Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 18 (1999), No. 3, 793–799

Some Inequalities of the Incomplete Gamma and Related Functions

Feng Qi and Jia-Qiang Mei

Dedicated to our advisor Prof. Sen-Lin Xu at USTC

Abstract. In the article, many inequalities of the integrals

∫_{x}^{∞} e^{−t^{p}}dt,  ∫_{0}^{x} e^{−t^{p}}dt,  ∫_{0}^{x} e^{t^{p}}dt

for p > 0, which are related to the incomplete gamma function, are established. The approach used in the paper could yield more particular inequalities of the above functions. Some known results are generalized, extended or refined.

Keywords: Inequalities, incomplete gamma function, complementary error function

AMS subject classification: Primary 33 B 20, secondary 26 D 07, 26 D 15

1. Introduction

It is well-known that the incomplete gamma function is defined for Re z > 0 by

Γ(z, x) = ∫_{x}^{∞}t^{z−1}e^{−t}dt,  γ(z, x) = ∫_{0}^{x} t^{z−1}e^{−t}dt,  (1)

and Γ(z, 0) = Γ(z) is called the gamma function, Γ(0, x) = E_{1}(x) the exponential integral. Notice that the integral ∫_{x}^{∞} e^{−t^{p}}dt can be expressed in terms of the incomplete gamma function as

∫_{x}^{∞}e^{−t^{p}}dt = \frac{1}{p}Γ(\frac{1}{p}, x^{p})  (p > 0).  (2)

Thus it is important and interesting to study the function ∫_{x}^{∞} e^{−t^{p}}dt and the related functions ∫_{0}^{x} e^{−t^{p}}dt and ∫_{0}^{x} e^{t^{p}}dt for p > 0.

Feng Qi: Jiaozuo Inst. Techn., Dept. Math., Jiaozuo City, Henan 454000, China; e-mail: qifeng@jzit.edu.cn. The author was partially supported by NSF of Henan Province, China Jia-Qiang Mei: Univ. Sci. & Techn. of China, Dept. Math., Hefei City, Anhui 230026, China

ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

In [2], A. Laforgia and S. Sismondi proved some monotonicity properties for the complementary error function

erfc (x) = \frac{2}{\sqrt{π}} ∫_{x}^{∞}e^{−t^{2}}dt (x > 0)

and as a consequence of these they established many inequalities for this function. Using the monotonicity and inequalities of the generalized weighted mean values with two parameters and the extended mean values defined and studied in [4, 8, 10], the first author [5] found some monotonicity results and inequalities for the gamma and incomplete gamma functions. Among other things, it is proved that the functions

(\frac{Γ(s)}{Γ(r)})^{\frac{1}{s−r}} ,  (\frac{Γ(s, x)}{Γ(r, x)})^{\frac{1}{s−r}} ,  (\frac{γ(s, x)}{γ(r, x)})^{\frac{1}{s−r}}

are increasing respect to r > 0, s > 0 and x > 0. Notice that the monotonicity of the function [\frac{Γ(s)}{Γ(r)}]^{1/(s−r)} respect to r > 0 and s > 0 is an old and well-known consequence of the logarithmical convexity of the gamma function Γ. By the Tchebycheff integral inequality, the first author, L.-H. Cui and S.-L. Xu in [7], inter alia, constructed many inequalities of the error function

erf (x) = \frac{2}{\sqrt{π}} ∫_{0}^{x} e^{−t^{2}}dt

and got the lower bound of the function ∫_{0}^{x} e^{−t^{α}}dt and the upper bound of the function ∫_{0}^{x} e^{t^{α}}dt for x > 0 and α > 1. From the Hermite-Hadamard inequality, considering the convexity of the function e^{t^{p}} and e^{−t^{p}}, the first author and S.-L. Guo in [9] obtained some inequalities of the functions ∫_{0}^{x} e^{−t^{p}}dt and ∫_{0}^{x} e^{t^{p}}dt for x > 0 and p > 0. More other inequalities for the gamma and incomplete gamma functions could be found in [1, 3].

In this article, motivated by [2] and [9], we further researched the integrals

∫_{x}^{∞} e^{−t^{α}}dt (α > 0),  ∫_{0}^{x} e^{t^{β}}dt (β > 0),  ∫_{0}^{x} e^{−t^{θ}}dt (θ > 0)

for x > 0, present some monotonicity properties that generalize one of the main results in [2] by A. Laforgia and S. Sismondi, and form some particular inequalities of the above integrals.

Some Inequalities 795

2. Main theorems

The following three theorems are elementary and important and they could yield more special results.

Theorem 1. For any given α > 0, let h_{α}(x) be defined for x > 0 by

h_{α}(x) = e^{x^{α}} ∫_{x}^{∞}e^{−t^{α}}dt + p_{α}(x)  (3)

where lim_{x→∞} e^{−x^{α}}p_{α}(x) = 0 and p_{α} satisfies

p′_{α}(x) − αx^{α−1}p_{α}(x) − 1 > 0.  (4)

Then h_{α}(x) < 0. If (4) is reversed, then h_{α}(x) > 0.

Proof. By direct calculation, we have

h′_{α}(x) = αx^{α−1}h_{α}(x) − αx^{α−1}p_{α}(x) + p′_{α}(x) − 1

and

e^{x^{α}}[e^{−x^{α}}h_{α}(x)]′ = h′_{α}(x) − αx^{α−1}h_{α}(x)
= p′_{α}(x) − αx^{α−1}p_{α}(x) − 1
> 0.

From this it follows that e^{−x^{α}}h_{α}(x) increases. Since lim_{x→∞} e^{−x^{α}}p_{α}(x) = 0, we have lim_{x→∞} e^{−x^{α}}h_{α}(x) = 0, hence h_{α}(x) < 0. The proof of Theorem 1 is completed

Theorerm 2. Let g_{β}(x) be defined for x > 0 by

g_{β}(x) = e^{−x^{β}} ∫_{0}^{x} e^{t^{β}}dt + q_{β}(x)  (5)

where lim_{x→0} q_{β}(x) = 0 and β > 0 is any given number. If q_{β} satisfies

q′_{β}(x) + βx^{β−1}q_{β}(x) + 1 > 0,  (6)

then g_{β}(x) > 0. If (6) is reversed, then g_{β}(x) < 0.

Proof. Easy calculation yields

g′_{β}(x) = −βx^{β−1}g_{β}(x) + βx^{β−1}q_{β}(x) + q′_{β}(x) + 1

and

\begin{aligned}[e^{x^{β}}g_{β}(x)]′ & = e^{x^{β}}[g′_{β}(x) + βx^{β−1}g_{β}(x)] \\ & = e^{x^{β}}[q′_{β}(x) + βx^{β−1}q_{β}(x) + 1]. \end{aligned}

Therefore, if inequality (6) is valid, the function e^{x^{β}}g_{β}(x) is increasing. This implies Theorem 2

796 Feng Qi and Jia-Qiang Mei

Similar arguments lead to the following

Theorem 3. Define f_{θ}(x) as

f_{θ}(x) = e^{x^{θ}} ∫_{0}^{x} e^{−t^{θ}}dt + r_{θ}(x)  (θ > 0)  (7)

where lim_{x→0} r_{θ}(x) = 0 and

r′_{θ}(x) − θx^{θ−1}r_{θ}(x) + 1 > 0  (8)

for x > 0. Then f_{θ}(x) > 0. If (8) is reversed, then f_{θ}(x) < 0.

3. Particular inequalities

3.1 Set p_{α}(x) = Ax^{1−α} (A < 0). Then

\begin{matrix}P_{α}(x) := & p′_{α}(x) − αx^{α−1}p_{α}(x) − 1 \\ = & (1 − α)Ax^{−α} − (1 + αA). \end{matrix}

For α > 1, if A < −\frac{1}{α}, we get P_{α}(x) > 0, thus h_{α}(x) < 0. Hence, for x > 0 and α > 1, the inequality

∫_{x}^{∞}e^{−t^{α}}dt < \frac{e^{−x^{α}}}{α x^{α−1}}  (9)

holds.

For 0 < α < 1, if A > −\frac{1}{α}, we have P_{α}(x) < 0 and h_{α}(x) > 0. Therefore, for x > 0 and 0 < α < 1, inequality (9) is reversed.

For α > 1, if 0 > A > −\frac{1}{α}, when x < (\frac{(1−α)A}{1+αA})^{\frac{1}{α}}, we obtain

∫_{x}^{∞} e^{−t^{α}}dt < −A\frac{e^{−x^{α}}}{x^{α−1}} ;  (10)

when x > (\frac{(1−α)A}{1+αA})^{\frac{1}{α}}, inequality (10) reverses.

For 0 < α < 1, if A < −\frac{1}{α}, when x < (\frac{(1−α)A}{1+αA})^{\frac{1}{α}}, inequality (10) is reversed; when x > (\frac{(1−α)A}{1+αA})^{\frac{1}{α}}, inequality (10) holds. 3.2 Set p_{α}(x) = \frac{C}{(x+B)^{α−1}} (B > 0, C < 0). Then

\begin{matrix}P_{α}(x) := & p′_{α}(x) − αx^{α−1}p_{α}(x) − 1 \\ = & \frac{(1 − α)C − αCx^{α−1}(x + B) − (x + B)^{α}}{(x + B)^{α}} . \end{matrix}

Some Inequalities 797

Let α = 3 and B = 1. Then

P_{3}(x) = \frac{−2C − 3Cx^{2}(x + 1) − (x + 1)^{3}}{(x + 1)^{3}} .

When −\frac{1}{3} < C < 0, then P_{α}(x) < 0, thus h_{α}(x) > 0, that is

∫_{x}^{∞}e^{−t^{3}}dt > \frac{e^{−x^{3}}}{3(x + 1)^{2}}  (x > 0).  (11)

Proposition 1. Denote S_{n}(x) = ∑_{k=0}^{n} \frac{x^{k}}{k!} (x ∈ ℝ), with n > 0 odd. If nβ > 1, then for x > 0 we have

\frac{(n − 1)![e^{x^{β}}S_{n−1}(−x^{β}) − 1]}{β x^{nβ−1}} < ∫_{0}^{x} e^{t^{β}}dt < \frac{n![e^{x^{β}}S_{n−1}(−x^{β}) − 1]}{x^{nβ−1}}.  (12)

If 0 < nβ < 1, then inequality (12) is reversed.

Proof. Choose

q_{β}(x) = \frac{E[e^{−x^{β}} − S_{n−1}(−x^{β})]}{x^{nβ−1}}

where E > 0 is an undetermined constant. Direct computation produces

q′_{β}(x) = \frac{E\{βx^{β}[S_{n−2}(−x^{β}) − e^{−x^{β}}] − (nβ − 1)[e^{−x^{β}} − S_{n−1}(−x^{β})]\}}{x^{nβ}}

and

\begin{matrix}Q_{β}(x) := & q′_{β}(x) + βx^{β−1}q_{β}(x) + 1 \\ = & 1 + \frac{(−1)^{n}β}{(n − 1)!}E − (nβ − 1)E(\frac{e^{−x^{β}} − S_{n−1}(−x^{β})}{x^{nβ}}). \end{matrix}

It is easy to prove that [6] the function \frac{e^{x}−S_{n}(x)}{x^{n+1}} is increasing for x ∈ (0, +∞) and

lim_{x→0} \frac{e^{x} − S_{n}(x)}{x^{n+1}} = \frac{1}{(n + 1)!}.

Therefore Q_{β}(x) is monotone and

lim_{x→0} Q_{β}(x) = 1 + \frac{(−1)^{n}}{n!} E and lim_{x→+∞} Q_{β}(x) = 1 + \frac{(−1)^{n}β}{(n − 1)!}E.

Now we have

Q_{β}(x) \{ \begin{matrix}> 0 & if \{ \begin{matrix}nβ > 1 and E < \frac{(n−1)!}{β} \\ or \\ 0 < nβ < 1 and E < n! \end{matrix} \\ < 0 & if \{ \begin{matrix}nβ > 1 and E > n! \\ or \\ 0 < nβ < 1 and E > \frac{(n−1)!}{β}. \end{matrix} \end{matrix}

Note that, if n is an odd number, we have e^{x} − S_{n}(x) ≥ 0 (x ∈ ℝ); if n is an even number, this inequality holds for x > 0 and reverses for x ≤ 0 (this conclusion can be found in [1: p. 357] and [6]). Hence, for nβ > 1, we have

\frac{(n − 1)![S_{n−1}(−x^{β}) − e^{−x^{β}}]}{β x^{nβ−1}e^{−x^{β}}} < ∫_{0}^{x} e^{t^{β}}dt < \frac{n![S_{n−1}(−x^{β}) − e^{−x^{β}}]}{x^{nβ−1}e^{−x^{β}}}.  (13)

For 0 < nβ < 1, the reversed inequality of (13) holds. The proof of Proposition 1 is completed

Proposition 2.

(i) For θ > 1, we have

∫_{0}^{x} e^{−t^{θ}}dt > \frac{1 − e^{−x^{θ}}}{x^{θ−1}}  (x > 0).  (14)

For 0 < θ < 1, inequality (14) reverses.

(ii) Let n ≥ 2 be an integer. If \frac{1}{n} < θ < \frac{(n−1)!}{n!−1}, the inequality

∫_{0}^{x} e^{−t^{θ}}dt ≥ \frac{(n − 1)!}{θ + (n − 1)! (1 − nθ)} · \frac{1 − e^{−x^{θ}}S_{n−1}(x^{θ})}{x^{nθ−1}}  (x > 0)  (15)

holds; if 0 < nθ < 1, the reversed inequality of (15) is sound.

Proof. Let r_{θ}(x) = D \frac{e^{x^{θ}} −1}{x^{θ−1}} (x > 0) and D < 0 an undetermined constant. Then

r′_{θ}(x) = D \frac{[θx^{θ}e^{x^{θ}} + (θ − 1)(1 − e^{x^{θ}})]}{x^{θ}}

and

\begin{matrix}R_{θ}(x) := & r′_{θ}(x) − θ x^{θ−1}r_{θ}(x) + 1 \\ = & 1 + θD + (θ − 1)D(\frac{1 − e^{x^{θ}}}{x^{θ}} ). \end{matrix}

When θ > 1, we get R_{θ}(x) > 1 + D; when 0 < θ < 1, we have R_{θ}(x) < 1 + D. Thus, inequality (14) is valid.

Let

r_{θ}(x) = F \frac{e^{x^{θ}} − S_{n−1}(x^{θ})}{x^{nθ−1}} (x > 0; F < 0, n ≥ 2).

Straightforwardly calculating leads to

r′_{θ}(x) = F \frac{θx^{θ}[e^{x^{θ}} − S_{n−2}(x^{θ})] − (nθ − 1)[e^{x^{θ}} − S_{n−1}(x^{θ})]}{x^{θ}}

Some Inequalities 799

and

\begin{matrix}R_{θ}(x) := & r′_{θ}(x) − θ x^{θ−1}r_{θ}(x) + 1 \\ = 1 + \frac{θ}{(n − 1)!}F + (1 − nθ)F (\frac{e^{x^{θ}} − S_{n−1}(x^{θ})}{x^{nθ}}). \end{matrix}

When nθ − 1 > 0, we have

R_{θ}(x) > 1 + \frac{[θ + (n − 1)! (1 − nθ)]}{(n − 1)!}F;

when nθ − 1 < 0, the above inequality is reversed. If

nθ − 1 > 0,  θ + (n − 1)! (1 − nθ) > 0,  F > \frac{(n − 1)!}{(n − 1)! (nθ − 1) − θ} ,

we obtain R_{θ}(x) > 0. From this, it is deduced that inequality (15) holds. The proof of Proposition 2 is completed

Remark. The results in this paper generalize, extend or refine those in [2, 7]. However, the results in this paper and those in [5, 9] are not included each other.

Acknowledgements. The authors are indebted to the referees for their helpful and valuable suggestions and corrections.

References

[1] Ji-Chang Kuang: Applied Inequalities (in Chinese), 2nd edition. Changsha (China): Hunan Education Press 1993.

[2] Laforgia, A. and S. Sismondi: Monotonicity results and inequalities for the gamma and error functions. J. Comp. Appl. Math. 23 (1988), 25 – 33.

[3] Mitrinović, D. S.: Analytic Inequalities. New York et al.: Springer-Verlag 1970.

[4] Feng Qi: Generalized weighted mean values with two parameters. Proc. Roy. Soc. London (Ser. A) 454 (1998), 2723 – 2732.

[5] Feng Qi: Monotonicity results and inequalities for the gamma and incomplete gamma functions. Math. Ineq. Appl. (2000) (to appear).

[6] Feng Qi: A method of constructing inequalities about e^{x}. Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 8 (1997), 16 – 23.

[7] Feng Qi, Li-Hong Cui and Sen-Lin Xu: Some inequalities constructed by Tchebycheff’s integral inequality. Math. Ineq. Appl. 1999 (to appear).

[8] Feng Qi and Bai-Ni Guo: Inequalities for generalized weighted mean values of convex function. Archiv Math. (to appear).

[9] Feng Qi and Sen-Lin Guo: Inequalities for the incomplete gamma and related functions. Math. Ineq. Appl. 2 (1999), 47 – 53.

[10] Feng Qi and Shi-Qin Zhang: Note on monotonicity of generalized weighted mean values. Proc. Roy. Soc. London (Ser. A) (1999) (to appear).

Received 17.09.1998