Degenerate exponential integral function and its properties

Nantomah, Kwara

doi:10.1108/AJMS-09-2021-0230

Purpose

In this paper, the author introduces a degenerate exponential integral function and further studies some of its analytical properties. The new function is a generalization of the classical exponential integral function and the properties established are analogous to those satisfied by the classical function.

Design/methodology/approach

The methods adopted in establishing the results are theoretical in nature.

Findings

A degenerate exponential integral function which is a generalization of the classical exponential integral function has been introduced and its properties investigated. Upon taking some limits, the established results reduce to results involving the classical exponential integral function.

Originality/value

The results obtained in this paper are new and have the potential of inspiring further research on the subject.

1. Introduction

The gamma function, also known as the Euler’s integral of second kind, is one of the most studied special functions. This is partly because of its numerous applications and its connection with other special functions. It is usually defined as

Γ (z) = \int_{0}^{\infty} t^{z - 1} e^{- t} d t

for z > 0. The upper incomplete gamma function is defined as

Γ (z, s) = \int_{s}^{\infty} t^{z - 1} e^{- t} d t

(1)

for s > 0 and z $\in$ (−∞, ∞) whilst the lower incomplete gamma function is defined as

γ (z, s) = \int_{0}^{s} t^{z - 1} e^{- t} d t

(2)

for z > 0.

The classical exponential integral function is defined by any of the following equivalent definitions [1, p. 228]

E (z) = \int_{z}^{\infty} \frac{e^{- t}}{t} d t,

(3)

= \int_{1}^{\infty} \frac{e^{- z t}}{t} d t,

(4)

= Γ (0, z),

(5)

= - E i (- z),

(6)

for z > 0 where the function Ei(z) is defined as

E i (z) = \int_{- \infty}^{z} \frac{e^{t}}{t} d t

(7)

for z ≠ 0. For z > 0, Ei(z) is interpreted as the Cauchy’s principal value of the integral (7) due to the singularity at zero. In some texts, Ei(z) is what is referred to as the exponential integral function. A generalized form of the function is defined as [2].

E_{k} (z) = z^{k - 1} \int_{z}^{\infty} \frac{e^{- t}}{t^{k}} d t

(8)

= \int_{1}^{\infty} \frac{e^{- z t}}{t^{k}} d t

(9)

= z^{k - 1} Γ (1 - k, z)

(10)

for z > 0 and k $\in$ (−∞, ∞). In some texts, the function E₁(z) is referred to as the Theis' well function [3]. The exponential integral function has many applications in areas such as transient groundwater flow, hydrological problems, mathematical physics, engineering, quantum mechanics and applied mathematics. Due to its practical importance, it has been studied in diverse ways and some generalizations given. For example see Refs. [3–15].

Among other things, Kim et al. [16] defined the modified degenerate gamma function as

Γ_{λ} (z) = \int_{0}^{\infty} t^{z - 1} {(1 + λ)}^{- \frac{t}{λ}} d t

(11)

where λ > 0 and $R (z) > 0$ ⁠. Motivated by this definition, the goal of this paper is to introduce a degenerate exponetial integral function and further study some of its analytical properties. This new function, which is a generalization of the classical exponential integral function may also be called, λ-analogue of the exponential integral function.

2. Generalized degenerate exponential integral function

In this section, the author defines the degenerate exponential function and further studies some of its properties. The author begins with the following auxiliary definition.

Definition 2.1.

For λ > 0, the author defines the upper incomplete degenerate gamma function as

Γ_{λ} (z, s) = \int_{s}^{\infty} t^{z - 1} {(1 + λ)}^{- \frac{t}{λ}} d t

(12)

where s > 0 and z

\in

(−∞, ∞) and the lower incomplete degenerate gamma function as

γ_{λ} (z, s) = \int_{0}^{s} t^{z - 1} {(1 + λ)}^{- \frac{t}{λ}} d t

(13)

where z > 0. Clearly,

γ_{λ} (z, s) + Γ_{λ} (z, s) = Γ_{λ} (z)

(14)

for z > 0 and as λ → 0, respectively return to .

Definition 2.2.

The generalized degenerate exponential integral function is defined as

E_{λ, k} (z) = \int_{1}^{\infty} \frac{{(λ + 1)}^{- \frac{z t}{λ}}}{t^{k}} d t,

(15)

= \int_{0}^{1} \frac{{(λ + 1)}^{- \frac{z}{λ t}}}{t^{2 - k}} d t,

(16)

= z^{k - 1} Γ_{λ} (1 - k, z),

(17)

= {(\frac{z \ln (λ + 1)}{λ})}^{k - 1} Γ (1 - k, \frac{z \ln (λ + 1)}{λ}),

(18)

for z > 0, λ > 0 and k

\in

(−∞, ∞), and satisfies the commutative diagram

Figure. Refer to the image caption for details.

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By this definition, the following identities are easily deduced.

E_{λ, k} (z) = E_{k} (\frac{z \ln (λ + 1)}{λ}),

(19)

Γ_{λ} (1 - k, z) = {(\frac{\ln (λ + 1)}{λ})}^{k - 1} Γ (1 - k, \frac{z \ln (λ + 1)}{λ}) .

(20)

It follows from (15) that

E_{λ, 0} (z) = {(\frac{z \ln (λ + 1)}{λ})}^{- 1} {(λ + 1)}^{- \frac{z}{λ}} .

(21)

Also, differentiating (15) for r number of times gives

E_{λ, k}^{(r)} (z) = {(- 1)}^{r} {(\frac{\ln (λ + 1)}{λ})}^{r} \int_{1}^{\infty} t^{r - k} {(λ + 1)}^{- \frac{z t}{λ}} d t

(22)

= {(- 1)}^{r} {(\frac{\ln (λ + 1)}{λ})}^{r} E_{λ, k - r} (z)

(23)

and in particular,

E_{λ, k}^{'} (z) = - (\frac{\ln (λ + 1)}{λ}) E_{λ, k - 1} (z) .

(24)

Additionally, it is deduced from (15) that

E_{λ, k} (z) > E_{λ, k + 1} (z)

(25)

and for k = 0, we obtain

E_{λ, 1} (z) < E_{λ, 0} (z) = {(\frac{z \ln (λ + 1)}{λ})}^{- 1} {(λ + 1)}^{- \frac{z}{λ}} .

(26)

Moreover, it follows from (22) that E_λ,k(z) is completely monotonic [17] and hence log-convex and decreasing.

Theorem 2.3.

The following identities are satisfied for z > 0, λ > 0 and k > 0.

k E_{λ, k + 1} (z) = {(λ + 1)}^{- \frac{z}{λ}} - \frac{z \ln (λ + 1)}{λ} E_{λ, k} (z),

(27)

= {(λ + 1)}^{- \frac{z}{λ}} + z E_{λ, k + 1}^{'} (z)

(28)

Proof.

The author employs the integration by parts technique. Let $u = {(λ + 1)}^{- \frac{z t}{λ}}$ such that $d u = - \frac{z \ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z t}{λ}} d t$ and dv = t^−k−1 such that $v = - \frac{t^{- k}}{k}$ ⁠. Then

\begin{array}{l} E_{λ, k + 1} (z) & = \int_{1}^{\infty} u d v = u v |_{1}^{\infty} - \int_{1}^{\infty} v d u \\ = \frac{{(λ + 1)}^{- \frac{z}{λ}}}{k} - \frac{z \ln (λ + 1)}{λ k} \int_{1}^{\infty} t^{- k} {(λ + 1)}^{- \frac{z t}{λ}} d t \\ = \frac{{(λ + 1)}^{- \frac{z}{λ}}}{k} - \frac{z \ln (λ + 1)}{λ k} E_{λ, k} (z) \end{array}

which gives . The identity follows directly from by applying . □

Theorem 2.4.

The double inequality

{(k + \frac{z \ln (λ + 1)}{λ})}^{- 1} {(λ + 1)}^{- \frac{z}{λ}} < E_{λ, k} (z) < {(k - 1 + \frac{z \ln (λ + 1)}{λ})}^{- 1} {(λ + 1)}^{- \frac{z}{λ}}

(29)

holds forz > 0,λ > 0 andk ≥ 1.

Proof.

By using (25) and (27), the author obtains

{(λ + 1)}^{- \frac{z}{λ}} - \frac{z \ln (λ + 1)}{λ} E_{λ, k} (z) = k E_{λ, k + 1} (z) < k E_{λ, k} (z)

which simplifies to give the left-hand side of . Likewise, , respectively imply that

- E_{λ, k - 1} (z) < - E_{λ, k} (z)

(30)

and

(k - 1) E_{λ, k} (z) = {(λ + 1)}^{- \frac{z}{λ}} - \frac{z \ln (λ + 1)}{λ} E_{λ, k - 1} (z) .

(31)

Now (30) and (31) imply that

(k - 1) E_{λ, k} (z) < {(λ + 1)}^{- \frac{z}{λ}} - \frac{z \ln (λ + 1)}{λ} E_{λ, k} (z)

which simplifies to give the right-hand side of .□

Theorem 2.5.

The inequality

k E_{λ, k + 1} (z) > (k - 1) E_{λ, k} (z)

(32)

holds forz > 0,λ > 0 andk ≥ 1.

Proof.

By using (27) and (30), the author obtains

\begin{array}{l} k E_{λ, k + 1} (z) & = {(λ + 1)}^{- \frac{z}{λ}} - \frac{z \ln (λ + 1)}{λ} E_{λ, k} (z) \\ > {(λ + 1)}^{- \frac{z}{λ}} - \frac{z \ln (λ + 1)}{λ} E_{λ, k - 1} (z) \\ = (k - 1) E_{λ, k} (z) \end{array}

which yields the desired result.□

Theorem 2.6.

Let r > 0, s > 0, a₁ > 1 and $\frac{1}{a_{1}} + \frac{1}{a_{2}} = 1$ ⁠. Then the inequality

E_{λ, \frac{r}{a_{1}} + \frac{s}{a_{2}}} (\frac{x}{a_{1}} + \frac{z}{a_{2}}) \leq {[E_{λ, r} (x)]}^{\frac{1}{a_{1}}} {[E_{λ, s} (z)]}^{\frac{1}{a_{2}}}

(33)

holds forx > 0,z > 0 andλ > 0.

Proof.

By using Holder’s integral inequality, the author obtains

\begin{array}{l} E_{λ, \frac{r}{a_{1}} + \frac{s}{a_{2}}} (\frac{x}{a_{1}} + \frac{z}{a_{2}}) & = \int_{1}^{\infty} \frac{{(λ + 1)}^{- \frac{x t}{λ a_{1}}}}{t^{\frac{r}{a_{1}}}} \frac{{(λ + 1)}^{- \frac{z t}{λ a_{2}}}}{t^{\frac{s}{a_{2}}}} d t \\ \leq {(\int_{1}^{\infty} \frac{{(λ + 1)}^{- \frac{x t}{λ}}}{t^{r}} d t)}^{\frac{1}{a_{1}}} {(\int_{1}^{\infty} \frac{{(λ + 1)}^{- \frac{z t}{λ}}}{t^{s}} d t)}^{\frac{1}{a_{2}}} \\ = {[E_{λ, r} (x)]}^{\frac{1}{a_{1}}} {[E_{λ, s} (z)]}^{\frac{1}{a_{2}}} \end{array}

which completes the proof.□

Remark 2.7.

Let a₁ = a₂ = 2, x = z, r = k + 1 and s = k − 1. Then inequality (33) reduces to the Turan-type inequality

E_{λ, k}^{2} (z) \leq E_{λ, k + 1} (z) E_{λ, k - 1} (z) .

(34)

If a₁ = a₂ = 2 and x = z, then (33) reduces to

E_{λ, \frac{r + s}{2}}^{2} (z) \leq E_{λ, r} (z) E_{λ, s} (z) .

(35)

Theorem 2.8.

For k > 0 and λ > 0, the function

B (z) = \frac{E_{λ, k + 1} (z)}{E_{λ, k} (z)}

(36)

is increasing on (0, ∞).

Proof.

Let z $\in$ (0, ∞). Then by using the (24) and (34), the author obtains

\begin{array}{l} E_{λ, k}^{2} (z) B^{'} (z) & = E_{λ, k + 1}^{'} (z) E_{λ, k} (z) - E_{λ, k + 1} (z) E_{λ, k}^{'} (z) \\ = (\frac{\ln (λ + 1)}{λ}) [E_{λ, k + 1} (z) E_{λ, k - 1} (z) - E_{λ, k}^{2} (z)] \geq 0 \end{array}

which completes the proof.□

The following Lemma 2.9 is a generalization of Lemma 2.1 of [9].

Lemma 2.9.

For k ≥ 1 and λ > 0, the function $P (z) = \frac{z E_{λ, k}^{'} (z)}{E_{λ, k}^{2} (z)}$ is strictly decreasing on (0, ∞).

Proof.

By using (18) along with the decreasing property of E_λ,k(z), we have

E_{λ, k}^{'} (z) = \frac{(k - 1) E_{λ, k} (z) - {(λ + 1)}^{- \frac{z}{λ}}}{z} < 0

which shows that

(k - 1) E_{λ, k} (z) < {(λ + 1)}^{- \frac{z}{λ}} .

(37)

Then

P (z) = \frac{(k - 1) E_{λ, k} (z) - {(λ + 1)}^{- \frac{z}{λ}}}{E_{λ, k}^{2} (z)}

and by differentiating and making use of , we obtain

\begin{array}{l} E_{λ, k}^{3} (z) P^{'} (z) & = [(k - 1) E_{λ, k}^{'} (z) + \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}}] E_{λ, k} (z) \\ - 2 [(k - 1) E_{λ, k} (z) - {(λ + 1)}^{- \frac{z}{λ}}] E_{λ, k}^{'} (z) \\ = - (k - 1) E_{λ, k}^{'} (z) E_{λ, k} (z) + \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k} (z) \\ + 2 {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k}^{'} (z) \\ < - {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k}^{'} (z) + \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k} (z) \\ + 2 {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k}^{'} (z) \\ = {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k}^{'} (z) + \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k} (z) \\ = - \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k - 1} (z) + \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}} E_{λ, k} (z) \\ = \frac{\ln (λ + 1)}{λ} {(λ + 1)}^{- \frac{z}{λ}} [E_{λ, k} (z) - E_{λ, k - 1} (z)] \\ < 0 . \end{array}

Thus, $P^{'} (z) < 0$ which completes the proof.□

Theorem 2.10.

For z > 0, k ≥ 1 and λ > 0, the inequality

\frac{2 E_{λ, k} (z) E_{λ, k} (1 / z)}{E_{λ, k} (z) + E_{λ, k} (1 / z)} \leq {(\frac{\ln (λ + 1)}{λ})}^{k - 1} Γ (1 - k, \frac{\ln (λ + 1)}{λ})

(38)

holds, and with equality ifz = 1.

Proof.

The case for z = 1 is easy to see. So let $Q (z) = \frac{2 E_{λ, k} (z) E_{λ, k} (1 / z)}{E_{λ, k} (z) + E_{λ, k} (1 / z)}$ for z $\in$ (0, 1) ∪ (1, ∞). Then direct computations results to

z \frac{Q^{'} (z)}{Q (z)} = z \frac{E_{λ, k}^{'} (z)}{E_{λ, k} (z)} - \frac{1}{z} \frac{E_{λ, k}^{'} (1 / z)}{E_{λ, k} (1 / z)} - \frac{z E_{λ, k}^{'} (z) - \frac{1}{z} E_{λ, k}^{'} (1 / z)}{E_{λ, k} (z) + E_{λ, k} (1 / z)}

which implies that

z [E_{λ, k} (z) + E_{λ, k} (1 / z)] \frac{Q^{'} (z)}{Q (z)} = z \frac{E_{λ, k}^{'} (z)}{E_{λ, k} (z)} E_{λ, k} (1 / z) - \frac{1}{z} \frac{E_{λ, k}^{'} (1 / z)}{E_{λ, k} (1 / z)} E_{λ, k} (z)

and this further implies that

\begin{array}{l} z [\frac{1}{E_{λ, k} (z)} + \frac{1}{E_{λ, k} (1 / z)}] \frac{Q^{'} (z)}{Q (z)} & = z \frac{E_{λ, k}^{'} (z)}{E_{λ, k}^{2} (z)} - \frac{1}{z} \frac{E_{λ, k}^{'} (1 / z)}{E_{λ, k}^{2} (1 / z)} \\ = K (z) \end{array}

Taking into account of Lemma 2.9, the author concludes that $K (z) > 0$ if z $\in$ (0, 1) and $K (z) < 0$ if z $\in$ (1, ∞). Accordingly, $Q (z)$ is increasing on (0, 1) and decreasing on (1, ∞). Therefore, for each of the cases, we have

Q (z) < \lim_{z \to 1} Q (z) = E_{λ, k} (1) = {(\frac{\ln (λ + 1)}{λ})}^{k - 1} Γ (1 - k, \frac{\ln (λ + 1)}{λ})

yielding the desired result.□

Remark 2.11.

Theorem 2.10 provides a far-reaching generalization of the results of the papers [8, 9].

In what follows, the author provides some results for the particular case where k = 1.

Remark 2.12.

The following identities hold.

E_{λ}^{'} (z) = - \frac{{(λ + 1)}^{- \frac{z}{λ}}}{z},

(39)

{E_{λ}}^{″} (z) = [\frac{1}{z} + \frac{\ln (λ + 1)}{λ}] \frac{{(λ + 1)}^{- \frac{z}{λ}}}{z},

(40)

= - [\frac{1}{z} + \frac{\ln (λ + 1)}{λ}] E_{λ}^{'} (z),

(41)

{E_{λ}}^{‴} (z) = - [{(\frac{z \ln (λ + 1)}{λ})}^{2} + \frac{2 z \ln (λ + 1)}{λ} + 2] \frac{{(λ + 1)}^{- \frac{z}{λ}}}{z^{3}},

(42)

= [{(\frac{\ln (λ + 1)}{λ})}^{2} + \frac{2 \ln (λ + 1)}{λ z} + \frac{2}{z^{2}}] E_{λ}^{'} (z)

(43)

Theorem 2.13 and Theorem 2.14 below, respectively generalizes Lemma 1 and Lemma 2 of [8].

Theorem 2.13.

For z > 0 and λ > 0, the inequality

E_{λ} (z) + E_{λ} (1 / z) \geq 2 Γ (0, \frac{\ln (λ + 1)}{λ})

(44)

holds, with equality whenz = 1.

Proof.

The case for z = 1 is easy to see. So let $U (s) = E_{λ} (z) + E_{λ} (1 / z)$ for z $\in$ (0, 1) ∪ (1, ∞). Then by differentiating and applying (39), the author obtains

\begin{array}{l} z U^{'} (z) & = z E_{λ}^{'} (z) - \frac{1}{z} E_{λ}^{'} (1 / z) \\ = {(λ + 1)}^{- \frac{1}{λ z}} - {(λ + 1)}^{- \frac{z}{λ}} \\ = f (z). \end{array}

If z $\in$ (0, 1) then f(z) < 0 and if z $\in$ (1, ∞) then f(z) > 0. It follows that $U (z)$ is decreasing on (0, 1) and increasing on (1, ∞). For each of the cases, the author has

U (z) > \lim_{z \to 1} U (z) = 2 E_{λ} (1) = 2 Γ (0, \frac{\ln (λ + 1)}{λ})

which completes the proof.□

Theorem 2.14.

For z > 0 and λ > 0, the inequality

E_{λ} (z) E_{λ} (1 / z) \leq Γ^{2} (0, \frac{\ln (λ + 1)}{λ})

(45)

holds, with equality whenz = 1.

Proof.

The case for z = 1 is easy to see. Hence let $W (s) = E_{λ} (z) E_{λ} (1 / z)$ for z $\in$ (0, 1) ∪ (1, ∞). Then

\begin{array}{l} z W^{'} (s) & = z E_{λ}^{'} (z) E_{λ} (1 / z) - \frac{1}{z} E_{λ} (z) E_{λ}^{'} (1 / z) \\ = {(λ + 1)}^{- \frac{1}{λ z}} E_{λ} (z) - {(λ + 1)}^{- \frac{z}{λ}} E_{λ} (1 / z) \end{array}

which implies that

\begin{array}{l} z {(λ + 1)}^{\frac{z}{λ}} {(λ + 1)}^{\frac{1}{λ z}} W^{'} (s) & = {(λ + 1)}^{\frac{z}{λ}} E_{λ} (z) - {(λ + 1)}^{\frac{1}{λ z}} E_{λ} (1 / z) \\ = β (z) \end{array}

Let $g (z) = {(λ + 1)}^{\frac{z}{λ}} E_{λ} (z)$ ⁠. Then by using (26), the authors arrives at

{(λ + 1)}^{- \frac{z}{λ}} g^{'} (z) = \frac{\ln (λ + 1)}{λ} E_{λ} (z) - \frac{{(λ + 1)}^{- \frac{z}{λ}}}{z} < 0

which implies that g(z) is decreasing on (0, ∞). If z $\in$ (0, 1) then β(z) > 0 and if z $\in$ (1, ∞) then β(z) < 0. It then follows that, $W (z)$ is increasing on (0, 1) and decreasing on (1, ∞). For each of the cases, the author has

W (z) < \lim_{z \to 1} W (z) = E_{λ}^{2} (1) = Γ^{2} (0, \frac{\ln (λ + 1)}{λ})

which completes the proof.□

Theorem 2.15.

Let λ > 0, u > 0, v > 0 such that u ≠ v. Then

\frac{v - u}{v} {(λ + 1)}^{- \frac{v}{λ}} < E_{λ} (u) - E_{λ} (v) < \frac{v - u}{u} {(λ + 1)}^{- \frac{u}{λ}}

(46)

holds.

Proof.

With no loss of generality, let u < v. Then by the classical mean value theorem, there exist μ $\in$ (u, v) such that

\frac{E_{λ} (u) - E_{λ} (v)}{u - v} = E_{λ}^{'} (z)

Since $E_{λ}^{'} (y)$ is increasing for y > 0, it follows that

E_{λ}^{'} (u) < \frac{E_{λ} (u) - E_{λ} (v)}{u - v} < E_{λ}^{'} (v)

and by using (39), the author obtains the inequality (46).□

Remark 2.16.

Upon letting u = z and v = z + 1 in (46), the author obtains

\frac{{(λ + 1)}^{- \frac{z + 1}{λ}}}{z + 1} < E_{λ} (z) - E_{λ} (z + 1) < \frac{{(λ + 1)}^{- \frac{z}{λ}}}{z} .

(47)

The author wishes to thank the anonymous reviewers for carefully reading the paper and for their comments and suggestions, which helped to improve the quality of this paper.

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Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this license may be seen at http://creativecommons.org/licences/by/4.0/legalcode

Degenerate exponential integral function and its properties

1. Introduction

2. Generalized degenerate exponential integral function

References

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1. Introduction

2. Generalized degenerate exponential integral function

References

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Degenerate exponential integral function and its properties