# Some Remarks on the Self-Exponential Function: Minimum Value, Inverse Function, and Indefinite Integral

Authors: J. L. González-Santander; G. Martín
Journal: International Journal of Analysis
 (2014)

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## Abstract

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


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## Body

## 1. Introduction

Let us consider the following real function of real variable,f:R→R:
(1)y=fx=xx=exp⁡xlog⁡x,
and let us pose the following questions.(1)
What is the minimum value offx?(2)
Can its inverse function be expressed in closed form?(3)
Is its indefinite integral known?The functionfx is termed asself-exponential function in [1, Section 26:14] andcoupled exponential function in [2, Equation 01.20], using in the latter the notation fx=cxtx. Probably, the most well-known property of fx is just its great growth rate. In fact, the rate of increase of fx as x→∞ is greater than the exponential function or the factorial function [3, Chapter I. section 5]
(2)xx≻x!≻ex.Regarding the domain offx, in [1, Section 26:14], fx is only defined as a real function for positive values of x, and [2, p. 10] states that, for arguments less than zero, fx is complex except for negative integers. However, [4] says that, for x<0, fx is only defined if x can be written as -p/q, where p and q are positive integers and q is odd. We will use this fact later on in order to answer the first question.This paper is organized such that each section is devoted to each of the questions raised above.


## 2. Minimum Value

The usual way to answer the first question is just to solve the equationf′x=0; that is,
(3)exp⁡xlog⁡xlog⁡x+1=0,
so
(4)x=1e.Since(5)f′′1e=e1-e-1>0,
then (4) is a local minimum. Moreover, since there are no more local extrema and f is a smooth function, (4) is the global minimum; thus,
(6)min⁡fx=f1e=e-1/e≈0.692201.Nonetheless, this reasoning fails, because it does not take into account negative values ofx. Therefore, we need to define first the domain of fx for negative values of x. Despite the fact that this is essentially done in [4], in order to answer the first question, we provide the following derivation.


### 2.1. Domain offx

Let us consider first the casex=0, where the function fx does not exist. However, applying L’Hôpital’s rule, the following limit is finite:
(7)lim⁡x→0+xx=1.Nevertheless, the right derivative offx at x=0 is infinite:
(8)lim⁡x→0+f′x=lim⁡x→0+xxlog⁡x+1=-∞.Forx≠0, we may rewrite fx by using thesignum function sgn⁡x as
(9)fx=exp⁡xlog⁡sgn⁡xx=xxexp⁡xlog⁡sgn⁡x.Now, forx≠0 and n∈Z, we have
(10)log⁡sgn⁡x=0,x>0iπ+i2πn-1,x<0=iπ2n-1θ-x,
where θx is the Heaviside function. Therefore, by applying Euler’s formula eiθ=cos⁡θ+isinθ, we obtain
(11)fx=xxexp⁡iπ2n-1θ-xx=xx,x>0xxcos⁡π2n-1x+isinπ2n-1x,x<0.Sincefx is a real function, the complex part of (11) has to be zero. For x>0, fx is never complex, and for x<0 the complex part of fx is zero when
(12)sinπ2n-1x=0⟶x=-m2n-1<0,n,m∈Z+.Therefore, substituting (12) into (11), the function fx is given by
(13)fx=xx,x>0-1mxx,x=-m2n-1<0,n,m∈Z+
and its domain is
(14)Dom⁡f=R+∪x=-m2n-1,n,m∈Z+.Notice that, despite the fact we have considered thelog⁡x function as a multivalued function in (10), fx is a single-valued function in (13), because we are considering fx as a real function. Figure 1 shows the plot of fx. According to (13), for x<0, the plot of f(x) lies on the following curves:
(15)g±x=±xx=±exp⁡xlog⁡-x,
with a numerable infinite number of points. Notice that + and − signs in (15) occur for m even and odd positive integers in (13), respectively.Figure 1
Plot ofxx as a real function of real variable.


### 2.2. Minimum Value offx

In order to calculate the minimum value offx, for x<0, let us solve the equation g±′x=0; that is,
(16)±exp⁡xlog⁡-xlog⁡-x+1=0.
Thus,
(17)x=-1e.Notice that(18)g±′′-1e=∓e1+e-1.
So, g+x has a maximum and g-x has a minimum in (17), which agrees with Figure 1. However, -1/e∉Dom⁡f, so we have to get the best rational approximation -p/q to -1/e in such a way that -p/q∈Dom⁡f. Moreover, since the minimum lies on the g-x curve, p and q must be both odd positive integers. In order to do so, let us consider the sequence
(19)akx0=-10kx0+10kx0mod⁡2+110k-1,x0∈R,k∈N,
where x denotes thefloor function. Notice that the numerator and the denominator of (19) are both odd, so akx=-p/q irreducible, with p,q being odd positive integers. Therefore, akx0 is a sequence of rational numbers for which fakx0 lies on the curve g-x. Also, akx0 is a monotonic decreasing sequence that satisfies
(20)lim⁡k→∞akx0=-x0.Defining now the rational number,r∈Q,
(21)r=min⁡k∈Nak-1e.
Then, the minimum value of fx is
(22)min⁡x∈Dom⁡ffx=fr≈-1.44467,
which is different from (6), as aforementioned.


## 2.1. Domain offx

Let us consider first the casex=0, where the function fx does not exist. However, applying L’Hôpital’s rule, the following limit is finite:
(7)lim⁡x→0+xx=1.Nevertheless, the right derivative offx at x=0 is infinite:
(8)lim⁡x→0+f′x=lim⁡x→0+xxlog⁡x+1=-∞.Forx≠0, we may rewrite fx by using thesignum function sgn⁡x as
(9)fx=exp⁡xlog⁡sgn⁡xx=xxexp⁡xlog⁡sgn⁡x.Now, forx≠0 and n∈Z, we have
(10)log⁡sgn⁡x=0,x>0iπ+i2πn-1,x<0=iπ2n-1θ-x,
where θx is the Heaviside function. Therefore, by applying Euler’s formula eiθ=cos⁡θ+isinθ, we obtain
(11)fx=xxexp⁡iπ2n-1θ-xx=xx,x>0xxcos⁡π2n-1x+isinπ2n-1x,x<0.Sincefx is a real function, the complex part of (11) has to be zero. For x>0, fx is never complex, and for x<0 the complex part of fx is zero when
(12)sinπ2n-1x=0⟶x=-m2n-1<0,n,m∈Z+.Therefore, substituting (12) into (11), the function fx is given by
(13)fx=xx,x>0-1mxx,x=-m2n-1<0,n,m∈Z+
and its domain is
(14)Dom⁡f=R+∪x=-m2n-1,n,m∈Z+.Notice that, despite the fact we have considered thelog⁡x function as a multivalued function in (10), fx is a single-valued function in (13), because we are considering fx as a real function. Figure 1 shows the plot of fx. According to (13), for x<0, the plot of f(x) lies on the following curves:
(15)g±x=±xx=±exp⁡xlog⁡-x,
with a numerable infinite number of points. Notice that + and − signs in (15) occur for m even and odd positive integers in (13), respectively.Figure 1
Plot ofxx as a real function of real variable.


## 2.2. Minimum Value offx

In order to calculate the minimum value offx, for x<0, let us solve the equation g±′x=0; that is,
(16)±exp⁡xlog⁡-xlog⁡-x+1=0.
Thus,
(17)x=-1e.Notice that(18)g±′′-1e=∓e1+e-1.
So, g+x has a maximum and g-x has a minimum in (17), which agrees with Figure 1. However, -1/e∉Dom⁡f, so we have to get the best rational approximation -p/q to -1/e in such a way that -p/q∈Dom⁡f. Moreover, since the minimum lies on the g-x curve, p and q must be both odd positive integers. In order to do so, let us consider the sequence
(19)akx0=-10kx0+10kx0mod⁡2+110k-1,x0∈R,k∈N,
where x denotes thefloor function. Notice that the numerator and the denominator of (19) are both odd, so akx=-p/q irreducible, with p,q being odd positive integers. Therefore, akx0 is a sequence of rational numbers for which fakx0 lies on the curve g-x. Also, akx0 is a monotonic decreasing sequence that satisfies
(20)lim⁡k→∞akx0=-x0.Defining now the rational number,r∈Q,
(21)r=min⁡k∈Nak-1e.
Then, the minimum value of fx is
(22)min⁡x∈Dom⁡ffx=fr≈-1.44467,
which is different from (6), as aforementioned.


## 3. Inverse Function

About the second question, a closed form expression for the inverse function does not seem to be found in the usual literature (see [2, Chapter 2]). However, by using theLambertWfunction [5], fx is very easily inverted. The Lambert W function is defined as the inverse function of xex and it is implemented in MATHEMATICA by the commandsProductLog orLambertW. The Lambert W function is a multivalued function that presents, for real arguments, two branches: W0x (principal branch) and W-1x. Figure 2 shows the plot of both branches.Figure 2
Branches of the LambertW function for real arguments.Let us consider now on a little more general function than (1), but, for simplicity, only for positive arguments; that is,
(23)fβx=y=x-βx,β∈R,x>0.Figure3 shows the plot of fβx for different values of β. It is easy to prove that
(24)fβ1=1,lim⁡x→0+fβx=1,fβ′x=0⟷x=1e,β≠0,(25)lim⁡x→0+fβ′x=+∞,β>0,0,β=0,-∞,β<0,
which agrees with Figure 3.Figure 3
Plot ofx-βx for different values of β.Solving (23), we have
(26)-log⁡yβ=xlog⁡x=elog⁡xlog⁡x.Applying now the LambertW function and taking into account (26), we obtain
(27)W-log⁡yβ=log⁡x=-log⁡yβx,
and thus
(28)fβ-1x=exp⁡W-log⁡xβ(29)ji=-log⁡x/βW-log⁡x/β.According to Figure3, notice that, depending on the value of x, fβ-1x sometimes is a double-valued function, so we have two real values of the Lambert W function in (29), that is, W0x and W-1x. In this latter case, we have used the notation W-1,0x. Also, from Figure 3, we can see that fβ-1x is a single-valued function for x>1 when β<0 and for x∈0,1 when β>0. Therefore, taking into account (24) and Figure 3, we can consider the following cases.(i)
Considerβ<0,
(30)fβ-1x=-log⁡x/βW-1,0-log⁡x/β,x¯≤x≤1,-log⁡x/βW0-log⁡x/β,x>1.(ii)
Considerβ>0,
(31)fβ-1x=-log⁡x/βW-1,0-log⁡x/β,1≤x≤x¯,-log⁡x/βW0-log⁡x/β,0<x<1,where we have defined
(32)x¯=fβ1e=eβ/e.By settingβ=-1, in (30), we obtain the inverse function of (1)(33)f-1x=log⁡xW-1,0log⁡x,e-1/e≤x≤1,log⁡xW0log⁡x,x>1.Curiously, (33) is just the closed expression given in [6] for the following power tower:
(34)gx=x1/x1/x⋯,
which converges if and only if
(35)e-1/e≤x≤ee.In order to see this, consider the power tower(36)hx=xxx⋯,
which converges if and only if [7, 8]
(37)e-e≤x≤e1/e.Taking in (36) the logarithm of both sides and plugging back the hx function definition, we obtain
(38)log⁡h=xxx⋯log⁡x=hlog⁡x.Performing now the change of variablesh=e-u, we get
(39)ueu=-log⁡x.Now, by using the principal branch of LambertW function, we can solve, for hx,
(40)hx=-W0-log⁡xlog⁡x,
but, from (34) and (40), we arrive at (33):
(41)gx=1h1/x=log⁡xW0log⁡x,
which, according to (37), converges if and only if
(42)e-e≤1x≤e1/e,
that is, the interval given in (35).In [2, Equation 02.03], f-1x is termed ascoupled root function. Since the latter reference is unaware of the closed expression (33), it performs numerically the following limit [2, Equation 02.07]:
(43)lim⁡x→∞f-1xlog⁡x=0,
in order to show that f-1x goes to infinity at a lower rate than logarithms. In fact, (43) is quite easily proved applying (33) and the property
(44)lim⁡x→∞xex=∞⟶lim⁡x→∞W0x=∞,
so that
(45)lim⁡x→∞f-1xlog⁡x=lim⁡x→∞1W0log⁡x=0.


## 4. Indefinite Integral

### 4.1. Indefinite Integral offβ-1

Forβ<0, according to (30), let us calculate
(46)∫1z‍fβ-1xdx=∫1z‍-log⁡x/βW0-log⁡x/βdx,z>1.Performing the change of variablesu=-log⁡x/β, we have
(47)∫1z‍fβ-1xdx=-β∫0-log⁡z/β‍ue-βuW0udu,
and expanding the exponential function e-βu, in (47), we obtain
(48)∫1z‍fβ-1xdx=∑n=0∞‍-βn+1n!∫0-log⁡z/β‍un+1W0udu.Performing now the change of variablesW0u=s (i.e., u=ses), we get
(49)∫1z‍fβ-1xdx=∑n=0∞‍-βn+1n!∫0α‍sns+1en+2sds,
where we have set
(50)α:=W0-log⁡zβ.The integral given in (49) can be expressed in terms of the lower incomplete gamma function [9, Equation 8.2.1]. Consider
(51)γa,z:=∫0z‍ta-1e-tdt,Rea>0;
thus, performing the change of variables -t=ks, we get
(52)∫0a‍smeksds=-k-1-m∫0-ka‍tme-tdt=-k-1-mγm+1,-ka.Therefore, applying (52) in (49), we obtain
(53)∫1z‍fβ-1xdx=∑n=0∞‍βn+1n!n+2n+1×γn+1,-n+2α-γn+2,-n+2αn+2.Applying now the property [9, Equation 8.8.1]
(54)γa+1,z=aγa,z-zae-z,
we may rewrite (53) as
(55)∫1z‍fβ-1xdx=∑n=0∞‍βn+1n!γn+1,-n+2αn+2n+2+-αn+1n+2en+2α.In order to compute the lower incomplete gamma function given in (55), we may use [1, Equation 45:4:1]
(56)γn,x=n-1!1-en-1xe-x,
where enx is theexponential polynomial and it is given by the power-series expansion of the exponential function by truncation after the nth term [1, Equation 26:12:2]:
(57)enx=∑k=0n‍xkk!.Therefore, we finally obtain(58)∫1z‍fβ-1xdx=∑n=0∞‍βn+1n+2×-αn+1n!-en-n+2αn+2n+11n+2n+1+-αn+1n!-en-n+2αn+2n+1en+2α,
where α=W0-log⁡z/β, z>1 and β<0.Notice as well that, forβ>0, according to (31), we have to calculate
(59)∫z1‍fβ-1xdx=∫z1‍-log⁡x/βW0-log⁡x/βdx,z∈0,1,
but, according to (58), we have
(60)∫z1‍fβ-1xdx=-∑n=0∞‍βn+1n+2×1n+2n+1+-αn+1n!-en-n+2αn+2n+1en+2α,
where α=W0-log⁡z/β, z∈0,1 and β>0.


### 4.2. Indefinite Integral offβ

The indefinite integral(61)∫‍xxdx
cannot be expressed in terms of a finite number of elementary functions [10]. Moreover, closed expression for (61) in the usual literature does not seem to be found. However, it can be expressed in closed form by using the upper incomplete gamma function [9, Equation 8.2.1]:
(62)Γa,z:=∫z∞‍ta-1e-tdt,Rea>0.Notice that ifz=0 and a=n is a positive integer, then we recover the usual gamma function:
(63)Γn,0=Γn=n-1!,n∈Z+.We can generalize (61), calculating the integral
(64)∫0z‍fβxdx=∫0z‍exp⁡-βxlog⁡xdx,z>0,β∈R.Expanding in power-series the exponential function given in (64) and integrating term by term, we get
(65)∫0z‍fβxdx=∑n=0∞‍-βnn!∫0z‍xlog⁡xndx.Performing now the change of variablesxn+1=e-t, we obtain
(66)∫0z‍fβxdx=∑n=0∞‍-βn-1n+1n+1!n+1n∫∞-n+1log⁡z‍e-ttndt=∑n=1∞‍βn-1n!nn-1∫-nlog⁡z∞‍e-ttn-1dt.By using now the definition of the upper incomplete gamma function (62), we arrive at
(67)∫0z‍x-βxdx=∑n=1∞‍βn-1n-1!nnΓn,-nlog⁡z.In [11, Lemma 10.6], we find a similar expression for the indefinite integral of the power tower
(68)∫‍xxx⋯dx=∑n=1∞‍-1n-1nn-2n-1!Γn,-log⁡x.Takingz=1 in (67) and using (63), we recover the expression given by [12, Equation 3.342]:
(69)∫01‍x-βxdx=∑n=1∞‍βn-1nn.Moreover, takingβ=±1, in (69), we recover the expressions given for the sophomore’s dream [13, pp. 4, 44], discovered in 1697 by Johann Bernoulli, as follows:
(70)∫01‍x∓xdx=∑n=1∞‍±1n-1nn.In order to compute the upper incomplete gamma function given in (67), we may use [1, Equation 45:4:2]:
(71)Γn,x=n-1!e-xen-1x,
where enx is the exponential polynomial (57). Therefore, (67) can be rewritten as
(72)∫0z‍x-βxdx=1β∑n=1∞‍βznnen-1-nlog⁡z.Let us now proceed to give another expression for the indefinite integral offβ by using the results given in (58) and (60). First, let us consider the cases β<0 and z>1, splitting (64) into two summands, as follows:
(73)∫0z‍x-βxdx=∫01‍x-βxdx+∫1z‍x-βxdx=∑n=1∞‍βn-1nn+∫1z‍x-βxdx,
where the first integral in (73) has been substituted by (69) and the second integral can be computed by knowing that fβx is an increasing function for x>1 when β<0.Indeed, according to Figure4, we have
(74)z×z-βz=1+∫1z‍fβxdx+∫1z-βz‍fβ-1xdx.
So, from (73) and (74) and taking into account (58), we get
(75)∫0z‍x-βxdx=z1-βz-1+∑n=1∞‍βn-1nn-∑n=0∞‍βn+1n+21n+2n+1+-αn+1n!-en-n+2αn+2n+1en+2α.Figure 4
Scheme for the integration offβ(x),  β<0, beyond z>1.Since the following series is a telescoping series:(76)∑n=0∞‍βnn+1n+1-βn+1n+2n+2=1,
we can simplify (75), obtaining
(77)∫0z‍x-βxdx=z1-βz-eα∑n=0∞‍βeαn+1n+2-αn+1n!-en-n+2αn+2n+1,
where α=W0zlog⁡z, z>1 and β<0.For the case ofβ>0, according to Figure 5, we have
(78)∫1z‍x-βxdx=∫z-βz1‍fβ-1xdx-1-z-βz︸A+z-1z-βz.Figure 5
Scheme for the integration offβ(x),  β>0, for z>1.Therefore, substituting (78) in (73) and taking into account (60) and (76), we arrive again at (77), but for β>0. Moreover, the range z>1 can be extended up to the point where fβ is a monotonic increasing or decreasing function. So, according to (24), we can say that z>1/e. Then, collecting all these results, we can conclude that
(79)∫0z‍x-βxdx=z1-βz-eα∑n=0∞‍βeαn+1n+2-αn+1n!-en-n+2αn+2n+1,
where α=W0zlog⁡z, z>1/e, and β>0.


## 4.1. Indefinite Integral offβ-1

Forβ<0, according to (30), let us calculate
(46)∫1z‍fβ-1xdx=∫1z‍-log⁡x/βW0-log⁡x/βdx,z>1.Performing the change of variablesu=-log⁡x/β, we have
(47)∫1z‍fβ-1xdx=-β∫0-log⁡z/β‍ue-βuW0udu,
and expanding the exponential function e-βu, in (47), we obtain
(48)∫1z‍fβ-1xdx=∑n=0∞‍-βn+1n!∫0-log⁡z/β‍un+1W0udu.Performing now the change of variablesW0u=s (i.e., u=ses), we get
(49)∫1z‍fβ-1xdx=∑n=0∞‍-βn+1n!∫0α‍sns+1en+2sds,
where we have set
(50)α:=W0-log⁡zβ.The integral given in (49) can be expressed in terms of the lower incomplete gamma function [9, Equation 8.2.1]. Consider
(51)γa,z:=∫0z‍ta-1e-tdt,Rea>0;
thus, performing the change of variables -t=ks, we get
(52)∫0a‍smeksds=-k-1-m∫0-ka‍tme-tdt=-k-1-mγm+1,-ka.Therefore, applying (52) in (49), we obtain
(53)∫1z‍fβ-1xdx=∑n=0∞‍βn+1n!n+2n+1×γn+1,-n+2α-γn+2,-n+2αn+2.Applying now the property [9, Equation 8.8.1]
(54)γa+1,z=aγa,z-zae-z,
we may rewrite (53) as
(55)∫1z‍fβ-1xdx=∑n=0∞‍βn+1n!γn+1,-n+2αn+2n+2+-αn+1n+2en+2α.In order to compute the lower incomplete gamma function given in (55), we may use [1, Equation 45:4:1]
(56)γn,x=n-1!1-en-1xe-x,
where enx is theexponential polynomial and it is given by the power-series expansion of the exponential function by truncation after the nth term [1, Equation 26:12:2]:
(57)enx=∑k=0n‍xkk!.Therefore, we finally obtain(58)∫1z‍fβ-1xdx=∑n=0∞‍βn+1n+2×-αn+1n!-en-n+2αn+2n+11n+2n+1+-αn+1n!-en-n+2αn+2n+1en+2α,
where α=W0-log⁡z/β, z>1 and β<0.Notice as well that, forβ>0, according to (31), we have to calculate
(59)∫z1‍fβ-1xdx=∫z1‍-log⁡x/βW0-log⁡x/βdx,z∈0,1,
but, according to (58), we have
(60)∫z1‍fβ-1xdx=-∑n=0∞‍βn+1n+2×1n+2n+1+-αn+1n!-en-n+2αn+2n+1en+2α,
where α=W0-log⁡z/β, z∈0,1 and β>0.


## 4.2. Indefinite Integral offβ

The indefinite integral(61)∫‍xxdx
cannot be expressed in terms of a finite number of elementary functions [10]. Moreover, closed expression for (61) in the usual literature does not seem to be found. However, it can be expressed in closed form by using the upper incomplete gamma function [9, Equation 8.2.1]:
(62)Γa,z:=∫z∞‍ta-1e-tdt,Rea>0.Notice that ifz=0 and a=n is a positive integer, then we recover the usual gamma function:
(63)Γn,0=Γn=n-1!,n∈Z+.We can generalize (61), calculating the integral
(64)∫0z‍fβxdx=∫0z‍exp⁡-βxlog⁡xdx,z>0,β∈R.Expanding in power-series the exponential function given in (64) and integrating term by term, we get
(65)∫0z‍fβxdx=∑n=0∞‍-βnn!∫0z‍xlog⁡xndx.Performing now the change of variablesxn+1=e-t, we obtain
(66)∫0z‍fβxdx=∑n=0∞‍-βn-1n+1n+1!n+1n∫∞-n+1log⁡z‍e-ttndt=∑n=1∞‍βn-1n!nn-1∫-nlog⁡z∞‍e-ttn-1dt.By using now the definition of the upper incomplete gamma function (62), we arrive at
(67)∫0z‍x-βxdx=∑n=1∞‍βn-1n-1!nnΓn,-nlog⁡z.In [11, Lemma 10.6], we find a similar expression for the indefinite integral of the power tower
(68)∫‍xxx⋯dx=∑n=1∞‍-1n-1nn-2n-1!Γn,-log⁡x.Takingz=1 in (67) and using (63), we recover the expression given by [12, Equation 3.342]:
(69)∫01‍x-βxdx=∑n=1∞‍βn-1nn.Moreover, takingβ=±1, in (69), we recover the expressions given for the sophomore’s dream [13, pp. 4, 44], discovered in 1697 by Johann Bernoulli, as follows:
(70)∫01‍x∓xdx=∑n=1∞‍±1n-1nn.In order to compute the upper incomplete gamma function given in (67), we may use [1, Equation 45:4:2]:
(71)Γn,x=n-1!e-xen-1x,
where enx is the exponential polynomial (57). Therefore, (67) can be rewritten as
(72)∫0z‍x-βxdx=1β∑n=1∞‍βznnen-1-nlog⁡z.Let us now proceed to give another expression for the indefinite integral offβ by using the results given in (58) and (60). First, let us consider the cases β<0 and z>1, splitting (64) into two summands, as follows:
(73)∫0z‍x-βxdx=∫01‍x-βxdx+∫1z‍x-βxdx=∑n=1∞‍βn-1nn+∫1z‍x-βxdx,
where the first integral in (73) has been substituted by (69) and the second integral can be computed by knowing that fβx is an increasing function for x>1 when β<0.Indeed, according to Figure4, we have
(74)z×z-βz=1+∫1z‍fβxdx+∫1z-βz‍fβ-1xdx.
So, from (73) and (74) and taking into account (58), we get
(75)∫0z‍x-βxdx=z1-βz-1+∑n=1∞‍βn-1nn-∑n=0∞‍βn+1n+21n+2n+1+-αn+1n!-en-n+2αn+2n+1en+2α.Figure 4
Scheme for the integration offβ(x),  β<0, beyond z>1.Since the following series is a telescoping series:(76)∑n=0∞‍βnn+1n+1-βn+1n+2n+2=1,
we can simplify (75), obtaining
(77)∫0z‍x-βxdx=z1-βz-eα∑n=0∞‍βeαn+1n+2-αn+1n!-en-n+2αn+2n+1,
where α=W0zlog⁡z, z>1 and β<0.For the case ofβ>0, according to Figure 5, we have
(78)∫1z‍x-βxdx=∫z-βz1‍fβ-1xdx-1-z-βz︸A+z-1z-βz.Figure 5
Scheme for the integration offβ(x),  β>0, for z>1.Therefore, substituting (78) in (73) and taking into account (60) and (76), we arrive again at (77), but for β>0. Moreover, the range z>1 can be extended up to the point where fβ is a monotonic increasing or decreasing function. So, according to (24), we can say that z>1/e. Then, collecting all these results, we can conclude that
(79)∫0z‍x-βxdx=z1-βz-eα∑n=0∞‍βeαn+1n+2-αn+1n!-en-n+2αn+2n+1,
where α=W0zlog⁡z, z>1/e, and β>0.



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*Source: 195329-2014-11-23.xml*