sqrt{x} e^{x / 2}} M_{κ, μ} (x) - \frac{x^{μ} I_{1} (κ, μ; x)}{Γ (μ + κ + \frac{1}{2})}, \end{matrix}

(80)

while from (42) and (79) we have

\begin{matrix} H_{2} (κ, μ; x) \\ = & \frac{Γ (μ + κ + \frac{1}{2}) ψ (μ - κ + \frac{1}{2})}{Γ (1 + 2 μ) \sqrt{x} e^{x / 2}} M_{κ, μ} (x) + \frac{e^{- x} x^{μ} I_{1} (κ, μ; x)}{Γ (μ - κ + \frac{1}{2})} . \end{matrix}

(81)

Corollary 7.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following infinite integrals holds true for

x \in R

\begin{matrix} \int_{0}^{\infty} \frac{e^{- t} ln t}{t^{(1 + ℓ) / 2}} I_{2 m + 1 - ℓ} (2 \sqrt{x t}) d t \\ = & H_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) \\ = & \frac{1}{m!} \{{(- 1)}^{m - ℓ} (H_{m} - γ) x^{- m + (ℓ - 1) / 2} [e^{x} P (- ℓ, m - ℓ, - x) - P (ℓ, m, x)] \\ - x^{m + (1 - ℓ) / 2} [e^{x} F (- ℓ, m - ℓ, - x) - F (ℓ, m, x)]\} . \end{matrix}

(82)

and

\begin{matrix} \int_{0}^{\infty} \frac{e^{- t} ln t}{t^{(1 - ℓ) / 2}} J_{2 m + 1 - ℓ} (2 \sqrt{x t}) d t \\ = & H_{2} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) \\ = & \frac{1}{(m - ℓ)!} \{{(- 1)}^{m - ℓ} (H_{m - ℓ} - γ) x^{- m + (ℓ - 1) / 2} [P (- ℓ, m - ℓ, - x) - e^{- x} P (ℓ, m, x)] \\ + x^{m + (1 - ℓ) / 2} [F (- ℓ, m - ℓ, - x) - e^{- x} F (ℓ, m, x)]\} . \end{matrix}

(83)

Proof.

Substitute the results provided in (54) and (68) into (80) and (81) and apply (26). □

3.3. Derivative with Respect to the Second Parameter $\partial M_{κ, μ} (x) / \partial μ$

In order to calculate the first derivative of

M_{κ, μ} (x)

with respect to parameter

μ

, we introduce the following finite logarithmic integrals.

Definition 5.

\begin{matrix} J_{1} (κ, μ; x) & = & \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln [t (1 - t)] d t, \end{matrix}

(84)

\begin{matrix} J_{2} (κ, μ; x) & = & \int_{0}^{1} e^{- x t} t^{μ + κ - 1 / 2} {(1 - t)}^{μ - κ - 1 / 2} ln [t (1 - t)] d t, \end{matrix}

(85)

\begin{matrix} J_{3} (κ, μ; x) & = & \int_{- 1}^{1} e^{x t / 2} {(1 + t)}^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln (1 - t^{2}) d t, \end{matrix}

(86)

\begin{matrix} J_{4} (κ, μ; x) & = & \int_{- 1}^{1} e^{- x t / 2} {(1 + t)}^{μ + κ - 1 / 2} {(1 - t)}^{μ - κ - 1 / 2} ln (1 - t^{2}) d t . \end{matrix}

(87)

Differentiation of (37) and (38) with respect to the

μ

parameter provides us with

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial μ} \\ = & [ln x - ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2}) + 2 ψ (2 μ + 1)] M_{κ, μ} (x) \end{matrix}

(88)

\begin{matrix} + \frac{x^{μ + 1 / 2} e^{- x / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} J_{1} (κ, μ; x) \\ = & [ln x - ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2}) + 2 ψ (2 μ + 1)] M_{κ, μ} (x) \\ + \frac{x^{μ + 1 / 2} e^{x / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} J_{2} (κ, μ; x) . \end{matrix}

(89)

For the other integral representations provided in (45) and (46), we have

\begin{matrix} \frac{\partial M_{κ, μ} (x)}{\partial μ} \\ = & [ln (x / 4) - ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2}) + 2 ψ (2 μ + 1)] M_{κ, μ} (x) \end{matrix}

(90)

\begin{matrix} + \frac{2^{- 2 μ} x^{μ + 1 / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} J_{3} (κ, μ; x) \\ = & [ln (x / 4) - ψ (μ - κ + \frac{1}{2}) - ψ (μ + κ + \frac{1}{2}) + 2 ψ (2 μ + 1)] M_{κ, μ} (x) \\ + \frac{2^{- 2 μ} x^{μ + 1 / 2}}{B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2})} J_{4} (κ, μ; x) . \end{matrix}

(91)

From (88)–(91), we obtain the following interrelationships:

\begin{matrix} J_{2} (κ, μ; x) & = & e^{- x} J_{1} (κ, μ; x), \\ J_{3} (κ, μ; x) & = & 2^{2 μ} [e^{- x / 2} J_{1} (κ, μ; x) + \frac{ln 4}{x^{μ + 1 / 2}} B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2}) M_{κ, μ} (x)], \\ J_{4} (κ, μ; x) & = & J_{3} (κ, μ; x) . \end{matrix}

Because

J_{2} (κ, μ; x)

J_{3} (κ, μ; x)

, and

J_{4} (κ, μ; x)

are reduced to the calculation of

J_{1} (κ, μ; x)

, we next calculate the latter integral.

Theorem 10.

According to the notation introduced in (6) and (7), the following integral holds true:

\begin{matrix} J_{1} (κ, μ; x) \\ = & B (μ + κ + \frac{1}{2}, μ - κ + \frac{1}{2}) \\ \{[ψ (\frac{1}{2} + μ + κ) + ψ (\frac{1}{2} + μ - κ) - 2 ψ (2 μ + 1)]_{1} F_{1} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x) \\ + G^{(1)} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x) + 2 H^{(1)} (\begin{matrix} \frac{1}{2} + μ - κ \\ 1 + 2 μ \end{matrix}| x)\} . \end{matrix}

(92)

Proof.

Comparing (88) to (32) while taking into account (1), we arrive at (92), as we wanted to prove. □

Theorem 11.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following integral holds true for

x \in R

J_{1} (\frac{ℓ}{2}, m + \frac{1 - ℓ}{2}; x) = e^{x} F (- ℓ, m - ℓ, - x) + F (ℓ, m, x) .

(93)

Proof.

From the definition of

J_{1} (κ, μ; x)

provided in (84), we have

\begin{matrix} J_{1} (κ, μ; x) & = & \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln t d t \\ + \int_{0}^{1} e^{x t} t^{μ - κ - 1 / 2} {(1 - t)}^{μ + κ - 1 / 2} ln (1 - t) d t . \end{matrix}

By performing a change of variables

τ = 1 - t

in the second integral above, we arrive at

J_{1} (κ, μ; x) = e^{x} I_{1} (- κ, μ; - x) + I_{1} (κ, μ; x),

(94)

where we follow the notation in (57) for the integral

I_{1} (κ, μ; x)

. According to the results obtained in (58) and (67), we arrive at (93), as we wanted to prove. □

Theorem 12.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following reduction formula holds true for

x \in R

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial μ}|}_{κ = ℓ / 2, μ = m + (1 - ℓ) / 2} \\ = & (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) x^{ℓ / 2 - m} e^{- x / 2} \\ \{{(- 1)}^{m - ℓ} (ln x + 2 H_{2 m - ℓ + 1} - H_{m - ℓ} - H_{m}) [e^{x} P (- ℓ, m - ℓ, - x) - P (ℓ, m, x)] \\ + x^{2 m + 1 - ℓ} [e^{x} F (- ℓ, m - ℓ, - x) + F (ℓ, m, x)]\} . \end{matrix}

(95)

Proof.

Insert (68) and (93) into (88) and apply (26). □

Table 7 shows the first derivative of

M_{κ, μ} (x)

with respect to the

μ

parameter for particular values of

κ

and

μ

and for

x \in R

, which are calculated from (95) and are not contained in Table 3 and Table 4.

Corollary 8.

For

ℓ \in Z

and

m = 0, 1, 2, \dots

, with

m \geq ℓ

, the following reduction formula holds true for

x \in R

\begin{matrix} H^{(1)} (\begin{matrix} m + 1 - ℓ \\ 2 (m + 1) - ℓ \end{matrix}| x) \\ = & (2 m - ℓ + 1) (\binom{2 m - ℓ}{m}) \\ \{{(- 1)}^{m - ℓ} x^{ℓ - 2 m - 1} (H_{2 m - ℓ + 1} - H_{m}) [e^{x} P (- ℓ, m - ℓ, - x) - P (ℓ, m, x)] \\ + e^{x} F (- ℓ, m - ℓ, - x)\} . \end{matrix}

(96)

Proof.

Take

κ = \frac{ℓ}{2}

and

μ = m + \frac{1 - ℓ}{2}

in (32), and substitute the results provided in (68), (73), and (95). After simplification, we arrive at (96), as we wanted to prove. □

3.4. Application to the Calculation of Finite Integrals

Theorem 13.

For

μ \geq 0

and

x \in R

, the following finite integral holds true:

\begin{matrix} \int_{0}^{1} e^{x t} {[t (1 - t)]}^{μ - 1 / 2} ln [t (1 - t)] d t \\ = & J_{1} (0, μ; x) \\ = & B (μ + \frac{1}{2}, μ + \frac{1}{2}) {(\frac{4}{|x|})}^{μ} e^{x / 2} Γ (1 + μ) \\ \{I_{μ} (\frac{|x|}{2}) [ψ (μ + \frac{1}{2}) - ln |x|] + \frac{\partial I_{μ} (|x| / 2)}{\partial μ}\}, \end{matrix}

(97)

where

\partial I_{μ} (x) / \partial μ

is provided by (35) or (36).

Proof.

First, consider that

x > 0

. Take

κ = 0

in (88) and substitute (34) to arrive at

\begin{matrix} {\frac{\partial M_{κ, μ} (x)}{\partial μ}|}_{κ = 0} \\ = & 4^{μ} Γ (1 + μ) \sqrt{x} I_{μ} (\frac{x}{2}) [ln x - 2 ψ (μ + \frac{1}{2}) + 2 ψ (2 μ + 1)] \\ + \frac{x^{μ + 1 / 2} e^{- x / 2}}{B (μ + \frac{1}{2}, μ + \frac{1}{2})} J_{1} (0, μ; x) \end{matrix}

(98)

Next, equate (98) to the expression provided in (33), and solve for

J_{1} (0, μ; x)

to obtain

\begin{matrix} J_{1} (0, μ; x) \\ = & B (μ + \frac{1}{2}, μ + \frac{1}{2}) {(\frac{4}{x})}^{μ} e^{x / 2} Γ (1 + μ) \\ = & \{I_{μ} (\frac{x}{2}) [ln (\frac{4}{x}) + ψ (1 + μ) + 2 ψ (μ + \frac{1}{2}) - 2 ψ (2 μ + 1)] + \frac{\partial I_{μ} (x / 2)}{\partial μ}\} . \end{matrix}

(99)

Now, apply the property ([8], Equation 5.5.8)

ψ (2 z) = \frac{1}{2} [ψ (z) + ψ (z + \frac{1}{2})] + ln 2

for

z = μ + \frac{1}{2}

to simplify (99) as

\begin{matrix} J_{1} (0, μ; x) \\ = & B (μ + \frac{1}{2}, μ + \frac{1}{2}) {(\frac{4}{x})}^{μ} e^{x / 2} Γ (1 + μ) \\ = & \{I_{μ} (\frac{x}{2}) [ψ (μ + \frac{1}{2}) - ln x] + \frac{\partial I_{μ} (x / 2)}{\partial μ}\}, \end{matrix}

(100)

where (100) holds true for

x > 0

. Finally, note that by performing the change of variables

τ = 1 - t

in (84) we obtain the reflection formula

J_{1} (0, μ; x) = e^{x} J_{1} (0, μ; - x),

(101)

thus, from (100) and (101) we arrive at (97), as we wanted to prove. □

Theorem 14.

For

μ \geq 0

and

x \in R

, the following finite integral holds true:

\begin{matrix} \int_{- 1}^{1} e^{x t / 2} {[t (1 - t)]}^{μ - 1 / 2} ln [t (1 - t)] d t \\ = & J_{3} (0, μ; x) \\ = & B (μ + \frac{1}{2}, μ + \frac{1}{2}) Γ (1 + μ) {(\frac{16}{|x|})}^{μ} \\ \{I_{μ} (\frac{|x|}{2}) [ψ (μ + \frac{1}{2}) + ln (\frac{4}{|x|})] + \frac{\partial I_{μ} (|x| / 2)}{\partial μ}\}, \end{matrix}

(102)

where

\partial I_{μ} (x) / \partial μ

is provided by (35) or (36).

Proof.

Consider

x > 0

. Take

κ = 0

in (90) and susbtitute (34) to obtain

\begin{matrix} J_{3} (0, μ; x) \\ = & 2^{2 μ} e^{- x / 2} J_{1} (0, μ; x) \\ + 2^{4 μ} \frac{ln 4}{x^{μ}} B (μ + \frac{1}{2}, μ + \frac{1}{2}) Γ (1 + μ) I_{μ} (\frac{x}{2}) . \end{matrix}

(103)

Now, insert in (103) the result in (100) and simplify to obtain the following for

x > 0

\begin{matrix} J_{3} (0, μ; x) \\ = & B (μ + \frac{1}{2}, μ + \frac{1}{2}) Γ (1 + μ) {(\frac{16}{x})}^{μ} \\ \{I_{μ} (\frac{x}{2}) [ψ (μ + \frac{1}{2}) + ln (\frac{4}{x})] + \frac{\partial I_{μ} (x / 2)}{\partial μ}\} . \end{matrix}

(104)

Finally, note that by performing the change of variables

τ = - t

in (86) we obtain the reflection formula

J_{3} (0, μ; x) = J_{3} (0, μ; - x),

(105)

thus, from (104) and (105) we arrive at (102), as we wanted to prove. □

Table 8 shows the integral

J_{1} (κ, μ; x)

for particular values of the parameters

κ

and

μ

and for

x \in R

obtained from (92), (93), and (97) with the aid of the MATHEMATICA program.

4. Conclusions

The Whittaker function

M_{κ, μ} (x)

is defined in terms of the Kummer confluent hypergeometric function; hence, its derivative with respect to the parameters

κ

and

μ

can be expressed as infinite sums of quotients of the digamma and gamma functions. In addition, parameter differentiation of the integral representations of

M_{κ, μ} (x)

leads to finite and infinite integrals of elementary functions. These sums and integrals have been calculated for particular values of the parameters

κ

and

μ

in closed form. As an application of these results, we have obtained several reduction formulas for the derivatives of the confluent Kummer function with respect to the parameters, i.e.,

G^{(1)} (a, b; x)

and

H^{(1)} (a, b; x)

. Additionally, we have calculated finite integrals containing a combination of the exponential, logarithmic, and algebraic functions, as well as several infinite integrals involving the exponential, logarithmic, algebraic, and Bessel functions. It is worth noting that all the results presented in this paper have been checked both numerically and symbolically with the MATHEMATICA program.

In Appendix A, we obtain the first derivative of the incomplete gamma functions in closed form. These results allow us to calculate a finite logarithmic integral, which is used to calculate one of the integrals appearing in the body of the paper.

In Appendix B, we calculate new reduction formulas for the integral Whittaker functions

{Mi}_{κ, μ} (x)

and

{mi}_{κ, μ} (x)

from two reduction formulas of the Whittaker function

M_{κ, μ} (x)

. One of the latter seems to have not been previously reported in the literature.

Finally, in Appendix C, we collect a number of reduction formulas for the Whittaker function

M_{κ, μ} (x)

Author Contributions

Conceptualization, A.A. and J.L.G.-S.; Methodology, A.A. and J.L.G.-S.; Resources, A.A.; Writing—original draft, A.A. and J.L.G.-S.; Writing—review and editing, A.A. and J.L.G.-S. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

We are grateful to Francesco Mainardi from the Department of Physics and Astronomy, University of Bologna, Bologna, Italy, for his kind encouragement and interest in our work.

Conflicts of Interest

The authors declare no conflict of interest.

Appendix A. Parameter Differentiation of the Incomplete Gamma Functions

Definition A1.

The lower incomplete gamma function is defined as follows [7]:

γ (ν, x) = \int_{0}^{x} t^{ν - 1} e^{- t} d t .

(A1)

Definition A2.

The upper incomplete gamma function is defined as follows ([7], Equation 45:3:2)

Γ (ν, x) = \int_{x}^{\infty} t^{ν - 1} e^{- t} d t .

(A2)

The relation between both functions is

Γ (ν) = γ (ν, x) + Γ (ν, x) .

(A3)

The lower incomplete gamma function has the following series expansion ([7], Equation 45:6:1):

γ (ν, x) = e^{- x} \sum_{k = 0}^{\infty} \frac{x^{k + ν}}{{(ν)}_{k + 1}} .

(A4)

In addition, the following integral representations in terms of infinite integrals hold true ([8], Equations 8.6.3 and 8.6.7) for

Re z > 0

\begin{matrix} γ (ν, z) & = & z^{ν} \int_{0}^{\infty} exp (- ν t - z e^{- t}) d t, \\ Γ (ν, z) & = & z^{ν} \int_{0}^{\infty} exp (ν t - z e^{- t}) d t . \end{matrix}

From (A1), the derivative of the lower incomplete gamma function with respect to the order

ν

has the following integral representation:

\frac{\partial γ (ν, x)}{\partial ν} = \int_{0}^{x} t^{ν - 1} e^{- t} ln t d t

(A5)

Theorem A1.

The parameter derivative of the lower incomplete gamma function is

\frac{\partial γ (ν, x)}{\partial ν} = γ (ν, x) ln x - \frac{x^{ν}}{ν^{2}}_{2} F_{2} (\begin{matrix} ν, ν \\ ν + 1, ν + 1 \end{matrix}| - x) .

(A6)

Proof.

According to (A1) and (A4), the derivative of the lower incomplete gamma function with respect to the parameter

ν

\begin{matrix} \frac{\partial γ (ν, x)}{\partial ν} & = & e^{- x} \sum_{k = 0}^{\infty} \frac{x^{k + ν} [ln x + ψ (ν) - ψ (k + 1 + ν)]}{{(ν)}_{k + 1}} \\ = & [ln x + ψ (ν)] γ (ν, x) - e^{- x} \sum_{k = 0}^{\infty} \frac{x^{k + ν - 1}}{{(ν)}_{k}} ψ (k + ν) . \end{matrix}

Now, we apply the sum formula ([26], Equation 6.2.1(63))

\begin{matrix} \sum_{k = 0}^{\infty} \frac{t^{k}}{{(a)}_{k}} ψ (k + a) \\ = & ψ (a) + e^{t} [t^{1 - a} ψ (a) γ (a, t) + \frac{t}{a^{2}}_{2} F_{2} (\begin{matrix} a, a \\ a + 1, a + 1 \end{matrix}| - t)], \end{matrix}

to arrive at (A6), as we wanted to prove. □

Theorem A2.

The parameter derivative of the upper incomplete gamma function is

\begin{matrix} \frac{\partial Γ (ν, x)}{\partial ν} \\ = & Γ (ν) ψ (ν) - γ (ν, x) ln x + \frac{x^{ν}}{ν^{2}}_{2} F_{2} (\begin{matrix} ν, ν \\ ν + 1, ν + 1 \end{matrix}| - x) . \end{matrix}

(A7)

Proof.

Differentiate (A3) with respect to the parameter

ν

and apply the result provided in (A6). □

Corollary A1.

From (A5) and (A6), we can calculate the following integral:

\int_{0}^{x} t^{ν - 1} e^{- t} ln t d t = γ (ν, x) ln x - \frac{x^{ν}}{ν^{2}}_{2} F_{2} (\begin{matrix} ν, ν \\ ν + 1, ν + 1 \end{matrix}| - x) .

(A8)

Corollary A2.

The following integral holds true for

x \in R

\int_{0}^{1} e^{x t} t^{ν - 1} ln t d t = - \frac{1}{ν^{2}}_{2} F_{2} (\begin{matrix} ν, ν \\ ν + 1, ν + 1 \end{matrix}| x) .

(A9)

Proof.

Perform the change of variables

t = z τ

in the integral provided in (A8), split the result in two integrals, and apply the change of variables

t = x τ

again to the first integral:

\begin{matrix} \int_{0}^{x} t^{ν - 1} e^{- t} ln t d t & = & x^{ν} [ln x \int_{0}^{1} τ^{ν - 1} e^{- x τ} d τ + \int_{0}^{1} t^{ν - 1} e^{- x τ} ln τ d τ] \\ = & ln x \underset{γ (ν, x)}{\underset{︸}{\int_{0}^{x} t^{ν - 1} e^{- t} d t}} + x^{ν} \int_{0}^{1} τ^{ν - 1} e^{- x τ} ln τ d τ . \end{matrix}

(A10)

Comparing (A8) to (A10), we obtain (A9), as we wanted to prove. □

Corollary A3.

According to the notation provided in (7), the following reduction formula holds true for

x \in R

H^{(1)} (\begin{matrix} 1 \\ b \end{matrix}| x) = - \frac{x e^{x}}{b^{2}}_{2} F_{2} (\begin{matrix} b, b \\ b + 1, b + 1 \end{matrix}| - x) .

(A11)

Proof.

Knowing that ([7], Equation 47:4:6)

_{1} F_{1} (\begin{matrix} 1 \\ b \end{matrix}| z) = 1 + z^{1 - b} e^{z} γ (b, z)

and applying (A6), we can calculate (A11), as we wanted to prove. □

Appendix B. Reduction Formulas for Integral Whittaker Functions Mi_κ,μ and mi_κ,μ

In [24], we found reduction formulas for the integral Whittaker function

{Mi}_{κ, μ} (x)

. Next, we derive new reduction formulas for

{Mi}_{κ, μ} (x)

and

{mi}_{κ, μ} (x)

from reduction formulas of the Whittaker function

M_{κ, μ} (x)

Theorem A3.

The following reduction formula holds true for

x \in R

n = 0, 1, 2, \dots

and

κ > 0

{Mi}_{κ + n, κ - 1 / 2} (x) = 2^{κ} \sum_{m = 0}^{n} (\binom{n}{m}) \frac{{(- 2)}^{m}}{{(2 κ)}_{m}} γ (κ + m, x / 2),

(A12)

where

γ (ν, z)

denotes the lower incomplete gamma function.

Proof.

Next, we can apply to the definition of the Whittaker function (1) the following reduction formula ([9], Equation 7.11.1(17)):

_{1} F_{1} (\begin{matrix} - n \\ b \end{matrix}| z) = \frac{n!}{{(b)}_{n}} L_{n}^{(b - 1)} (z)

from which we obtain ([8], Equation 13.18.17)

M_{κ + n, κ - 1 / 2} (x) = \frac{n! e^{- x / 2} x^{κ}}{{(2 κ)}_{n}} L_{n}^{(2 κ - 1)} (x),

(A13)

where ([27], Equation 4.17.2)

L_{n}^{(α)} (x) = \sum_{m = 0}^{n} \frac{Γ (n + α + 1)}{Γ (m + α + 1)} \frac{{(- x)}^{m}}{m! (n - m)!}

(A14)

denotes the Laguerre polynomials. We can now insert (A14) in (A13) and integrate term by term according to the definition of the integral Whittaker function (4) to obtain

\begin{matrix} {Mi}_{κ + n, κ - 1 / 2} (x) \\ = & \sum_{m = 0}^{n} (\binom{n}{m}) \frac{{(- 1)}^{m}}{{(2 κ)}_{m}} \int_{0}^{x} e^{- t / 2} t^{κ + m - 1} d t . \end{matrix}

Finally, taking into account the definition of the lower incomplete gamma function (A1), we can simplify the result to arrive at (A12), as we wanted to prove. □

Remark A1.

Taking

n = 0

in (A12), we recover the formula provided in [24].

Theorem A4.

The following reduction formula holds true for

x > 0

n = 0, 1, 2, \dots

and

κ \in R

{mi}_{κ + n, κ - 1 / 2} (x) = 2^{κ} \sum_{m = 0}^{n} (\binom{n}{m}) \frac{{(- 2)}^{m}}{{(2 κ)}_{m}} Γ (κ + m, x / 2),

(A15)

where

Γ (ν, z)

denotes the upper incomplete gamma function.

Proof.

Following similar steps as in the previous theorem, here we instead consider the definition of the upper incomplete gamma function (A2). □

Theorem A5.

The following reduction formula holds true for

x \in R

n = 0, 1, 2, \dots

, and

κ > 0

{Mi}_{- κ - n, κ - 1 / 2} (x) = {(- 1)}^{- sign (x) κ} 2^{κ} \sum_{m = 0}^{n} (\binom{n}{m}) \frac{{(- 2)}^{m}}{{(2 κ)}_{m}} γ (κ + m, - x / 2) .

(A16)

Proof.

From the property for

x > 0

([7], Equation 48:13:3)

M_{κ, μ} (- x) = {(- 1)}^{μ + 1 / 2} M_{- κ, μ} (x),

for

x \in R

we have

M_{- κ, μ} (x) = {(- 1)}^{- sign (x) (μ + 1 / 2)} M_{κ, μ} (- x),

(A17)

We can apply (A17) to (A13) to obtain

M_{- κ - n, - κ - 1 / 2} (x) = {(- 1)}^{- sign (x) κ} \frac{n! e^{x / 2} {(- x)}^{κ}}{{(2 κ)}_{n}} L_{n}^{(2 κ - 1)} (- x) .

(A18)

Now, by inserting (A14) in (A13) and integrating term by term according to the definition of the integral Whittaker function (4), we obtain

\begin{matrix} {Mi}_{- κ - n, κ - 1 / 2} (x) \\ = & {(- 1)}^{- sign (x) κ} \sum_{m = 0}^{n} \frac{1}{{(2 κ)}_{m}} (\binom{n}{m}) \int_{0}^{x} e^{t / 2} t^{m - 1} {(- t)}^{κ} d t . \end{matrix}

Finally, takeing into account the definition of the lower incomplete gamma function (A1) and simplifying the result, we arrive at (A16), as we wanted to prove. □

Remark A2.

It is worth noting here that we could not locate the reduction Formula (A18) in the existing literature.

Appendix C. Reduction Formulas for the Whittaker Function M_κ,μ(x)

For the convenience of readers, reduction formulas for the Whittaker function

M_{κ, μ} (x)

are presented in their explicit forms in Table A1 for

x \in R

Table A1. Whittaker function

M_{κ, μ} (x)

for particular values of

κ

and

μ

Table A1. Whittaker function

M_{κ, μ} (x)

for particular values of

κ

and

μ

$κ$	$μ$	$M_{κ, μ} (x)$
$- \frac{1}{4}$	$\frac{1}{4}$	$\frac{\sqrt{π}}{2} e^{x / 2} x^{1 / 4} \erf (\sqrt{x})$
$- \frac{1}{2}$	$\frac{1}{2}$	$x [I_{0} (\frac{x}{2}) + I_{1} (\frac{x}{2})]$
$- \frac{1}{2}$	$\frac{1}{6}$	$2^{- 2 / 3} x Γ (\frac{2}{3}) [I_{- 1 / 3} (\frac{x}{2}) + I_{2 / 3} (\frac{x}{2})]$
$- \frac{1}{2}$	1	$x^{- 1 / 2} e^{- x / 2} [2 e^{x} (x - 1) + 2]$
0	0	$\sqrt{x} I_{0} (\frac{x}{2})$
0	$\frac{1}{2}$	$2 sinh (\frac{x}{2})$
0	1	$4 \sqrt{x} I_{1} (\frac{x}{2})$
0	$\frac{3}{2}$	$12 [cosh (\frac{x}{2}) - \frac{2}{x} sinh (\frac{x}{2})]$
0	$\frac{5}{2}$	$120 x^{- 2} [(x^{2} + 12) sinh (\frac{x}{2}) - 6 x cosh (\frac{x}{2})]$
$\frac{1}{6}$	0	$\sqrt{x} e^{- x / 2} L_{- 1 / 3} (x)$
$\frac{1}{4}$	$- \frac{1}{4}$	$x^{1 / 4} e^{- x / 2}$
$\frac{1}{4}$	$\frac{1}{4}$	$x^{1 / 4} e^{x / 2} F (\sqrt{x})$
$\frac{1}{3}$	0	$\sqrt{x} e^{- x / 2} L_{- 1 / 6} (x)$
$\frac{1}{2}$	$\frac{1}{6}$	$2^{- 2 / 3} x Γ (\frac{2}{3}) [I_{- 1 / 3} (\frac{x}{2}) - I_{2 / 3} (\frac{x}{2})]$
$\frac{1}{2}$	$\frac{1}{4}$	$2^{- 1 / 2} x Γ (\frac{3}{4}) [I_{- 1 / 4} (\frac{x}{2}) - I_{3 / 4} (\frac{x}{2})]$
$\frac{1}{2}$	$\frac{1}{2}$	$x [I_{0} (\frac{x}{2}) - I_{1} (\frac{x}{2})]$
$\frac{1}{2}$	1	$2 x^{- 1 / 2} e^{- x / 2} (e^{x} - x - 1)$
$\frac{1}{2}$	2	$12 x^{- 3 / 2} e^{- x / 2} [2 e^{x} (x - 3) + x^{2} + 4 x + 6]$
1	$- \frac{3}{2}$	$e^{- x / 2} (\frac{x}{2} + 1 + \frac{1}{x})$
1	1	$\frac{4}{3} \sqrt{x} [x I_{0} (\frac{x}{2}) - (x + 1) I_{1} (\frac{x}{2})]$
1	$\frac{3}{2}$	$x^{- 1} e^{- x / 2} (6 e^{x} - 3 x^{2} - 6 x - 6)$
1	2	$\frac{32}{5} x^{- 1 / 2} [(x^{2} + 4 x + 12) I_{1} (\frac{x}{2}) - (x^{2} + 3 x) I_{0} (\frac{x}{2})]$
2	2	$\frac{32}{35} x^{- 1 / 2} [x (2 x^{2} + 2 x + 3) I_{0} (\frac{x}{2}) - 2 (x^{3} + 2 x^{2} + 4 x + 6) I_{1} (\frac{x}{2})]$

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Table 1. Derivative of

M_{κ, μ}

with respect to

κ

using (13).

Table 1. Derivative of

M_{κ, μ}

with respect to

κ

using (13).

$κ$	$μ$	$\frac{\partial M_{κ, μ} (x)}{\partial κ}$
$- \frac{3}{4}$	$\frac{1}{4}$	$- \frac{2}{3} x^{7 / 4} e^{x / 2}_{2} F_{2} (1, 1; \frac{5}{2}, 2; - x)$
$- \frac{1}{2}$	0	$- \sqrt{x} e^{x / 2} [γ + ln x + Shi (x) - Chi (x)]$
$- \frac{1}{4}$	$- \frac{1}{4}$	$- 2 x^{5 / 4} e^{x / 2}_{2} F_{2} (1, 1; \frac{3}{2}, 2; - x)$
$- \frac{1}{6}$	$- \frac{1}{3}$	$- 3 x^{7 / 6} e^{x / 2}_{2} F_{2} (1, 1; \frac{4}{3}, 2; - x)$
0	$\frac{1}{2}$	$e^{- x / 2} [Shi (x) + Chi (x) - ln x - γ] - e^{x / 2} [Shi (x) - Chi (x) + ln x + γ]$
$\frac{1}{6}$	$- \frac{2}{3}$	$3 x^{5 / 6} e^{x / 2}_{2} F_{2} (1, 1; \frac{2}{3}, 2; - x)$
$\frac{1}{2}$	1	$\begin{matrix} - \frac{2}{\sqrt{x}} \{e^{x / 2} [γ + 1 + ln x + Shi (x) - Chi (x)] \\ + e^{- x / 2} (x + 1) [γ - 1 + ln x - Shi (x) - Chi (x)]\} \end{matrix}$
1	$\frac{3}{2}$	$\begin{matrix} - \frac{3}{x} \{e^{- x / 2} [(x^{2} + 2 x + 2) (ln x - Shi (x) - Chi (x) + γ)] \\ + e^{x / 2} [2 ln x + 2 Shi (x) - 2 Chi (x) + x + 2 γ + 3]\} \end{matrix}$

Table 2. Derivative of

M_{κ, μ}

with respect to

κ

using (22).

Table 2. Derivative of

M_{κ, μ}

with respect to

κ

using (22).

$κ$	$μ$	$\frac{\partial M_{κ, μ} (x)}{\partial κ}$
1	$\frac{1}{2}$	$x e^{- x / 2} [ln \|x\| - Ei (x) + γ - 1] + 2 sinh (\frac{x}{2})$
2	$\frac{1}{2}$	$\frac{1}{2} x e^{- x / 2} \{(2 - x) [ln \|x\| - Ei (x) + γ - \frac{3}{2}] - e^{x} + 3\} + sinh (\frac{x}{2})$
3	$\frac{1}{2}$	$\begin{matrix} \frac{1}{6} x e^{- x / 2} [(x^{2} - 6 x + 6) (ln \|x\| - Ei (x) + γ - \frac{11}{6}) \\ + (e^{x} - 5) (x - 2) - 3 e^{x} + 4] + \frac{2}{3} sinh (\frac{x}{2}) \end{matrix}$

Table 3. Derivative of

M_{κ, μ}

with respect to

μ

using (29).

Table 3. Derivative of

M_{κ, μ}

with respect to

μ

using (29).

$κ$	$μ$	$\frac{\partial M_{κ, μ} (x)}{\partial μ}$
$- \frac{3}{2}$	1	$\frac{1}{\sqrt{x}} \{e^{x / 2} [x^{2} (Chi (x) - Shi (x) - γ) + \frac{3}{2} x^{2} - 2 x + 1] + e^{- x / 2} (x - 1)\}$
$- 1$	$\frac{1}{2}$	$x e^{x / 2} [Chi (x) - Shi (x) - γ + 1] - 2 sinh (\frac{x}{2})$
$- \frac{3}{4}$	$\frac{1}{4}$	$e^{x / 2} x^{3 / 4} [ln x - \frac{2}{3} x_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, \frac{5}{2} \end{matrix}\| - x)]$
$- \frac{1}{2}$	0	$e^{x / 2} \sqrt{x} [Chi (x) - Shi (x) - γ]$
$- \frac{1}{4}$	$- \frac{1}{4}$	$e^{x / 2} x^{1 / 4} [ln x - 2 x_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, \frac{3}{2} \end{matrix}\| - x)]$
$- \frac{1}{6}$	$- \frac{1}{3}$	$e^{x / 2} x^{1 / 6} [ln x - 3 x_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, \frac{4}{3} \end{matrix}\| - x)]$
$\frac{1}{6}$	$- \frac{2}{3}$	$e^{x / 2} x^{- 1 / 6} [ln x + 3 x_{2} F_{2} (\begin{matrix} 1, 1 \\ 2, \frac{2}{3} \end{matrix}\| - x)]$

Table 4. Derivative of

M_{κ, μ}

with respect to

μ

using (33).

Table 4. Derivative of

M_{κ, μ}

with respect to

μ

using (33).

$κ$	$μ$	$\frac{\partial M_{κ, μ} (x)}{\partial μ}$
0	$- \frac{1}{2}$	$[Chi (x) - γ] cosh (\frac{x}{2}) - \frac{2}{x} {sinh}^{3} (\frac{x}{2})$
0	0	$\sqrt{x} [(ln 4 - γ) I_{0} (\frac{x}{2}) - K_{0} (\frac{x}{2})]$
0	$\frac{1}{4}$	$\begin{matrix} \frac{x^{3 / 4}}{15}_{0} F_{1} (; \frac{5}{4}; \frac{x^{2}}{16}) [x^{2}_{3} F_{4} (1, 1, \frac{3}{2}; \frac{7}{4}, 2, 2, \frac{9}{4}; \frac{x^{2}}{4}) + 15 (ln x + 2)] \\ - \frac{2 π x}{Γ (\frac{1}{4})} I_{- \frac{1}{4}} (\frac{x}{2})_{2} F_{3} (\frac{1}{4}, \frac{3}{4}; \frac{5}{4}, \frac{5}{4}, \frac{3}{2}; \frac{x^{2}}{4}) \end{matrix}$
0	$\frac{1}{3}$	$\begin{matrix} \frac{x^{5 / 6}}{128} \{_{0} F_{1} (; \frac{4}{3}; \frac{x^{2}}{16}) [9 x^{2}_{3} F_{4} (1, 1, \frac{3}{2}; \frac{5}{3}, 2, 2, \frac{7}{3}; \frac{x^{2}}{4}) + 64 (2 ln x + 3)] \\ - 192_{0} F_{1} (; \frac{2}{3}; \frac{x^{2}}{16})_{2} F_{3} (\frac{1}{3}, \frac{5}{6}; \frac{4}{3}, \frac{4}{3}, \frac{5}{3}; \frac{x^{2}}{4})\} \end{matrix}$
0	$\frac{1}{2}$	$2 [Chi (x) - γ + 2] sinh (\frac{x}{2}) - 2 Shi (x) cosh (\frac{x}{2})$
0	$\frac{2}{3}$	$\begin{matrix} \frac{x^{7 / 6}}{80} \{_{0} F_{1} (; \frac{5}{3}; \frac{x^{2}}{16}) [9 x^{2}_{3} F_{4} (1, 1, \frac{3}{2}; \frac{4}{3}, 2, 2, \frac{8}{3}; \frac{x^{2}}{4}) + 80 ln x + 60] \\ - 60_{0} F_{1} (; \frac{1}{3}; \frac{x^{2}}{16})_{2} F_{3} (\frac{2}{3}, \frac{7}{6}; \frac{5}{3}, \frac{5}{3}, \frac{7}{3}; \frac{x^{2}}{4})\} \end{matrix}$
0	$\frac{3}{4}$	$\begin{matrix} \frac{x^{5 / 4}}{21}_{0} F_{1} (; \frac{7}{4}; \frac{x^{2}}{16}) [3 x^{2}_{3} F_{4} (1, 1, \frac{3}{2}; \frac{5}{4}, 2, 2, \frac{11}{4}; \frac{x^{2}}{4}) + 21 ln x + 14] \\ - \frac{π x^{2}}{4 Γ (\frac{7}{4})} I_{- \frac{3}{4}} (\frac{x}{2})_{2} F_{3} (\frac{3}{4}, \frac{5}{4}; \frac{7}{4}, \frac{7}{4}, \frac{5}{2}; \frac{x^{2}}{4}) \end{matrix}$
0	1	$\begin{matrix} 4 \sqrt{x} \{I_{1} (\frac{x}{2}) [1 - γ + ln 4 - \frac{1}{2 \sqrt{π}} G_{1, 3}^{2, 1} (\frac{x^{2}}{4}; \frac{1}{2}; 0, 0, - 1)] \\ - K_{1} (\frac{x}{2}) [I_{0}^{2} (\frac{x}{2}) - I_{1}^{2} (\frac{x}{2}) - 1]\} \end{matrix}$
0	$\frac{3}{2}$	$\begin{matrix} \frac{4}{x} \{sinh (\frac{x}{2}) [6 γ - 6 Chi (x) - 3 x Shi (x) - 28] \\ + cosh (\frac{x}{2}) [(3 Chi (x) + 8 - 3 γ) x + 6 Shi (x)]\} \end{matrix}$
0	2	$\begin{matrix} 32 \sqrt{x} \{I_{2} (\frac{x}{2}) [\frac{3}{2} - γ + ln 4 - \frac{1}{\sqrt{π}} G_{2, 4}^{3, 1} (\frac{x^{2}}{4}; \frac{1}{2}, 1; 0, 0, 2, - 2)] \\ + K_{2} (\frac{x}{2}) [2_{1} F_{2} (\frac{1}{2}; 1, 3; \frac{x^{2}}{4}) -_{2} F_{3} (\frac{1}{2}, 2; 1, 1, 3; \frac{x^{2}}{4}) - 1]\} \end{matrix}$

Table 5. Integral

I_{1} (κ, μ; x)

for particular values of

κ

and

μ

Table 5. Integral

I_{1} (κ, μ; x)

for particular values of

κ

and

μ

$κ$	$μ$	$I_{1} (κ, μ; x)$
$- \frac{1}{2}$	1	$\frac{1}{x^{2}} \{e^{x} (1 - x) [ln x + γ + Shi (x) - Chi (x)] + ln x + γ - Chi (x) - Shi (x)\}$
$- \frac{1}{2}$	$μ$	$- \frac{\sqrt{π}}{2} Γ (μ) \{\frac{e^{x / 2} x^{1 / 2 - μ}}{μ} [I_{μ - 1 / 2} (\frac{x}{2}) + I_{μ + 1 / 2} (\frac{x}{2})] + \frac{2^{1 - 2 μ}}{Γ (μ + \frac{1}{2})} G^{(1)} (μ + 1; 2 μ + 1; x)\}$
$\frac{1}{2}$	1	$\frac{1}{x^{2}} \{(x + e^{x} + 1) [Chi (x) - ln x - γ] + (x - e^{x} + 1) Shi (x)\}$
$\frac{1}{2}$	$μ$	$\frac{\sqrt{π}}{2} Γ (μ) \{\frac{e^{x / 2} x^{1 / 2 - μ}}{μ} [I_{μ - 1 / 2} (\frac{x}{2}) - I_{μ + 1 / 2} (\frac{x}{2})] - \frac{2^{1 - 2 μ}}{Γ (μ + \frac{1}{2})} G^{(1)} (μ; 2 μ + 1; x)\}$
1	$μ$	$\begin{matrix} Γ (μ - \frac{1}{2}) \{\frac{4 \sqrt{π} μ e^{x / 2} x^{- μ}}{4 μ^{2} - 1} [(2 μ - x + 1) I_{μ} (\frac{x}{2}) + x I_{μ + 1} (\frac{x}{2})] \\ - \frac{Γ (μ + \frac{2}{3})}{Γ (2 μ + 1)} G^{(1)} (μ - \frac{1}{2}; 2 μ + 1; x)\} \end{matrix}$
$κ$	0	$π sec (π κ) [π tan (π κ) L_{κ - 1 / 2} (x) - G^{(1)} (\frac{1}{2} - κ; 1; x)]$
$κ$	$\frac{1}{2}$	$- π csc (π κ) \{[π κ cot (π κ) - 1]_{1} F_{1} (1 - κ; 2; x) + κ G^{(1)} (1 - κ; 2; x)\}$
$κ$	$κ$	$\sqrt{π} \frac{Γ (2 κ + \frac{1}{2})}{Γ (2 κ + 1)} \{[H_{2 κ - 1 / 2} + 2 ln 2]_{1} F_{1} (\frac{1}{2}; 2 κ + 1; x) - G^{(1)} (\frac{1}{2}; 2 κ + 1; x)\}$
$\frac{1}{4}$	$\frac{1}{4}$	$\frac{4 e^{x} ln 2}{\sqrt{x}} F (\sqrt{x}) - 2 G^{(1)} (\frac{1}{2}; \frac{3}{2}; x)$

Table 6. Derivative of

M_{κ, μ}

with respect to

κ

using (72).

Table 6. Derivative of

M_{κ, μ}

with respect to

κ

using (72).

$κ$	$μ$	$\frac{\partial M_{κ, μ} (x)}{\partial κ}$
$- \frac{3}{2}$	2	$\begin{matrix} - \frac{4}{x^{3 / 2}} \{e^{x / 2} [(x^{3} - 3 x^{2} + 6 x - 6) (Shi (x) - Chi (x) + ln x + γ) - \frac{11}{6} x^{3} + \frac{15}{2} x^{2} - 15 x + 11] \\ + e^{- x / 2} [6 (Chi (x) + Shi (x) - ln x - γ) - x^{2} + 4 x - 11]\} \end{matrix}$
$- 1$	$\frac{3}{2}$	$\begin{matrix} \frac{3}{2 x} \{e^{x / 2} [(2 x^{2} - 4 x + 4) (Chi (x) - Shi (x) - ln x - γ) + 3 x^{2} - 8 x + 6] \\ + 2 e^{- x / 2} [2 Chi (x) + 2 Shi (x) + x - 2 ln x - 2 γ - 3]\} \end{matrix}$
$- \frac{1}{2}$	1	$\begin{matrix} \frac{2}{\sqrt{x}} \{e^{x / 2} (x - 1) (Chi (x) - Shi (x) - ln x - γ + 1) \\ + e^{- x / 2} (ln x - Chi (x) - Shi (x) + γ + 1)\} \end{matrix}$
$- \frac{1}{2}$	2	$\begin{matrix} \frac{6}{x^{3 / 2}} \{e^{x / 2} [(x^{2} - 4 x + 6) (2 Chi (x) - 2 Shi (x) - 2 ln x - 2 γ + 3) - 12] \\ + e^{- x / 2} [6 (x - 1) - 4 (x + 3) (ln x - Chi (x) - Shi (x) + γ)]\} \end{matrix}$
0	$\frac{3}{2}$	$\begin{matrix} \frac{6}{x} \{e^{x / 2} [(x - 2) (Chi (x) - Shi (x) - ln x - γ) + x] \\ + e^{- x / 2} [(x + 2) (ln x - Chi (x) - Shi (x) + γ) - x]\} \end{matrix}$
$\frac{1}{2}$	2	$\begin{matrix} \frac{6}{x^{3 / 2}} \{e^{x / 2} [6 (x + 1) - 4 (x - 3) (ln x + Shi (x) - Chi (x) + γ)] \\ + e^{- x / 2} [(x^{2} + 4 x + 6) (2 ln x - 2 Chi (x) - 2 Shi (x) + 2 γ - 3) + 12]\} \end{matrix}$

Table 7. Derivative of

M_{κ, μ}

with respect to

μ

using (95).

Table 7. Derivative of

M_{κ, μ}

with respect to

μ

using (95).

$κ$	$μ$	$\frac{\partial M_{κ, μ} (x)}{\partial μ}$
$- \frac{3}{2}$	2	$\begin{matrix} \frac{4}{x^{3 / 2}} \{e^{x / 2} [(x^{3} - 3 x^{2} + 6 x - 6) (Chi (x) - Shi (x) - γ) + \frac{7}{3} x^{3} - 11 x^{2} + 28 x - 36] \\ + e^{- x / 2} [6 (Chi (x) + Shi (x) - γ) + x^{2} - 4 x + 36]\} \end{matrix}$
$- 1$	$\frac{3}{2}$	$\begin{matrix} \frac{1}{x} \{e^{x / 2} [3 (x^{2} - 2 x + 2) (Chi (x) - Shi (x) - γ) + \frac{13}{2} x^{2} - 22 x + 31] \\ + e^{- x / 2} [3 (x - 2 Chi (x) - 2 Shi (x) + 2 γ) - 31]\} \end{matrix}$
$- \frac{1}{2}$	1	$\begin{matrix} \frac{2}{\sqrt{x}} \{e^{x / 2} [(x - 1) (Chi (x) - Shi (x) - γ + 2) - 2] \\ + e^{- x / 2} (Chi (x) + Shi (x) - γ + 4)\} \end{matrix}$
$- \frac{1}{2}$	2	$\begin{matrix} \frac{8}{x^{3 / 2}} \{e^{x / 2} [3 (\frac{1}{2} x^{2} - 2 x + 3) (Chi (x) - Shi (x) - γ) + 4 x^{2} - 22 x + 48] \\ - e^{- x / 2} [3 (x + 3) (Chi (x) + Shi (x) - γ) + 8 (x + 6)]\} \end{matrix}$
$\frac{1}{2}$	1	$\begin{matrix} \frac{2}{\sqrt{x}} \{e^{x / 2} (Chi (x) - Shi (x) - γ + 4) \\ - e^{- x / 2} [(x + 1) (Chi (x) + Shi (x) - γ + 2) + 2]\} \end{matrix}$
$\frac{1}{2}$	2	$\begin{matrix} \frac{4}{x^{3 / 2}} \{e^{x / 2} [6 (x - 3) (Chi (x) - Shi (x) - γ) + 16 (x - 6)] \\ + e^{- x / 2} [3 (x^{2} + 4 x + 6) (Chi (x) + Shi (x) - γ) + 8 x^{2} + 44 x + 96]\} \end{matrix}$

Table 8. Integral

J_{1} (κ, μ; x)

for particular values of

κ

and

μ

Table 8. Integral

J_{1} (κ, μ; x)

for particular values of

κ

and

μ

$κ$	$μ$	$J_{1} (κ, μ; x)$
$- 1$	0	$π \{2 e^{x / 2} (ln 4 - 2) [(x + 1) I_{0} (\frac{x}{2}) + x I_{1} (\frac{x}{2})] - G^{(1)} (\frac{3}{2}; 1; x) - 2 H^{(1)} (\frac{3}{2}; 1; x)\}$
$- \frac{1}{2}$	1	$x^{- 2} \{e^{x} [(x - 1) (Chi (x) - Shi (x) - ln x - γ) - 2] + Chi (x) + Shi (x) - ln x - γ + 2\}$
$- \frac{1}{3}$	0	$2 π \{G^{(1)} (\frac{5}{6}; 1; x) + 2 H^{(1)} (\frac{5}{6}; 1; x) - ln (432) L_{- 5 / 6} (x)\}$
0	0	$- π e^{x / 2} \{K_{0} (\frac{\|x\|}{2}) + [ln (4 \|x\|) + γ] I_{0} (\frac{\|x\|}{2})\}$
0	$\frac{1}{2}$	$x^{- 1} \{e^{x} [Chi (x) - Shi (x) - ln x - γ] - Chi (x) - Shi (x) + ln x + γ\}$
0	1	$\begin{matrix} \{I_{1} (\frac{\|x\|}{2}) [I_{1} (\frac{\|x\|}{2}) K_{1} (\frac{\|x\|}{2}) - ln (4 \|x\|) - γ + 2 - \frac{1}{2 \sqrt{π}} G_{1, 3}^{2, 1} (\frac{x^{2}}{4}; 1 / 2; 0, 0, - 1)] \\ + K_{1} (\frac{\|x\|}{2}) [1 - I_{0}^{2} (\frac{\|x\|}{2})]\} \frac{π}{2 \|x\|} e^{x / 2} \end{matrix}$
$\frac{1}{3}$	0	$2 π \{G^{(1)} (\frac{1}{6}; 1; x) + 2 H^{(1)} (\frac{1}{6}; 1; x) - ln (432) L_{- 1 / 6} (x)\}$
$\frac{1}{2}$	$\frac{1}{2}$	$\frac{π}{2} \{G^{(1)} (\frac{1}{2}; 2; x) + 2 H^{(1)} (\frac{1}{2}; 2; x) - 2 e^{x / 2} ln 4 [I_{0} (\frac{x}{2}) - I_{1} (\frac{x}{2})]\}$
$\frac{1}{2}$	1	$x^{- 2} \{e^{x} [Chi (x) - Shi (x) - ln x - γ + 2] - (x + 1) [Chi (x) + Shi (x) - ln x - γ] - 2\}$

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Apelblat, A.; González-Santander, J.L. Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I. Axioms 2023, 12, 381. https://doi.org/10.3390/axioms12040381

AMA Style

Apelblat A, González-Santander JL. Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I. Axioms. 2023; 12(4):381. https://doi.org/10.3390/axioms12040381

Chicago/Turabian Style

Apelblat, Alexander, and Juan Luis González-Santander. 2023. "Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I" Axioms 12, no. 4: 381. https://doi.org/10.3390/axioms12040381

APA Style

Apelblat, A., & González-Santander, J. L. (2023). Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I. Axioms, 12(4), 381. https://doi.org/10.3390/axioms12040381

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Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I

Abstract

1. Introduction

2. Parameter Differentiation of $M_{κ, μ}$ via Kummer Function $_{1} F_{1}$

2.1. Derivative with Respect to the First Parameter $\partial M_{κ, μ} (x) / \partial κ$

2.2. Derivative with Respect to the Second Parameter $\partial M_{κ, μ} (x) / \partial μ$

3. Parameter Differentiation of $M_{κ, μ}$ via Integral Representations

3.1. Derivative with Respect to the First Parameter $\partial M_{κ, μ} (x) / \partial κ$

3.2. Application to the Calculation of Infinite Integrals

3.3. Derivative with Respect to the Second Parameter $\partial M_{κ, μ} (x) / \partial μ$

3.4. Application to the Calculation of Finite Integrals

4. Conclusions

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Acknowledgments

Conflicts of Interest

Appendix A. Parameter Differentiation of the Incomplete Gamma Functions

Appendix B. Reduction Formulas for Integral Whittaker Functions Mi_κ,μ and mi_κ,μ

Appendix C. Reduction Formulas for the Whittaker Function M_κ,μ(x)

References

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Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker Mκ,μ(x) Function I

Abstract

1. Introduction

2. Parameter Differentiation of M κ , μ via Kummer Function 1 F 1

2.1. Derivative with Respect to the First Parameter ∂ M κ , μ x / ∂ κ

2.2. Derivative with Respect to the Second Parameter ∂ M κ , μ x / ∂ μ

3. Parameter Differentiation of M κ , μ via Integral Representations

3.1. Derivative with Respect to the First Parameter ∂ M κ , μ x / ∂ κ

3.2. Application to the Calculation of Infinite Integrals

3.3. Derivative with Respect to the Second Parameter ∂ M κ , μ x / ∂ μ

3.4. Application to the Calculation of Finite Integrals

4. Conclusions

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Acknowledgments

Conflicts of Interest

Appendix A. Parameter Differentiation of the Incomplete Gamma Functions

Appendix B. Reduction Formulas for Integral Whittaker Functions Miκ,μ and miκ,μ

Appendix C. Reduction Formulas for the Whittaker Function Mκ,μ(x)

References

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Infinite Series and Logarithmic Integrals Associated to Differentiation with Respect to Parameters of the Whittaker M_κ_,_μ(x) Function I

2. Parameter Differentiation of $M_{κ, μ}$ via Kummer Function $_{1} F_{1}$

2.1. Derivative with Respect to the First Parameter $\partial M_{κ, μ} (x) / \partial κ$

2.2. Derivative with Respect to the Second Parameter $\partial M_{κ, μ} (x) / \partial μ$

3. Parameter Differentiation of $M_{κ, μ}$ via Integral Representations

3.1. Derivative with Respect to the First Parameter $\partial M_{κ, μ} (x) / \partial κ$

3.3. Derivative with Respect to the Second Parameter $\partial M_{κ, μ} (x) / \partial μ$

Appendix B. Reduction Formulas for Integral Whittaker Functions Mi_κ,μ and mi_κ,μ

Appendix C. Reduction Formulas for the Whittaker Function M_κ,μ(x)