On Some Inequalities for the Incomplete Gamma Function

MATHEMATICS OF COMPUTATION Volume 66, Number 218, April 1997, Pages 771–778 S 0025-5718(97)00814-4

ON SOME INEQUALITIES FOR THE INCOMPLETE GAMMA FUNCTION

HORST ALZER

Abstract. Let p = 1 be a positive real number. We determine all real numbers α = α(p)and6 β=β(p) such that the inequalities x βxp 1/p 1 tp αxp 1/p [1 e− ] < e− dt < [1 e− ] − Γ(1 + 1/p) − Z0 are valid for all x>0. And, we determine all real numbers a and b such that t ax ∞ e− bx log(1 e− ) dt log(1 e− ) − − ≤ t ≤− − Zx hold for all x>0.

1. Introduction In 1955, J. T. Chu [1] presented sharp upper and lower bounds for the error 2 x t2 function erf(x)= √π 0 e− dt. He proved that the inequalities rx2 1/2 sx2 1/2 (1.1) R[1 e− ] erf(x) [1 e− ] − ≤ ≤ − are valid for all x 0 if and only if 0 r 1ands 4/π. The right-hand inequality of (1.1)≥ (with s =4/π) was proved≤ ≤ independently≥ by J. D. Williams (1946) and G. P´olya (1949); see [1]. An interesting survey on inequalities involving the complementary error function 2 2 ∞ t erfc(x)= √π x e− dt and related functions is given in [4, pp. 177–181]. In 2 2 x /2 ∞ t /2 particular, oneR can ﬁnd inequalities for Mills’ ratio e x e− dt, derived by several authors. In 1959, W. Gautschi [3] provided upper and lower boundsR for the more general expression

xp ∞ tp (1.2) Ip(x)=e e− dt. Zx He established that the double-inequality

1 p 1/p p 1/p (1.3) [(x +2) x] 1andx 0. It has been pointed out in [3] that the integral in (1.2) for p = 3 occurs in heat≥ transfer problems, and for p = 4 in the study of electrical discharge through gases. We note

Received by the editor May 10, 1995 and, in revised form, April 5, 1996. 1991 Mathematics Subject Classiﬁcation. Primary 33B20; Secondary 26D07, 26D15. Key words and phrases. Incomplete gamma function, exponential integral, inequalities.

c 1997 American Mathematical Society

771

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 772 HORST ALZER

p ∞ t that the integral x e− dt can be expressed in terms of the incomplete gamma function R ∞ a 1 t Γ(a, x)= t − e− dt, Zx namely,

p 1 1 ∞ e t dt = Γ ,xp . − p p Zx Gautschi [3] showed that the inequalities (1.3) can be used to derive bounds for the exponential integral E1(x)=Γ(0,x). Indeed, if p tends to , then (1.3) leads to ∞ 1 2 x 1 (1.4) log 1+ e E1(x) log 1+ (0 1, but also for p (0, 1). In particular, we obtain an extension of Chu’s double-inequality (1.1). Moreover,∈ we provide sharp inequalities for the exponential integral E1(x). Finally, in Section 3 we compare our bounds with those given in (1.3) and (1.4).

2. Main results First, we generalize the inequalities (1.1). Theorem 1. Let p =1be a positive real number, and let α = α(p) and β = β(p) be given by 6 p α =1,β= [Γ(1 + 1/p)]− , if 0 1. Then we have for all positive real x: x βxp 1/p 1 tp αxp 1/p (2.1) [1 e− ] < e− dt < [1 e− ] . − Γ(1 + 1/p) − Z0 Proof. We have to show that the functions x tp xp 1/p Fp(x)= e− dt Γ(1 + 1/p)[1 e− ] − − Z0 and x tp axp 1/p p Gp(x)= e− dt +Γ(1+1/p)[1 e− ] (a = [Γ(1 + 1/p)]− ) − − Z0 are both positive on (0, ), if p>1, and are both negative on (0, ), if 0

xp ∂ (1 p)/p e Fp(x)=1 Γ(1 + 1/p)[L(z(x))] − , ∂x −

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON SOME INEQUALITIES FOR THE INCOMPLETE GAMMA FUNCTION 773

where xp L(z)=(z 1)/ log(z)andz(x)=e− . − xp ∂ Setting fp(x)=e ∂xFp(x), we obtain

∂ p 1 2+1/p d d (2.2) fp(x)= − Γ(1 + 1/p)(L(z(x)))− z(x) L(z) . ∂x p dx dz |z=z(x) Since d L(z) = [log(z) 1+1/z]/(log(z))2 > 0(01, ∂x p and ∂ f (x) > 0, if 0 1, then we have

fp(0) = 1 Γ(1 + 1/p) > 0 and lim fp(x)= , x − →∞ −∞ which implies that there exists a number x0 > 0 such that fp(x) > 0forx (0,x0) ∈ and fp(x) < 0forx (x0, ). Hence, the function x Fp(x) is strictly increasing ∈ ∞ 7→ on [0,x0] and strictly decreasing on [x0, ). Since Fp(0) = limx Fp(x)=0,we ∞ →∞ obtain Fp(x) > 0 for all x>0. If 0

fp(0) = 1 Γ(1 + 1/p) < 0 and lim fp(x)=1. x − →∞ This implies that there exists a number x1 > 0 such that x Fp(x) is strictly de- 7→ creasing on [0,x1] and strictly increasing on [x1, ). From Fp(0) = limx Fp(x)= ∞ →∞ 0 we conclude that Fp(x) < 0 for all x>0. Next, we consider Gp(x). Diﬀerentiation leads to

xp ∂ 1 1/a (1 p)/p (2.3) e Gp(x)= 1+(y(x)) − [L(y(x))] − , ∂x − where axp L(y)=(y 1)/ log(y)andy(x)=e− − p ∂ with a = a(p) = [Γ(1 + 1/p)]− . To determine the sign of ∂xGp(x) we need the inequalities 1 p 1 (2.4) 0 < 1 < for 0 0for0

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 774 HORST ALZER

To establish (2.5) we deﬁne for x>0: x +1 g(x)=logΓ(x+1) xlog . − 2 Then we have d2 d x+2 ∞ 1 1 g(x)= ψ(x +1) = dx2 dx (x+1)2 (x + n)2 x +1 − n=2 − X ∞ dt 1 < =0. (x + t)2 − x +1 Z1 Thus, g is strictly concave on [0, ). Since g(0) = g(1) = 0, we conclude that g is positive on (0, 1) and negative on∞ (1, ). This implies (2.5). Let 0

2 1 r ∂ (y 1) y − hr(y)=[(r 1)y r] log(y)+y 1 − ∂y − − − =ϕr(y), say. Since ∂2 r 1 r ϕr(y)= − y , ∂y2 y2 −1 r − r r it follows that ϕr is strictly convex on (0, 1 r ) and strictly concave on ( 1 r , 1). − − From limy 0 ϕr(y)= , → ∞ ∂ ∂2 ϕr(1) = ϕr(y) y=1 =0 and ϕr(y) y=1 =2r 1<0, ∂y | ∂y2 | −

we conclude that there exists a number y0 (0, 1) such that ϕr is positive on (0,y0) ∈ and negative on (y0, 1).Thisimpliesthaty hr(y) is strictly increasing on (0,y0) 7→ and strictly decreasing on (y0, 1). Since limy 0 hr(y) = 0 and limy 1 hr(y)=1,we → → conclude that there exists a number y1 (0, 1) such that hr(y) < 1fory (0,y1) ∈ axp ∈ and hr(y) > 1fory (y1,1). The function y(x)=e− is strictly decreasing on ∈ [0, ). Since y(0) = 1 and limx y(x) = 0, there exists a number x∗ > 0such that∞ →∞

0 1, and, if x∗ 1, then ∂ ∂ Gp(x) > 0forx (0,x∗)and Gp(x) < 0forx (x∗, ); ∂x ∈ ∂x ∈ ∞

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON SOME INEQUALITIES FOR THE INCOMPLETE GAMMA FUNCTION 775

and, if 0 0forx (x∗, ). ∂x ∈ ∂x ∈ ∞

Since Gp(0) = limx Gp(x) = 0, we conclude that →∞ Gp(x) > 0forx (0, ), if p>1, ∈ ∞ and

Gp(x) < 0forx (0, ), if 0

Remark. It is natural to ask whether the double-inequality (2.1) can be reﬁned by replacing α by a positive number which is smaller than

p 1, if 0 1, or by replacing β by a number which is greater than

p p [Γ(1 + 1/p)]− , if 0 1. We show that the answer is “no”! Let α>0 be a real number such that the right-hand inequality of (2.1) holds for all x>0. This implies that the function x tp αxp 1/p Fp(x)= e− dt Γ(1 + 1/p)[1 e− ] − − Z0 is negative on (0, e). Since Fp(0) = 0, we obtain ∞ ∂ 1/p Fp(xe) x=0 =1 α Γ(1 + 1/p) 0, ∂x | − ≤ p which leads to α [Γ(1 + 1/p)]− .Ifα (0, 1), then we conclude from ≥ e ∈ xp ∂ lim e Fp(x)= x ∂x →∞ −∞ that there exists a number x>0 suche that x Fp(x) is negative and strictly 7→ decreasing on [x, ). This contradicts limx Fp(x)=0.Thus,wehaveα ∞ p →∞ ≥ max 1, [Γ(1 + 1/p)]− . e { } Next, we suppose that β>0 is a real numbere such that the ﬁrst inequality of (2.1) is valid for all x>0. This implies x tp βxp 1/p Gp(x)= e− dt +Γ(1+1/p)[1 e− ] < 0 − − Z0 for all x>0. Sincee Gp(0) = 0, we obtain

∂ 1/p e Gp(x) x=0 = β Γ(1 + 1/p) 1 0, ∂x | − ≤ p which yields β [Γ(1 + 1/p)]− .Ifβ>1, then we get ≤ e xp ∂ lim e Gp(x)= 1. x ∂x →∞ − e

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 776 HORST ALZER

This implies that there exists a numberx> ˜ 0 such that x Gp(x) is negative and 7→ strictly decreasing on [˜x, ). This contradicts limx Gp(x) = 0. Hence, we get p∞ →∞ β min 1, [Γ(1 + 1/p)]− . e ≤ { } As an immediate consequence of Theorem 1, the Remark,e and the representation p x p ∞ t t x e− dt =Γ(1+1/p) 0 e− dt, we obtain the following sharp bounds for the p −p ratio ∞ e t dt/ ∞ e t dt. R x − 0 − R Corollary.R Let pR=1be a positive real number. The inequalities 6 αxp 1/p 1 ∞ tp βxp 1/p (2.6) 1 [1 e− ] < e− dt < 1 [1 e− ] − − Γ(1 + 1/p) − − Zx are valid for all positive x if and only if p p α max 1, [Γ(1 + 1/p)]− and 0 β min 1, [Γ(1 + 1/p)]− . ≥ { } ≤ ≤ { } Now, we provide new upper and lower bounds for the exponential integral e t ∞ − E1(x)= x t dt. TheoremR 2. The inequalities ax bx (2.7) log(1 e− ) E1(x) log(1 e− ) − − ≤ ≤− − are valid for all positive real x if and only if a eC and 0 0) is strictly decreasing on (0, ). Therefore, it suﬃces to7→ prove − (2.7)− with a = eC and b =1.Letp>1; from (2.6)∞ p p with α = [Γ(1 + 1/p)]− , β =1,andxinstead of x ,weobtain

αx 1/p ∞ 1+1/p t x 1/p Γ(1/p)[1 (1 e− ) ] < t− e− dt < Γ(1/p)[1 (1 e− ) ]. − − − − Zx If p tends to ,thenweget ∞ ax x log(1 e− ) E1(x) log(1 e− ) − − ≤ ≤− − with a = eC . We assume that there exists a real number b>1 such that bx E1(x) log(1 e− ) ≤− − holds for all x>0. Since n x k 1 k e E1(x)= ( 1) − (k 1)! x− + rn(x)(x>0) − − k=1 X with n 1 rn(x)

∞ xn E (x)= C log(x) ( 1)n (x>0) 1 n! n − − − n=1 − X

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON SOME INEQUALITIES FOR THE INCOMPLETE GAMMA FUNCTION 777

(see [2, p. 674]), we conclude from the left-hand inequality of (2.7) that

x ∞ xn log C ( 1)n . 1 e ax n! n − ≤− −n=1 − − X If x tends to 0, then we obtain log(1/a) C or a eC . ≤− ≥ The proof of Theorem 2 is complete.

3. Concluding remarks In the ﬁnal part of this paper we want to compare the bounds for the integrals p e t ∞ t ∞ − x e− dt (p>1) and x t dt which are given in (1.3), (2.6) and (1.4), (2.7), p respectively. First, we consider the bounds for ∞ e t dt. We deﬁne R R x − xp αxp 1/pR e− p 1/p Rp(x)=Γ(1+1/p) 1 [1 e− ] [(x +2) x] { − − }− 2 − with p α = [Γ(1 + 1/p)]− and p>1. Then we have 1+1/p Rp(0) = Γ(1 + 1/p) 2− > 0, − xp ∂ lim Rp(x) = 0 and lim e Rp(x)=1. x x ∂x →∞ →∞ This implies

Rp(x) > 0 for all suﬃciently small x>0, and

Rp(x) < 0 for all suﬃciently large x. Let xp 1/p xp p 1/p Sp(x)=Γ(1+1/p) 1 [1 e− ] ce− [(x +1/c) x] { − − }− − with p/(p 1) c = [Γ(1 + 1/p)] − and p>1.

From Sp (0) = 0,

∂ p/(p 1) lim Sp(x) = [Γ(1 + 1/p)] − Γ(1 + 1/p) < 0, x 0 ∂x → − xp ∂ lim Sp(x) = 0 and lim e Sp(x)= , x x ∂x →∞ →∞ −∞ we conclude

Sp(x) < 0 for all suﬃciently small x>0, and

Sp(x) > 0 for all suﬃciently large x. p ∞ t Hence, for small x>0 the bounds for x e− dt (p>1) which are given in (2.6) are better than those presented in (1.3), whereas for large values of x the opposite is true. R

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 778 HORST ALZER

Next, we compare the bounds for the exponential integral E1(x). First, we show that for all x>0 the upper bound given in (1.4) is better than the upper bound given in (2.7). This means, we have to prove that x x (3.1) e− log(1 + 1/x) < log(1 e− ) − − for all x>0. Using the extended Bernoulli inequality (1 + z)t 1+tz (t>1; z> 1) ≥ − (see [4, p. 34]), and the elementary inequality et > 1+t (t = 0), we obtain for x>0: 6 x 1 e ex 1 1 1+ 1+ =1+ >1+ , ex 1 ≥ ex 1 1 e x x − − − − which leads immediately to (3.1). Finally, we compare the lower bounds for E1(x) given in (2.7) and (1.4). Let x e− ax T (x)= log(1 + 2/x) + log(1 e− ) 2 − C with a = e . Since limx 0 T (x)= ,weobtainT(x)<0 for all suﬃciently small x. And, from → −∞ d lim T (x) = 0 and lim eax T (x)= , x x dx →∞ →∞ −∞ we conclude that T (x) > 0 for all suﬃciently large x. Thus, for small x>0the lower bound for E1(x) which is given in (2.7) is better than the bound given in (1.4), while for large values of x the latter is better. Acknowledgement I thank the referee for helpful comments. References

1. J. T. Chu, On bounds for the normal integral, Biometrika 42 (1955), 263–265. MR 16:838f 2. G. M. Fichtenholz, Diﬀerential- und Integralrechnung, II, Dt. Verlag Wissensch., Berlin, 1979. MR 80f:26001 3. W. Gautschi, Some elementary inequalities relating to the gamma and incomplete gamma function,J.Math.Phys.38 (1959), 77–81. MR 21:2067 4. D. S. Mitrinovi´c, Analytic inequalities, Springer-Verlag, New York, 1970. MR 43:448

Morsbacher Str. 10, 51545 Waldbrol,¨ Germany

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use