# The Multiple Gamma-Functions and the Log-Gamma Integrals

Authors: X.-H. Wang; Y.-L. Lu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

In this paper, which is a companion paper to [W],
starting from the Euler integral which appears in a generalization
of Jensen’s formula, we shall give a closed form for the integral
of logΓ(1±t). This enables us to locate the genesis of two new
functions A1/a and C1/a considered by Srivastava and Choi. We
consider the closely related function A(a) and the Hurwitz zeta function,
which render the task easier than working with the A1/a functions themselves. We shall also give a direct proof of Theorem
4.1, which is a consequence of [CKK, Corollary 1.1], though.


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## Body

## 1. Introduction

Iff(z) is analytic in a domain D containing the circle C:|z|=r and has no zero on the circle, then the Gauss mean value theorem
(1.1)log|f(0)|=12π∫02π‍log|f(reiθ)|dθ
is true. In [1, page 207] the case is considered where f(z) has a zero reiθ0 on the circle, and (1.1) turns out that the Euler integral
(1.2)∫0π/2‍logsinxdx=-π2log2
which is essential in proving a generalization of Jensen's formula [1, pages 207-208].LetG denote the Catalan constant defined by the absolutely convergent series
(1.3)G=∑n=1∞‍(-1)n-1(2n-1)2=L(2,χ4),
where χ4 is the nonprincipal Dirichlet character mod 4.As a next step from (1.2) the relation
(1.4)∫0π/4‍logsintdt=-π4log2-12G
holds true. In this connection, in [2] we obtained some results on G viewing it as an intrinsic value to the Barnes G-function. The Barnes G-function (which is Γ2-1 in the class of multiple gamma functions) is defined as the solution to the difference equation (cf. (2.3))
(1.5)logG(z+1)-logG(z)=logΓ(z)
with the initial condition
(1.6)logG(1)=0
and the asymptotic formula to be satisfied
(1.7)logG(z+N+2)=N+1+z2log2π+12(N2+2N+1+B2+z2+2(N+1)z)logN-34N2-N-Nz-logA+112+O(N-1),N→∞, where Γ(s) indicates the Euler gamma function (cf., e.g., [3]).Invoking the reciprocity relation for the gamma function(1.8)Γ(s)sinπs=πΓ(1-s),
it is natural to consider the integrals of logΓ(α+t) or of multiple gamma functions Γr (cf., e.g., [4, 5]). Barnes’ theorem [6, page 283] reads
(1.9)∫0a‍logΓ(α+t)dt=-logG(α+a)G(α)-(1-α)logΓ(α+a)Γ(α)+alogΓ(α+a)-12a2+12(log2π+1-2α)a
valid for nonintegral values of a.In this paper, motivated by the above, we proceed in another direction to developing some generalizations of the above integrals considered by Srivastava and Choi [7]. For q-analogues of the results, compare the recent book of the same authors [8]. Our main result is Theorem 2.1 which gives a closed form for ∫0a‍logΓ(1-t)dt and locates its genesis. A slight modification of Theorem 2.1 gives the counterpart of Barnes’ formula (1.9) which reads.Corollary 1.1.
Except for integral values ofa, one has
(1.10)∫0a‍logΓ(α-t)dt=logG(α-a)G(α)+(1-α)logΓ(α-a)Γ(α)+alogΓ(α-a)+12a2+12(log2π+1-2α)a.Srivastava and Choi introduced two functionslogA1/a and logC1/a by (2.9) and (2.9) with formal replacement of 1/a by -1/a, respectively. They state C1/a=A-1/a, which is rather ambiguous as to how we interpret the meaning because (2.9) is defined for a>0 [7, page 347, l.11]. They use this C1/a function to express the integral ∫0a‍logΓ(1-t)dt, without giving proof. This being the case, it may be of interest to locate the integral of logΓ(1-t) [7, (13), page 349], thereby logC1/a [7, page 347].For this purpose we use a more fundamental functionA(a) than A1/a defined by
(1.11)logA(a)=-ζ′(-1,a)+112,
where ζ(s,a) is the Hurwitz zeta-function
(1.12)ζ(s,a)=∑n=0∞‍1(n+a)s,Res=σ>1
in the first instance. For its theory, compare, for instance, [3], [9, Chapter 3].We shall prove the following corollary which gives the right interpretation of the functionC1/a.Corollary 1.2.
For0<a<1,
(1.13)logC1/a=logA(1-a)-14a2,
or
(1.14)logC1/a=logA1-1/a+14(1-a)2+(1-a)log(1-a)-14a2.


## 2.  Barnes Formula

There is a generalization of (1.4) as well as (1.2) in the form [7, equation (28), page 31]:
(2.1)∫0a‍logsinπtdt=alogsinπa2π+logG(1+a)G(1-a),a∉ℤ.
Equation (2.1) is Barnes’ formula [6, page 279] which is equivalent to Kinkelin's 1860 result [10] [7, equation (26), page 30]:
(2.2)∫0z‍πtcotπtdt=logG(1-z)G(1+z)+zlog2π.
Since (1.5) is equivalent to
(2.3)G(z+1)=G(z)Γ(z),
it follows that
(2.4)∫0a‍logsinπtdt=alogsinπa2π+logG(a)G(1-a)+logΓ(a).
Putting a=1/2, we obtain
(2.5)π-1∫0π/2‍logsinxdx=∫01/2‍logsinπtdt=-12log2π+logΓ(12)=-12log2,
which is (1.2).The counterpart of (2.1) follows from the reciprocity relation (1.8), known as Alexeievsky's Theorem [7, equation (42), page 32]. 						(2.6)∫0a‍logΓ(1+t)dt=12(log2π-1)a-a22+alogΓ(a+1)-logG(a+1),
which in turn is a special case of (1.9).Indeed, in [7, page 207], only (1.9) and the integral of logG(t+α) are in closed form and the integral of logΓ3(t+α) is not. A general formula is given by Barnes [4] with constants to be worked out. We shall state a concrete form for this integral in Section 3, using the relation [7, equation (455), page 210] between logΓ3(t+α) and the integral of ψ and appealing to a closed form for the latter in [11].Formula (2.6) is stated in the following form [7, equation (12), page 349]:
(2.7)∫0a‍logΓ(1+t)dt=12(log2π-1)a-34a2+logA-logA1/a,
where logA is the Glaisher-Kinkelin constant defined by [7, equation (2), page 25]
(2.8)logA=limN→∞(∑n=1N‍nlogn-12(N2+N+B2)logN+14N2),
and logA1/a is defined by [7, equation (9), page 347]
(2.9)logA1/a=limN→∞(∑n=1N‍(n+a)log(n+a)-12(N2+(2a+1)N+a2+a+B2)log(N+a)+14N2+a2N∑n=1N‍),
for a>0.Comparing (2.6) and (2.7), we immediately obtain
(2.10)logA1/a=logG(a+1)-alogΓ(a+1)+logA-a24=logG(a)+(1-a)logΓ(a)+loga-a24-aloga,
on using the difference relation Γ(a+1)=aΓ(a).Thus, in a sense we have located the genesis of the functionlogA1/a. although they prove (2.7) by an elementary method [7, page 348].Indeed,A1/a and A(a) are almost the same:
(2.11)logA1/a=logA(a)-14a2-aloga,
a proof being given below. However, logA(a) is more directly connected with ζ'(-1,a) for which we have rich resources of information as given in [9, Chapter 3].We prove the following theorem which gives a closed form for∫0a‍logΓ(1-t)dt, thereby giving the genesis of the constant C1/a.Theorem 2.1.
Fora∉ℤ, one has
(2.12)∫0a‍logΓ(1-t)dt=logG(1-a)+alogΓ(1-a)+12a2+12(log2π-1)a.
If 0<a<1, then
(2.13)∫0a‍logΓ(1-t)dt=logA(1-a)-logA+12a2+12(log2π-1)a.Proof.
We evaluate the integral(2.14)I=∫0a‍logΓ(1+t)sinπtdt
in two ways. First,
(2.15)I=alogπ+aloga-a-∫0a‍logΓ(1-t)dt.
On the other hand, noting that I is the sum of (2.1) and (2.7), we deduce that
(2.16)I=alogsinπa2π+logG(a+1)+logA-logG(1-a)+12(log2π-1)a-34a2-logA1/a.
Substituting (1.5), we obtain
(2.17)I=alogsinπa2π+alogΓ(a)+logA(a)-logA1/a-logG(1-a)+12(log2π-1)a-34a2.
The first two terms on the right of (2.17) become
(2.18)alogΓ(a)sinπa2π=alog12Γ(1-a)=-a(log2+logΓ(1-a)),
while the 3rd and the 4th terms give, in view of (2.11), (1/4)a2+aloga.
Hence, altogether(2.19)I=-alog2-alogΓ(1-a)-logG(1-a)+aloga-12a2+12(log2π-1)a.
Comparing (2.15) and (2.19) proves (2.12), completing the proof.Comparing (2.13) and [7, equation (13), page 349]
(2.20)∫0a‍logΓ(1-t)dt=logA(1-a)-logA+34a2+12(log2π-1)a,
we prove Corollary 1.2.Hence the relation betweenC1/a and A1/a is (1.14), that is, one between C1/a and A1-1/a rather than C1/a=A-1/a as Srivastava and Choi state.At this point we shall dwell on the underlying integral representation for (the derivative of) the Hurwitz zeta-function, which makes the argument rather simple and lucid as in [12] and gives some consequences.Proof of (2.11).
Consider that(2.21)ζ′(s,a)-112=-12a2loga-14a2-12aloga-B22loga-13!∫0∞‍B¯3(t)(t+a)-2dt
[9, (3.15), page 59], where the last integral may be also expressed as
(2.22)-12!∫0∞‍B¯2(t)(t+a)-1dt,
and where B¯k(t) is the kth periodic Bernoulli polynomial. Then
(2.23)-ζ′(-1,a)=∑0≤n≤x‍(n+a)log(n+a)-12(x+a)2log(x+a)+14(x+a)2+B¯1(x)(x+a)-12B¯2(x)(x+a)+O(x-1logx);
whence in particular, we have the generic formula for ζ'(-1,a) and consequently for logA(a) through (1.11):
(2.24)logA(a)=limN→∞(∑n=0∞‍(n+a)log(n+a)-12log(N+a)×((N+a)2+N+a+B2)+14(N+a)2∑n=0∞‍).
This may be slightly modified in the form
(2.25)logA(a)=limN→∞(∑n=0N‍(n+a)log(n+a)-12(N2+(2a+1)N+a2+a+B2)log(N+a)+14N2+12aN∑n=0N‍)+14a2+aloga.
Comparing (2.9) and (2.25), we verify (2.11).The merit of usingA(a) is that by way of ζ'(-1,a), we have a closed form for it:
(2.26)logA(a)=12a2loga-14a2+12aloga+B22loga+12!∫0∞‍B¯2(t)(t+a)-1dt.
In the same way, via another important relation [7, equation (23), page 94],
(2.27)logG(a)=-(ζ′(-1,a)-112)-logA-(1-a)logΓ(a).
Equation (2.21) gives a closed form for logG(a), too. We also have from (1.11) and (2.27)
(2.28)logA(a)=logG(a)+(1-a)logΓ(a)+logA=logG(a+1)-alogΓ(a)+logA.There are some known expressions not so handy as given by (2.27). For example, [7, page 25] and [7, equation (440), page 206], one of which reads
(2.29)G′G(1+z)=∑n=1∞‍(nz+n-1+zn)+12(log2π-1)-(1+γ)z,
with γ designating the Euler constant. Equation (2.29) is a basis of (2.2) (cf. proof of [2, Lemma 1]).Remark 2.2.
The Glaisher-Kinkelin constantA is connected with A(1) and A1 as follows:
(2.30)logA=logA(1)=logA1+14.
This can also be seen from Vardi's formula [7, (31), page 97]:
(2.31)logA=-ζ′(-1)+112,
which is (1.11) with a=1.We may also give another direct proof of Corollary1.2.Proof of Corollary1.2 (another proof).
log
C
1
/
a is the limit of the expression
(2.32)SN=∑k=1N‍(k-1+α)log(k-1+α)-(12N2+(α-12)N+12B2(α))×log(N-1+α)+14N2+N2(α-1),
where α=1-a. Let N=M+1. Then
(2.33)SN=∑k=0M‍(k+α)log(k+α)-(12(M+1)2+(α-12)(M+1)+12B2(α))×log(M+α)+14(M+1)2+M+12α-M+12.
Hence, simplifying, we find that(2.34)SN=∑k=1M‍(k+α)log(k+α)-(12M2+(α+12)M+12(α2+α+B2))×log(M+α)+14M2+12αM+αlogα-(α-1)24+14α2.
Hence(2.35)logC1/a=logAα+αlogα-(α-1)24+14α2,
which is (1.14). This completes the proof.As an immediate consequence of Corollary1.2, we prove (2.36) as can be found in [7, pages 350–351]. 						(2.36)A1/a=(πasinπa)-aG(1+a)G(1-a)C1/a,0<a<1.Proof of (2.36).
From (2.28), (1.5), and (1.8), we obtain
(2.37)logA(a)-logA(1-a)=logG(1+a)G(1-a)-alogπsinπa.
On the other hand, by (2.11) and (1.13), we see that the left-hand side of (2.37) is
(2.38)logA1/aC1/a+aloga,
whence we conclude that
(2.39)logA1/aC1/a=logG(1+a)G(1-a)-alogπasinπa.
On exponentiating, (2.37) leads to (2.36).


## 3. Polygamma Function of Negative Order

In this section we introduce the functionA~k(q) [13]:
(3.1)A~k(q)=kζ′(1-k,q),
which is closely related to the polygamma function of negative order and states some simple applications. We recall some properties of A~k(q):
(3.2)A~2(q+1)=A~2(q)+2qlogq,A~2(12)=-ζ'(-1)-112log2,A~2(14)=-14ζ′(-1)+G2π,(3.3)A~2(34)=-12ζ'(-1)-A~2(14).Equation (3.3) is [2, equation (2.31)], which is used in proving [2, Theorem 2] and can be read off from the distribution property [9, equation (3.72), page 76] as follows:
(3.4)∑a=14‍ζ(s,a4)=4sζ(s).Differentiation gives(3.5)∑n=14‍ζ'(s,a4)=4s((log4)ζ(s)+ζ′(s)).
Putting s=-1, we obtain
(3.6)ζ′(-1)+ζ'(-1,12)+ζ'(-1,14)+ζ'(-1,34)=4-1((log4)ζ(-1)+ζ′(-1)),
which we solve in ζ'(-1,3/4):
(3.7)ζ′(-1,34)=14((2log2)ζ(-1)+ζ′(-1))-ζ′(-1)-12A~2(12)-ζ'(-1,14).Substituting (3.2) and ζ(-1)=-B2/2=-1/12 and simplifying, we conclude that
(3.8)ζ'(-1,34)=-14ζ′(-1)-ζ'(-1,14)
and that
(3.9)A~2(34)=2ζ'(-1,34)=-12ζ'(-1)-2ζ'(-1,14),
whence (3.3).Using these, we deduce from (2.37) the following.Example 3.1.
(3.10)
log
A
1
/
4
=
5
64
+
1
2
log
2
-
1
8
log
A
-
G
2
π
.Proof.
By (1.11) and (3.1), for q>0,
(3.11)logA(q)=-12A~2(q)+112.
SincelogA(1/4)-logA(3/4)=-1/2(A~2(1/4)-A~2(3/4)), it follows from (3.3) that the left-hand side of (2.37) is
(3.12)-A~2(14)-14ζ′(-1),
which is
(3.13)2logA(14)-16+14(logA-112)
where we used (2.31).
The right-hand side of (2.37), log(G(5/4)/G(3/4))-1/4log(π/sin(π/4)), becomes -(G/2π), in view of known values of G [7, page 30].
Hence, altogether, (2.37) with a=1/4 reads
(3.14)-G2π=2logA(14)+14logA-316.
Invoking (2.11), this becomes (3.10).We note that (3.14) gives a proof of the third equality in (3.2). Both (2.36) and (3.10) are contained in [14, 1999a] and are given as exercises in [7].


## 4. The Triple Gamma Function

For general material, we refer to [7, page 42]. As can been seen on [7, page 207], the important integral ∫0z‍logΓ3(t+a)dt is not in closed form. Recently, Chakraborty-Kanemitsu-Kuzumaki [5, Corollary 1.1] have given a general expressions for all the integrals in logΓr, by appealing to Barnes' original results.In this section, we shall give a direct derivation of a closed form by combining [7, (455), page 210] and [11, Corollary 3] (with λ=3). The first reads
(4.1)2∫0z‍logΓ3(t+a)dt=-∫0z‍t3ψ(t+a)dt+2zlogΓ3(z+a)-2(2a-3)logΓ3(z+a)logΓ3(a)+(3a2-9a+7)logG(z+a)logG(a)-(a-1)3logΓ(z+a)logΓ(a)+38z4+13(1-log2π)z3+(-34a2+74a-98+14(2a-3)log2π+logA)z2,+(a2-32a+14+12(a-2-3a+2)log2π+2(3-2a)logA)z,
while the second reads (cf. also [15])
(4.2)∫0z‍t3logψ(t+a)dt=-∑r=03‍C3(r,a)logΓr+1(a+z)Γr+1(a)-∑l=13‍(-1)l((3l)ζ′(l-3)+B4-l(a)l(4-l))zl+1124z4,
where C3(r,a) are defined by
(4.3)C3(r,a)=(-1)rr!∑m=r3‍(3m)(-1)mS(m,n)(a-1)3-m
and where S(m,n) are the Stirling numbers of the second kind [7, page 58]. To express the values of ζ′(l-3), we appeal to [7] (i)
ζ'(0)=-(1/2)log2π [7, (20), page 92],(ii)
ζ'(-2)=logB=ζ(3)/4π2 [7, pages 99-100]and (2.31). After some elementary but long calculations, we arrive at
(4.4)∫0z‍t3logψ(t+a)dt=-3!logΓ4(a+z)Γ4(a)-6(a-2)logΓ3(a+z)Γ3(a)-(3a2-9a+7)logΓ2(a+z)Γ2(a)-(a-1)3logΓ(a+z)Γ(a)1124z4+(-12log2π+13B1(a))z3-(14-3logA+14B2(a))z2+3(logB+13B3(a))z.
Combining we have the following.Theorem 4.1 (see [5, Example 2.3]).
Except for the singularities of the multiple gamma function, one has(4.5)∫0z‍logΓ3(t+a)dt=3logΓ4(a+z)Γ4(a)+zlogΓ3(z+a)+(a-3)logΓ3(a+z)Γ3(a)-124z4-16(a-32-12log2π)z3+18(-2a2+6a-103+(2a-3)log2π-8logA)z2-12(a3-52a2+2a-14+12(a2-3a+2)log2π+2(2a-3)logA+3logB52)z.This theorem enables us to put many formulas in [7] in closed form including, for instance, [7, (698), page 245]. Compare [5].



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