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Hindawi Publishing Corporation International Journal of Analysis Volume 2014, Article ID 195329, 7 pages http://dx.doi.org/10.1155/2014/195329

Research Article Some Remarks on the Self-Exponential : Minimum Value, , and Indefinite

J. L. González-Santander and G. Martín

Departamento de Ciencias Experimentales y Matematicas,´ Universidad CatolicadeValencia“SanVicenteM´ artir”,´ C/Guillem de Castro 96, 46001 Valencia, Spain

Correspondence should be addressed to J. L. Gonzalez-Santander;´ [email protected]

Received 9 September 2014; Accepted 10 November 2014; Published 23 November 2014

Academic Editor: Shamsul Qamar

Copyright © 2014 J. L. Gonzalez-Santander´ and G. Mart´ın. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

𝑥 Considering the function 𝑥 as a real function of real variable, what is its minimum value? Surprisingly, the minimum value is −𝛽𝑥 reached for a negative value of 𝑥. Furthermore, considering the function 𝑓𝛽(𝑥) = 𝑥 , 𝛽∈R and 𝑥>0, two different expressions −1 in closed form for the inverse function 𝑓𝛽 can be obtained. Also, two different series expansions for the indefinite integral of 𝑓𝛽 −1 and 𝑓𝛽 are derived. The latter does not seem to be found in the literature.

1. Introduction complex except for negative . However, [4]saysthat, for 𝑥<0, 𝑓(𝑥) is only defined if 𝑥 can be written as −𝑝/𝑞, Let us consider the following real function of real variable, where 𝑝 and 𝑞 arepositiveintegersand𝑞 is odd. We will use 𝑓:R → R : thisfactlateroninordertoanswerthefirstquestion. 𝑦=𝑓(𝑥) =𝑥𝑥 = (𝑥 𝑥) , This paper is organized such that each section is devoted exp log (1) to each of the questions raised above. and let us pose the following questions. 2. Minimum Value (1) What is the minimum value of 𝑓(𝑥)? The usual way to answer the first question is just to solve the (2) Can its inverse function be expressed in closed form? 󸀠 𝑓 (𝑥) =;thatis, 0 (3) Is its indefinite integral known? exp (𝑥 log 𝑥)(log 𝑥+1) =0, (3) The function 𝑓(𝑥) is termed as self- so in [1,Section26:14]andcoupled exponential function in [2, 1 Equation 01.20], using in the latter the notation 𝑓(𝑥) = 𝑥= . (4) cxt(𝑥). Probably, the most well-known property of 𝑓(𝑥) is 𝑒 just its great growth rate. In fact, the rate of increase of 𝑓(𝑥) Since as 𝑥→∞is greater than the exponential function or the 1 −1 𝑓󸀠󸀠 ( )=𝑒1−𝑒 >0, factorial function [3, Chapter I. section 5] 𝑒 (5) 𝑥 𝑥 𝑥 ≻𝑥!≻𝑒 . (2) then (4) is a local minimum. Moreover, since there are no more local extrema and 𝑓 is a smooth function, (4) is the Regarding the domain of 𝑓(𝑥),in[1,Section26:14],𝑓(𝑥) global minimum; thus, is only defined as a real function for positive values of 𝑥,and 1 𝑓 (𝑥) =𝑓( ) =𝑒−1/𝑒 ≈ 0.692201. [2, p. 10] states that, for arguments less than zero, 𝑓(𝑥) is min 𝑒 (6) 2 International Journal of Analysis

Nonetheless, this reasoning fails, because it does not take Notice that, despite the fact we have considered the log(𝑥) into account negative values of 𝑥. Therefore, we need to define function as a multivalued function in (10), 𝑓(𝑥) is a single- firstthedomainof𝑓(𝑥) for negative values of 𝑥.Despitethe valued function in (13), because we are considering 𝑓(𝑥) as a fact that this is essentially done in [4], in order to answer the real function. Figure 1 shows the plot of 𝑓(𝑥). According to first question, we provide the following derivation. (13),for𝑥<0,theplotof𝑓(𝑥) lies on the following curves: 𝑔 (𝑥) =±|𝑥|𝑥 =± (𝑥 (−𝑥)), 2.1. Domain of 𝑓(𝑥). Let us consider first the case 𝑥=0, ± exp log (15) where the function 𝑓(𝑥) does not exist. However, applying L’ H opital’sˆ rule, the following limit is finite: with a numerable infinite number of points. Notice that + and − signs in (15) occur for 𝑚 even and odd positive integers in 𝑥 lim 𝑥 =1. (13),respectively. 𝑥→0+ (7)

Nevertheless, the right of 𝑓(𝑥) at 𝑥=0is 2.2. Minimum Value of 𝑓(𝑥). In order to calculate the infinite: minimum value of 𝑓(𝑥),for𝑥<0,letussolvetheequation 𝑔󸀠 (𝑥) = 0 󸀠 𝑥 ± ;thatis, lim 𝑓 (𝑥) = lim 𝑥 (log 𝑥+1)=−∞. 𝑥→0+ 𝑥→0+ (8) ± exp (𝑥 log (−𝑥))[log (−𝑥) +1] =0. (16) For 𝑥 =0̸ ,wemayrewrite𝑓(𝑥) by using the signum function sgn(𝑥) as Thus, 1 𝑓 (𝑥) = exp (𝑥 log (sgn (𝑥) |𝑥|)) 𝑥=− . 𝑒 (17) 𝑥 (9) = |𝑥| exp (𝑥 log (sgn 𝑥)) . Notice that Now, for 𝑥 =0̸ and 𝑛∈Z,wehave 󸀠󸀠 −1 1+𝑒−1 𝑔 ( )=∓𝑒 . (18) 0, 𝑥>0 ± 𝑒 ( 𝑥) = { log sgn 𝑖𝜋 + 𝑖2𝜋 (𝑛−1) , 𝑥<0 (10) So, 𝑔+(𝑥) has a maximum and 𝑔−(𝑥) has a minimum in −1/𝑒 ∉ 𝑓 =𝑖𝜋(2𝑛 − 1) 𝜃 (−𝑥) , (17),whichagreeswithFigure 1.However, Dom , so we have to get the best rational approximation −𝑝/𝑞 to −1/𝑒 −𝑝/𝑞 ∈ 𝑓 where 𝜃(𝑥) is the Heaviside function. Therefore, by applying in such a way that Dom .Moreover,since 𝑖𝜃 the minimum lies on the 𝑔−(𝑥) curve, 𝑝 and 𝑞 must be both Euler’s formula 𝑒 = cos 𝜃+𝑖sin 𝜃,weobtain odd positive integers. In order to do so, let us consider the 𝑓 (𝑥)

= |𝑥|𝑥 (𝑖𝜋 (2𝑛 − 1) 𝜃 (−𝑥) 𝑥) 𝑘 󵄨 󵄨 𝑘 󵄨 󵄨 exp ⌊10 󵄨𝑥0󵄨⌋+⌊10 󵄨𝑥0󵄨⌋ (mod 2) +1 𝑎 (𝑥 )=− , 𝑘 0 10𝑘 −1 |𝑥|𝑥 , 𝑥>0 (19) ={ 𝑥 |𝑥| [cos (𝜋 (2𝑛 − 1) 𝑥) +𝑖sin (𝜋 (2𝑛 − 1) 𝑥)] , 𝑥<0. 𝑥0 ∈ R,𝑘∈N, (11) where ⌊𝑥⌋ denotes the floor function.Noticethatthenumer- Since 𝑓(𝑥) is a real function, the complex part of (11) has ator and the denominator of (19) are both odd, so 𝑎𝑘(𝑥) = to be zero. For 𝑥>0, 𝑓(𝑥) is never complex, and for 𝑥<0 −𝑝/𝑞 irreducible, with 𝑝, 𝑞 being odd positive integers. the complex part of 𝑓(𝑥) is zero when Therefore, 𝑎𝑘(𝑥0) is a sequence of rational numbers for which 𝑓(𝑎 (𝑥 )) 𝑔 (𝑥) 𝑎 (𝑥 ) 𝑚 𝑘 0 lies on the curve − .Also, 𝑘 0 is a monotonic (𝜋 (2𝑛 − 1) 𝑥) =0󳨀→𝑥=− <0, 𝑛,𝑚∈Z+. decreasing sequence that satisfies sin 2𝑛 − 1 󵄨 󵄨 (12) lim 𝑎𝑘 (𝑥0)=−󵄨𝑥0󵄨 . 𝑘→∞ 󵄨 󵄨 (20) Therefore, substituting (12) into (11),thefunction𝑓(𝑥) is 𝑟∈Q given by Defining now the rational number, , −1 𝑥𝑥,𝑥>0 { 𝑟=min {𝑎𝑘 ( )} . (21) 𝑓 (𝑥) = 𝑚 𝑘∈N 𝑒 {(−1)𝑚 |𝑥|𝑥 ,𝑥=− <0, 𝑛,𝑚∈Z+ 2𝑛 − 1 { 𝑓(𝑥) (13) Then, the minimum value of is min 𝑓 (𝑥) =𝑓(𝑟) ≈ −1.44467, and its domain is 𝑥∈Dom 𝑓 (22) 𝑚 𝑓=R+ ∪{𝑥=− ,𝑛,𝑚∈Z+}. Dom 2𝑛 − 1 (14) which is different from (6), as aforementioned. International Journal of Analysis 3

4 1 3

2

1 1 2 3

f(x) x

−3 −2 −112 −1 x −1 W(x)

−2 −2

𝑥 Figure 1: Plot of 𝑥 as a real function of real variable. −3 3. Inverse Function About the second question, a closed form expression for −4 the inverse function does not seem to be found in the W0(x) usual literature (see [2, Chapter 2]). However, by using W (x) the Lambert 𝑊 function [5], 𝑓(𝑥) is very easily inverted. −1 The Lambert 𝑊 function is defined as the inverse func- 𝑊 𝑥 Figure 2: Branches of the Lambert function for real arguments. tion of 𝑥𝑒 and it is implemented in MATHEMATICA by the commands ProductLog or LambertW.TheLambert𝑊 function is a multivalued function that presents, for real 3.5 arguments, two branches: 𝑊0(𝑥) () and 𝑊−1(𝑥). Figure 2 shows the plot of both branches. 3.0 Let us consider now on a little more general function than 2.5 (1), but, for simplicity, only for positive arguments; that is, 2.0 −𝛽𝑥 𝑓 (𝑥) =𝑦=𝑥 ,𝛽∈R, 𝑥>0. (x)

𝛽 (23) 𝛽

f 1.5 𝑓 (𝑥) 𝛽 Figure 3 shows the plot of 𝛽 for different values of . 1.0 It is easy to prove that 0.5 𝑓𝛽 (1) =1, 0.5 1.0 1.5 2.0 lim 𝑓𝛽 (𝑥) =1, 𝑥→0+ (24) x 1 𝑓󸀠 (𝑥) =0←→𝑥= ,𝛽=0,̸ 𝛽=−2 𝛽=1 𝛽 𝑒 𝛽=−1 𝛽=2 𝛽=0 +∞, 𝛽 > 0, { −𝛽𝑥 󸀠 Figure 3: Plot of 𝑥 for different values of 𝛽. lim 𝑓𝛽 (𝑥) = 0, 𝛽 = 0, (25) 𝑥→0+ { {−∞, 𝛽 < 0, which agrees with Figure 3. and thus Solving (23),wehave −1 log 𝑥 𝑦 𝑓𝛽 (𝑥) = exp (𝑊 (− )) (28) −log =𝑥 𝑥 𝛽 𝛽 log (26) 𝑥 − log 𝑥/𝛽 =𝑒log log 𝑥. = . (29) 𝑊(−log 𝑥/𝛽) Applying now the Lambert 𝑊 function and taking into account (26),weobtain According to Figure 3, notice that, depending on the 𝑥 𝑓−1(𝑥) log 𝑦 value of , 𝛽 sometimes is a double-valued function, so 𝑊(− )= 𝑥 𝑊 𝛽 log we have two real values of the Lambert function in (29), 𝑊 (𝑥) 𝑊 (𝑥) (27) that is, 0 and −1 .Inthislattercase,wehaveused log 𝑦 the notation 𝑊−1,0(𝑥).Also,fromFigure 3,wecanseethat =− , −1 𝛽𝑥 𝑓𝛽 (𝑥) is a single-valued function for 𝑥>1when 𝛽<0and 4 International Journal of Analysis for 𝑥 ∈ (0, 1) when 𝛽>0. Therefore, taking into account (24) but, from (34) and (40), we arrive at (33): and Figure 3, we can consider the following cases. 1 𝑥 𝑔 (𝑥) = = log , (i) Consider 𝛽<0, (41) ℎ (1/𝑥) 𝑊0 (log 𝑥) − 𝑥/𝛽 { log , 𝑥≤𝑥≤1, { which, according to (37),convergesifandonlyif {𝑊−1,0 (− log 𝑥/𝛽) 𝑓−1 (𝑥) = 𝛽 { (30) −𝑒 1 1/𝑒 { − log 𝑥/𝛽 𝑒 ≤ ≤𝑒 , { , 𝑥>1. 𝑥 (42) {𝑊0 (− log 𝑥/𝛽) that is, the interval given in (35). −1 (ii) Consider 𝛽>0, In [2,Equation02.03],𝑓 (𝑥) is termed as coupled root function. Since the latter reference is unaware of the closed − log 𝑥/𝛽 { ,1≤𝑥≤𝑥, expression (33), it performs numerically the following limit {𝑊 (− 𝑥/𝛽) −1 { −1,0 log [2, Equation 02.07]: 𝑓𝛽 (𝑥) = { (31) { − 𝑥/𝛽 { log , 0<𝑥<1, 𝑓−1 (𝑥) =0, {𝑊0 (− log 𝑥/𝛽) lim (43) 𝑥→∞ log 𝑥 where we have defined −1 in order to show that 𝑓 (𝑥) goes to infinity at a lower rate 1 𝛽/𝑒 than . In fact, (43) is quite easily proved applying 𝑥=𝑓𝛽 ( )=𝑒 . (32) 𝑒 (33) and the property By setting 𝛽=−1,in(30),weobtaintheinversefunction 𝑥 lim 𝑥𝑒 =∞󳨀→ lim 𝑊0 (𝑥) =∞, of (1) 𝑥→∞ 𝑥→∞ (44) 𝑥 { log ,𝑒−1/𝑒 ≤𝑥≤1, so that { {𝑊−1,0 (log 𝑥) −1 𝑓−1 (𝑥) = 𝑓 (𝑥) 1 { (33) lim = lim =0. (45) { 𝑥 𝑥→∞ 𝑥 𝑥→∞𝑊 ( 𝑥) { log ,𝑥>1. log 0 log {𝑊0 (log 𝑥) 4. Indefinite Integral Curiously, (33) is just the closed expression given in [6] for the following power tower: −1 4.1. Indefinite Integral of 𝑓𝛽 . For 𝛽<0, according to (30),let (1/𝑥)⋅⋅⋅ 𝑔 (𝑥) =𝑥(1/𝑥) , (34) us calculate 𝑧 𝑧 −1 − log 𝑥/𝛽 which converges if and only if ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 = ∫ 𝑑𝑥, 𝑧 >1. (46) 1 1 𝑊0 (− log 𝑥/𝛽) −1/𝑒 𝑒 𝑒 ≤𝑥≤𝑒. (35) Performing the change of variables 𝑢=−log 𝑥/𝛽,wehave In order to see this, consider the power tower 𝑧 − log 𝑧/𝛽 −𝛽𝑢 −1 𝑢𝑒 𝑥⋅⋅⋅ 𝑥 ∫ 𝑓𝛽 (𝑥) 𝑑𝑥=−𝛽∫ 𝑑𝑢, (47) ℎ (𝑥) =𝑥 , (36) 1 0 𝑊0 (𝑢)

−𝛽𝑢 which converges if and only if [7, 8] and expanding the exponential function 𝑒 ,in(47),we −𝑒 1/𝑒 obtain 𝑒 ≤𝑥≤𝑒 . (37) 𝑛+1 𝑧 ∞ (−𝛽) − log 𝑧/𝛽 𝑢𝑛+1 Taking in (36) the of both sides and plugging −1 ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 = ∑ ∫ 𝑑𝑢. (48) back the ℎ(𝑥) function definition, we obtain 1 𝑛=0 𝑛! 0 𝑊0 (𝑢) ⋅⋅⋅ 𝑥𝑥 log ℎ=𝑥 log 𝑥=ℎlog 𝑥. (38) Performing now the change of variables 𝑊0(𝑢) = 𝑠 (i.e., 𝑢=𝑠𝑒𝑠 −𝑢 ), we get Performing now the change of variables ℎ=𝑒 ,weget 𝑧 ∞ (−𝛽)𝑛+1 𝛼 𝑢𝑒𝑢 =− 𝑥. −1 𝑛 (𝑛+2)𝑠 log (39) ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 = ∑ ∫ 𝑠 (𝑠+1) 𝑒 𝑑𝑠, (49) 1 𝑛=0 𝑛! 0 Now, by using the principal branch of Lambert 𝑊 function, we can solve, for ℎ(𝑥), wherewehaveset

−𝑊0 (− log 𝑥) log 𝑧 ℎ (𝑥) = , (40) 𝛼:=𝑊0 (− ). (50) log 𝑥 𝛽 International Journal of Analysis 5

The integral given in (49) can be expressed in terms of Notice as well that, for 𝛽>0, according to (31),wehave the lower incomplete gamma function [9,Equation8.2.1]. to calculate Consider 1 1 −1 − log 𝑥/𝛽 𝑧 ∫ 𝑓 (𝑥) 𝑑𝑥 = ∫ 𝑑𝑥, 𝑧∈ (0, 1) , 𝑎−1 −𝑡 𝛽 𝑊 (− 𝑥/𝛽) (59) 𝛾 (𝑎,) 𝑧 := ∫ 𝑡 𝑒 𝑑𝑡, Re 𝑎>0; (51) 𝑧 𝑧 0 log 0 but, according to (58),wehave thus, performing the change of variables −𝑡 = 𝑘𝑠,weget 1 𝑎 −𝑘𝑎 −1 𝑚 𝑘𝑠 −1−𝑚 𝑚 −𝑡 ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 ∫ 𝑠 𝑒 𝑑𝑠 = (−𝑘) ∫ 𝑡 𝑒 𝑑𝑡 𝑧 0 0 (52) ∞ 𝛽𝑛+1 = (−𝑘)−1−𝑚 𝛾 (𝑚 + 1, )−𝑘𝑎 . =−∑ 𝑛=0 𝑛+2

Therefore, applying (52) in (49),weobtain 𝑛+1 1 (−𝛼) 𝑒 (− (𝑛+2) 𝛼) (𝑛+2)𝛼 𝑧 ×{ +( − 𝑛 )𝑒 }, −1 (𝑛+2)𝑛+1 𝑛! (𝑛+2)𝑛+1 ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 1 (60) ∞ 𝛽𝑛+1 = ∑ where 𝛼=𝑊0(− log 𝑧/𝛽), 𝑧 ∈ (0, 1) and 𝛽>0. 𝑛+1 𝑛=0 𝑛! (𝑛+2) 𝑓 𝛾 (𝑛+2,−(𝑛+2) 𝛼) 4.2. Indefinite Integral of 𝛽. The indefinite integral ×{𝛾(𝑛+1,−(𝑛+2) 𝛼) − }. 𝑛+2 ∫ 𝑥𝑥𝑑𝑥 (53) (61) Applying now the property [9,Equation8.8.1] cannotbeexpressedintermsofafinitenumberofelementary functions [10]. Moreover, closed expression for (61) in the 𝛾 (𝑎+1,𝑧) =𝑎𝛾(𝑎,) 𝑧 −𝑧𝑎𝑒−𝑧, (54) usual literature does not seem to be found. However, it can we may rewrite (53) as be expressed in closed form by using the upper incomplete gamma function [9,Equation8.2.1]: 𝑧 −1 ∞ ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 𝑎−1 −𝑡 1 Γ (𝑎,) 𝑧 := ∫ 𝑡 𝑒 𝑑𝑡, Re 𝑎>0. (62) 𝑧 ∞ 𝑛+1 𝑛+1 𝛽 𝛾 (𝑛+1,−(𝑛+2) 𝛼) (−𝛼) (𝑛+2)𝛼 = ∑ { + 𝑒 }. Notice that if 𝑧=0and 𝑎=𝑛is a positive , then 𝑛! (𝑛+2)𝑛+2 𝑛+2 𝑛=0 we recover the usual gamma function: (55) + Γ (𝑛,) 0 =Γ(𝑛) = (𝑛−1)!, 𝑛 ∈ Z . (63) In order to compute the lower incomplete gamma func- tion given in (55),wemayuse[1, Equation 45:4:1] We can generalize (61),calculatingtheintegral −𝑥 𝑧 𝑧 𝛾 (𝑛,) 𝑥 = (𝑛−1)![1−𝑒𝑛−1 (𝑥) 𝑒 ], (56) ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 = ∫ exp (−𝛽𝑥 log 𝑥) 𝑑𝑥, 0 0 (64) where 𝑒𝑛(𝑥) is the exponential and it is given by the power-series expansion of the exponential function by 𝑧>0, 𝛽∈R. truncation after the 𝑛th term [1, Equation 26:12:2]: Expanding in power-series the exponential function 𝑛 𝑥𝑘 given in (64) and integrating term by term, we get 𝑒𝑛 (𝑥) = ∑ . 𝑘! (57) 𝑛 𝑘=0 𝑧 ∞ (−𝛽) 𝑧 ∫ 𝑓 (𝑥) 𝑑𝑥 = ∑ ∫ (𝑥 𝑥)𝑛 𝑑𝑥. 𝛽 𝑛! log (65) Therefore, we finally obtain 0 𝑛=0 0 𝑧 𝑛+1 −𝑡 −1 Performing now the change of variables 𝑥 =𝑒 ,we ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 1 obtain 𝑧 ∞ 𝛽𝑛+1 1 = ∑ ×{ ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 𝑛+1 0 𝑛=0 𝑛+2 (𝑛+2) 𝑛 ∞ (−𝛽) (−1)𝑛+1 −(𝑛+1) log 𝑧 (−𝛼)𝑛+1 𝑒 (− (𝑛+2) 𝛼) = ∑ ∫ 𝑒−𝑡𝑡𝑛𝑑𝑡 +( − 𝑛 )𝑒(𝑛+2)𝛼}, (𝑛+1)! (𝑛+1)𝑛 (66) 𝑛! (𝑛+2)𝑛+1 𝑛=0 ∞ (58) ∞ 𝛽𝑛−1 ∞ = ∑ ∫ 𝑒−𝑡𝑡𝑛−1𝑑𝑡. 𝑛−1 where 𝛼=𝑊0(− log 𝑧/𝛽), 𝑧>1and 𝛽<0. 𝑛=1 𝑛!𝑛 −𝑛 log 𝑧 6 International Journal of Analysis

By using now the definition of the upper incomplete gamma function (62), we arrive at z−𝛽z ∞ 𝑛−1 𝑧 𝛽 1.5 ∫ 𝑥−𝛽𝑥𝑑𝑥 = ∑ Γ(𝑛,−𝑛 𝑧) . 𝑛 log (67) 0 𝑛=1 (𝑛−1)!𝑛

In [11, Lemma 10.6], we find a similar expression for the indefinite integral of the power tower 1.0 ⋅⋅⋅ ∞ 𝑛−1 𝑛−2 𝑥𝑥 (−1) 𝑛 (x) 𝛽

∫ 𝑥 𝑑𝑥 = ∑ Γ(𝑛,−log 𝑥) . (68) f 𝑛=1 (𝑛−1)!

Taking 𝑧=1in (67) and using (63),werecoverthe expression given by [12, Equation 3.342]: 0.5

1 ∞ 𝛽𝑛−1 ∫ 𝑥−𝛽𝑥𝑑𝑥 = ∑ . 𝑛 (69) 0 𝑛=1 𝑛 z

Moreover, taking 𝛽=±1,in(69),werecoverthe 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 expressions given for the sophomore’s dream [13,pp.4,44], x discovered in 1697 by Johann Bernoulli, as follows: Figure 4: Scheme for the integration of 𝑓𝛽(𝑥), 𝛽<0,beyond𝑧>1. 1 ∞ (±1)𝑛−1 ∫ 𝑥∓𝑥𝑑𝑥 = ∑ . 𝑛 (70) 0 𝑛=1 𝑛 So, from (73) and (74) andtakingintoaccount(58),weget In order to compute the upper incomplete gamma func- 𝑧 −𝛽𝑥 tion given in (67),wemayuse[1,Equation45:4:2]: ∫ 𝑥 𝑑𝑥 0 −𝑥 ∞ 𝑛−1 Γ (𝑛,) 𝑥 = (𝑛−1)!𝑒 𝑒𝑛−1 (𝑥) , (71) 𝛽 =𝑧1−𝛽𝑧 −1+∑ 𝑛 𝑛=1 𝑛 where 𝑒𝑛(𝑥) is the exponential polynomial (57). Therefore, (67) canberewrittenas ∞ 𝛽𝑛+1 1 − ∑ { 𝑛+1 𝑛=0 𝑛+2 (𝑛+2) 𝑧 1 ∞ 𝛽𝑧 𝑛 ∫ 𝑥−𝛽𝑥𝑑𝑥 = ∑ ( ) 𝑒 (−𝑛 𝑧) . 𝛽 𝑛 𝑛−1 log (72) (−𝛼)𝑛+1 𝑒 (− (𝑛+2) 𝛼) 0 𝑛=1 +( − 𝑛 )𝑒(𝑛+2)𝛼}. 𝑛! (𝑛+2)𝑛+1 Let us now proceed to give another expression for the (75) indefinite integral of 𝑓𝛽 by using the results given in (58) and (60). First, let us consider the cases 𝛽<0and 𝑧>1,splitting Since the following series is a telescoping series: (64) into two summands, as follows: ∞ 𝛽𝑛 𝛽𝑛+1 ∑ ( − ) =1, (76) 𝑧 1 𝑧 (𝑛+1)𝑛+1 (𝑛+2)𝑛+2 ∫ 𝑥−𝛽𝑥𝑑𝑥 = ∫ 𝑥−𝛽𝑥𝑑𝑥 + ∫ 𝑥−𝛽𝑥𝑑𝑥 𝑛=0 0 0 1 (73) we can simplify (75),obtaining ∞ 𝛽𝑛−1 𝑧 = ∑ + ∫ 𝑥−𝛽𝑥𝑑𝑥, 𝑧 𝑛 −𝛽𝑥 1−𝛽𝑧 𝑛=1 𝑛 1 ∫ 𝑥 𝑑𝑥=𝑧 0

∞ 𝛼 𝑛+1 𝑛+1 where the first integral in (73) has been substituted by (69) 𝛼 (𝛽𝑒 ) (−𝛼) and the second integral can be computed by knowing that −𝑒 ∑ { (77) 𝑛=0 𝑛+2 𝑛! 𝑓𝛽(𝑥) is an increasing function for 𝑥>1when 𝛽<0. Indeed, according to Figure 4,wehave 𝑒 (− (𝑛+2) 𝛼) − 𝑛 }, (𝑛+2)𝑛+1 𝑧 𝑧−𝛽𝑧 𝑧×𝑧−𝛽𝑧 =1+∫ 𝑓 𝑥 𝑑𝑥 + ∫ 𝑓−1 𝑥 𝑑𝑥. 𝛽 ( ) 𝛽 ( ) (74) 𝛼=𝑊(𝑧 𝑧) 𝑧>1 𝛽<0 1 1 where 0 log , and . International Journal of Analysis 7

1.4 [7] L. Euler, “De serie Lambertina Plurimisque eius insignibus pro- 1.2 prietatibus,” Acta Academiae Scientarum Imperialis Petropoliti- 1.0 nae, vol. 2, pp. 29–51, 1783, Reprinted in L. Euler, Opera Omnia, 0.8 Series Prima, Volume 6: Commentationes Algebraicae, pp. 350–

(x) 369. 𝛽 𝛼⋅⋅⋅ f 0.6 𝛼 [8] G. Eisenstein, “Entwicklung von 𝛼 ,” Journal fur¨ die Reine und 0.4 A z−𝛽z Angewandte Mathematik,vol.28,pp.49–52,1844. 0.2 z [9]F.W.J.Olver,D.W.Lozier,R.F.Boisvert,andC.W.Clark, 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Eds., NIST Handbook of Mathematical Functions, Cambridge University Press, New York, NY, USA, 2010. x [10] E. A. Marchisotto and G.-A. Zakeri, “An invitation to integra-

Figure 5: Scheme for the integration of 𝑓𝛽(𝑥), 𝛽>0,for𝑧>1. tion in finite terms,” College Journal,vol.25,pp. 295–308, 1994. [11] I. N. Galidakis, “On an application of Lambert’s W function to infinite exponentials,” Complex Variables. Theory and Applica- For the case of 𝛽>0, according to Figure 5,wehave tion,vol.49,no.11,pp.759–780,2004.

𝑧 1 [12] I. S. Gradshteyn and I. M. Ryzhik, Table of , Series, and −𝛽𝑥 −1 −𝛽𝑧 −𝛽𝑧 Products, Academic Press, New York, NY, USA, 7th edition, ∫ 𝑥 𝑑𝑥 = ∫ 𝑓𝛽 (𝑥) 𝑑𝑥 − (1 −𝑧 ) + (𝑧−1) 𝑧 . 1 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟𝑧−𝛽𝑧 2007. 𝐴 [13] J. Borwein, D. Bailey, and R. Girgensohn, Experimentation in (78) Mathematics: Computational Paths to Discovery, A K Peters, Wellesley, Mass, USA, 2004. Therefore, substituting (78) in (73) and taking into account (60) and (76),wearriveagainat(77),butfor𝛽>0. Moreover, the range 𝑧>1can be extended up to the point where 𝑓𝛽 is a monotonic increasing or decreasing function. So, according to (24),wecansaythat𝑧>1/𝑒.Then,collecting all these results, we can conclude that 𝑧 ∫ 𝑥−𝛽𝑥𝑑𝑥=𝑧1−𝛽𝑧 0

∞ (𝛽𝑒𝛼)𝑛+1 (−𝛼)𝑛+1 𝑒 (− (𝑛+2) 𝛼) −𝑒𝛼 ∑ { − 𝑛 }, 𝑛+1 𝑛=0 𝑛+2 𝑛! (𝑛+2) (79) where 𝛼=𝑊0(𝑧 log 𝑧), 𝑧>1/𝑒,and𝛽>0.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

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