# Exponential Polynomials, Stirling Numbers, and Evaluation of Some Gamma Integrals

Authors: Khristo N. Boyadzhiev
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

This article is a short elementary review of the exponential polynomials (also
called single-variable Bell polynomials) from the point of view of analysis. Some new properties
are included, and several analysis-related applications are mentioned. At the end of the paper one
application is described in details—certain Fourier integrals involvingΓ(a+it) and Γ(a+it)Γ(b−it) are evaluated in terms of Stirling numbers.


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## Body

## 1. Introduction

We review the exponential polynomialsϕn(x) and present a list of properties for easy reference. Exponential polynomials in analysis appear, for instance, in the rule for computing derivatives like (d/dt)neaet and the related Mellin derivatives:(1.1)(xddx)nf(x),(ddxx)nf(x).
Namely, we have
(1.2)(ddt)neaet=ϕn(aet)eaet,
or, after the substitution x=et,(1.3)(xddx)neax=ϕn(ax)eax.We also include in this review two properties relating exponential polynomials to Bernoulli numbers,Bk. One is the semiorthogonality(1.4)∫-∞0ϕn(x)ϕm(x)e2xdxx=(-1)n2n+m-1n+mBn+m,
where the right-hand side is zero if n+m is odd. The other property is (2.25).At the end we give one application. Using exponential polynomials we evaluate the integrals(1.5)∫ℝe-itλtnΓ(a+it)dt,∫ℝe-itλtnΓ(a+it)Γ(b-it)dt
for n=0,1,…, in terms of Stirling numbers.


## 2. Exponential Polynomials

The evaluation of the series(2.1)Sn=∑k=0∞knk!,n=0,1,2,…
has a long and interesting history. Clearly, S0=e, with the agreement that 00=1. Several reference books (e.g., [1]) provide the following numbers:(2.2)S1=e,S2=2e,S3=5e,S4=15e,S5=52e,S6=203e,S7=877e,S8=4140e
As noted by Gould in [2, page 93], the problem of evaluating Sn appeared in the Russian journal Matematicheskii   Sbornik, 3 (1868), page 62,  with solution ibid, 4 (1868-9), page 39. Evaluations are presented also in two papers by Dobinski and Ligowski. In 1877 Dobinski [3] evaluated the first eight series S1,…,S8 by regrouping
(2.3)S1=∑1∞kk!=1+22!+33!+⋯=1+11!+12!+⋯=e,S2=∑1∞k2k!=1+222!+323!+⋯=1+21!+32!+43!+⋯={1+11!+12!+13!+⋯}+{11!+22!+33!+⋯}=e+S1=2e,
and continuing like that to S8. For large n this method is not convenient. However, later that year Ligowski [4] suggested a better method, providing a generating function for the numbers Sn:(2.4)eez=∑k=0∞ekzk!=∑k=0∞∑n=0∞knk!znn!=∑n=0∞Snznn!.
Further, an effective iteration formula was found(2.5)Sn=∑j=0n-1(n-1j)Sj
by which every Sn can be evaluated starting from S1.These results were preceded, however, by the work [5] of Grunert (1797–1872), professor at Greifswalde. Among other things, Grunert obtained formula (2.9) from which the evaluation of (2.1) follows immediately.The structure of the seriesSn hints at the exponential function. Differentiating the expansion
(2.6)ex=∑k=0∞xkk!
and multiplying both sides by x we get(2.7)xex=∑k=0∞kxkk!,
which, for x=1, gives S1=e. Repeating the procedure, we find S2=2e from(2.8)x(xex)′=(x+x2)ex=∑k=0∞k2xkk!,
and continuing like that, for every n=0,1,2,…, we find the relation(2.9)(xddx)nex=ϕn(x)ex=∑k=0∞knxkk!,
where ϕn are polynomials of degree n. Thus,(2.10)Sn=ϕn(1)e,∀n≥0.
The polynomials ϕn deserve a closer look. From the defining relation (2.9) we obtain(2.11)x(ϕnex)′=x(ϕn′+ϕn)ex=ϕn+1ex,
that is,
(2.12)ϕn+1=x(ϕn′+ϕn),
which helps to find ϕn explicitly starting from ϕ0:(2.13)ϕ0(x)=1,ϕ1(x)=x,ϕ2(x)=x2+x,ϕ3(x)=x3+3x2+x,ϕ4(x)=x4+6x3+7x2+x,ϕ5(x)=x5+10x4+25x3+15x2+x,
and so on. Another interesting relation, easily proved by induction, is
(2.14)ϕn+1(x)=x∑k=0n(nk)ϕk(x).
From (2.12) and (2.14) one finds immediately(2.15)ϕn′(x)=∑k=0n-1(nk)ϕk(x).Obviously,x=0 is a zero for all ϕn,  n>0. It can be seen that all the zeros of ϕn are real, simple, and nonpositive. The nice and short induction argument belongs to Harper [6].The assertion is true forn=1. Suppose that for some n the polynomial ϕn has n distinct real nonpositive zeros (including x=0). Then the same is true for the function(2.16)fn(x)=ϕn(x)ex.
Moreover, fn is zero at -∞ and by Rolle's theorem its derivative(2.17)ddxfn=ddx(ϕn(x)ex)
has n distinct real negative zeros. It follows that the function(2.18)ϕn+1(x)ex=xddx(ϕn(x)ex)
has n+1 distinct real nonpositive zeros (adding here x=0).The polynomialsϕn can be defined also by the exponential generating function (extending Ligowski's formula)(2.19)ex(ez-1)=∑n=0∞ϕn(x)znn!.
It is not obvious, however, that the polynomials defined by (2.9) and (2.19) are the same, so we need the following simple statement.Proposition 2.1.
The  polynomialsϕn(x)  defined  by (2.9)  are  exactly  the  partial derivatives (∂/∂z)nex(ez-1) evaluated  at z=0.Equation (2.19) follows from (2.9) after expanding the exponential exez in double series and  changing the order of summation. A different proof will be given later.Settingz=2kπi, k=±1,±2,…, in the generating function (2.19) one finds(2.20)e2kπi=1,ex(ez-1)=e0=1,
which shows that the exponential polynomials are linearly dependent:(2.21)1=∑n=0∞ϕn(x)(2kπi)nn!or0=∑n=1∞ϕn(x)(2kπi)nn!,k=±1,±2,….
In particular, ϕn are not orthogonal for any scalar product on polynomials. (However, they have the semiorthogonality property mentioned in Section 1 and proved in Section 4.)Comparing coefficient forz in the equation(2.22)e(x+y)ez=exezeyez
yields the binomial identity(2.23)ϕn(x+y)=∑k=0n(nk)ϕk(x)ϕn-k(y).
With y=-x this implies the interesting “orthogonality’’ relation for n≥1:(2.24)∑k=0n(nk)ϕk(x)ϕn-k(-x)=0.Next, letBn, n=0,1,…, be the Bernoulli numbers. Then for p=0,1,…, we have
(2.25)∫0xϕp(t)dt=1p+1∑k=1p+1(p+1k)Bp+1-kϕk(x).
For proof see Example  4  in [7, page 51], or [9].Some Historical Notes
As already mentioned, formula (2.9) appears in the work of Grunert [5, page 260], where he gives also the representation (3.4) and computes explicitly the first six exponential polynomials. The polynomials ϕn were studied more systematically (and independently) by S. Ramanujan in his unpublished notebooks. Ramanujan's work is presented and discussed by Berndt in [7, Part 1, Chapter 3]. Ramanujan, for example, obtained (2.19) from (2.9) and also proved (2.14), (2.15), and (2.25). Later, these polynomials were studied by Bell [10] and Touchard [11, 12]. Both Bell and Touchard called them “exponential’’ polynomials, because of their relation to the exponential function, for example, (1.2), (1.3), (2.9), and (2.19). This name was used also by Rota [13]. As a matter of fact, Bell introduced in [10] a more general class of polynomials of many variables, Yn,k, including  ϕn as a particular case. For this reason ϕn are known also as the single-variable Bell polynomials [14–17]. These polynomials are also a special case of the actuarial polynomials introduced by Toscano [18] which, on their part, belong to the more general class of Sheffer polynomials [19]. The exponential polynomials appear in a number of papers and in different applications—see [9, 13, 20–24] and the references therein. In [25] they appear on page 524 as the horizontal generating functions of the Stirling numbers of the second kind (see (3.4)).The numbers(2.26)bn=ϕn(1)=1eSn
are sometimes called exponential numbers, but a more established name is Bell numbers. They have interesting combinatorial and analytical applications [15, 16, 18, 26–32]. An extensive list of 202 references for Bell numbers is given in [33].We note that (2.9) can be used to extend ϕn to ϕz for any complex number z by the formula(2.27)ϕz(x)=e-x∑k=0∞kzxkk!
(Butzer et al. [34, 35]). The function appearing here is an interesting entire function in both variables, x and z. Another possibility is to study the polyexponential function(2.28)es(x,λ)=∑n=0∞xnn!(n+λ)s,
where Reλ>0. When s is a negative integer, the polyexponential can be written as a finite linear combination of exponential polynomials (see [9]).


## 3. Stirling Numbers and Mellin Derivatives

The iteration formula (2.12) shows that all polynomials ϕn have positive integer coefficients. These coefficients are the Stirling numbers of the second kind {nk} (or S(n,k))—see [25, 28, 36–39]. Given a set of n elements, {nk} represents the number of ways by which this set can be partitioned into k nonempty subsets (0≤k≤n). Obviously, {n1}=1,  {nn}=1 and a short computation gives {n2}=2n-1-1. For symmetry one sets {00}=1, {n0}=0. The definition of  {nk} implies the property
(3.1){n+1k}=k{nk}+{nk-1}
(see [38, page 259]) which helps to compute all {nk} by iteration. For instance,(3.2){n3}=3n-1-2n+12.
A general formula for the Stirling numbers of the second kind is(3.3){nk}=1k!∑j=1k(-1)k-j(kj)jn.Proposition 3.1.
For  everyn=0,1,2,…(3.4)ϕn(x)={n0}+{n1}x+{n2}x2+⋯+{nn}xn=∑k=0n{nk}xk.The proof is by induction and is left to the reader. Setting herex=1 we come to the well-known representation for the numbers Sn(3.5)Sn=e({n0}+{n1}+{n2}+⋯+{nn}).It is interesting that formula (3.4) is very old—it was obtained by Grunert [5, page 260] together with the representation (3.3) for the coefficients which are called now Stirling numbers of the second kind. In fact, coefficients of the form(3.6){nk}k!
appear in the computations of Euler—see [37].Next we turn to some special differentiation formulas. LetD=d/dx.Mellin Derivatives
It is easy to see that the first equality in (2.9) extends to (1.3), where a is an arbitrary complex number, that is,
(3.7)(xD)neax=ϕn(ax)eax
by the substitution x→ax. Even further, this extends to
(3.8)(xD)neaxp=pnϕn(axp)eaxp
for any a, p and n=0,1,… (simple induction and (2.12)). Again by induction, it is easy to prove that
(3.9)(xD)nf(x)=∑k=0n{nk}xkDkf(x)
for any n-times differentiable function f. This formula was obtained by Grunert [5, pages 257-258] (see also [2, page 89], where a proof by induction is given).As we know the action ofxD on exponentials, formula (3.9) can be “discovered’’ by using Fourier transform. Let
(3.10)F[f](t)=∫ℝe-ixtf(x)dx
be the Fourier transform of some function f. Then
(3.11)f(x)=12π∫ℝeixtF[f](t)dx,(xD)nf(x)=12π∫ℝeixtϕn(ixt)F[f](t)dx=∑k=0n{nk}xkF-1[(it)kF[f]](x)=∑k=0n{nk}xkDkf(x).Next we turn to formula (1.2) and explain its relation to (1.3). If we set x=et, then for any differentiable function f(3.12)ddtf=(ddxf)dxdt=(ddxf)et=(xD)f,
and we see that (1.2) and (1.3) are equivalent:
(3.13)(ddt)neaet=(xD)neax=ϕn(ax)eax=ϕn(aet)eaet.Proof of Proposition2.1.
We apply (1.2) to the function fx(z)=ex(ez-1)=exeze-x:
(3.14)(ddz)nfx(z)=ϕn(xez)fx(z).
From here, with z=0(3.15)(ddz)nfx(z)|z=0=ϕn(x)
as needed.Now we list some simple operational formulas. Starting from the obvious relation(3.16)(xD)nxk=knxk,n=0,1,…,k∈ℝ
for any function of the form(3.17)f(t)=∑n=0∞antn,
we define the differential operator(3.18)f(xD)=∑n=0∞an(xD)n
with action on functions g(x):(3.19)f(xD)g(x)=∑n=0∞an(xD)ng(x).
When g(x)=xk, (3.16) and (3.19) show that(3.20)f(xD)xk=∑n=0∞anknxk=f(k)xk.
If now
(3.21)g(x)=∑k=0∞ckxk
is a function analytical in a neighborhood of zero, the action of f(xD) on this function is given by
(3.22)f(xD)g(x)=∑k=0∞ckf(k)xk,
provided that the series on the right side converges. When f is a polynomial, formula (3.22) helps to evaluate series like
(3.23)∑k=0∞ckf(k)xk
in a closed form. This idea was exploited by Schwatt [40] and more recently by the present author in [20]. For instance, when g(x)=ex, (3.22) becomes(3.24)∑k=0∞f(k)xkk!=ex∑n=0∞anϕn(x).
As shown in [20] this series transformation can be used for asymptotic series expansions of certain functions.Leibniz Rule
The higher-order Mellin derivative(xD)n satisfies the Leibniz rule
(3.25)(xD)n(fg)=∑k=0n(nk)[(xD)n-kf][(xD)kg].
The proof is easy, by induction, and is left to the reader. We shall use this rule to prove the following proposition.Proposition 3.2.
For alln,m=0,1,2,…(3.26)ϕn+m(x)=∑k=0n‍∑j=0m(nk){mj}jn-kxjϕk(x).Proof.
One has(3.27)ϕn+m(x)=(xD)n+mex=(xD)n(xD)mex=(xD)n(ϕm(x)ex),
which by the Leibniz rule (3.25) equals
(3.28)∑k=0n(nk)[(xD)n-kϕm(x)][(xD)kex].
Using (3.3) and (3.16) we write
(3.29)(xD)n-kϕm(x)=∑j=0m{mj}jn-kxj,
and since also
(3.30)(xD)kex=ϕk(x)ex,
we obtain (3.26) from (3.27). The proof is completed.Settingx=1 in (3.25) yields an identity for the Bell numbers:
(3.31)bn+m=∑k=0n‍∑j=0m(nk){mj}jn-kbk.
This identity was recently published by Spivey [32], who gave a combinatorial proof. After that Gould and Quaintance [16] obtained the generalization (3.26) together with two equivalent versions. The proof in [16] is different from the one above.Using the Leibniz rule forxD we can prove also the following extension of property (2.24).Proposition 3.3.
For any two integersn,m≥0(3.32)(xD)nϕm(x)=∑k=0m{mk}knxk=∑k=0n‍(nk)ϕm+k(x)ϕn-k(-x).The proof is simple. Just compute(3.33)(xD)nϕm(x)=(xD)n[(e-x)(ϕm(x)ex)]=∑k=0n(nk)[(xD)n-ke-x][(xD)k(ϕm(x)ex)]
and (3.32) follows from (1.3).For completeness we mention also the following three properties involving the operatorDx. Proofs and details are left to the reader:
(3.34)(Dx)neax=ϕn+1(ax)axeax,(Dx)nf(x)=∑k=0n{n+1k+1}xkDkf(x),f(Dx)g(x)=∑k=0∞ckf(k+1)xk,
analogous to (1.3), (3.9), and (3.22) correspondingly.For a comprehensive study of the Mellin derivative we refer to [41–43].More Stirling Numbers
The polynomialsϕn, n=0,1,…, form a basis in the linear space of all polynomials. Formula (3.4) shows how this basis is expressed in terms of the standard basis 1,x,x2,…,xn,…. We can solve for xk in (3.4) and express the standard basis in terms of the exponential polynomials
(3.35)1=ϕ0,x=ϕ1,x2=-ϕ1+ϕ2,x3=2ϕ1-3ϕ2+ϕ3,x4=-6ϕ1+11ϕ2-6ϕ3+ϕ4,
and so forth. The coefficients here are also special numbers. If we write
(3.36)xn=∑k=0n(-1)n-k[nk]ϕk,
then [nk] are the (absolute) Stirling numbers of first kind, as defined in [38]. (The numbers [nk] are nonnegative. The symbol s(n,k)=(-1)n-k[nk] is used for Stirling numbers of the first kind with changing sign—see [28, 33, 39] for more details.) [nk] is the number of ways to arrange n objects into k cycles. According to this interpretation,
(3.37)[nk]=(n-1)[n-1k]+[n-1k-1],n≥1.


## 4. Semiorthogonality ofϕn

Proposition 4.1.
For everyn,m=1,2,…, one has
(4.1)∫0∞ϕn(-x)ϕm(-x)e-2xdxx=(-1)n-12n+m-1n+mBn+m.
Here Bk are the Bernoulli numbers. Note that the right-hand side is zero when k+m is odd, as all Bernoulli numbers with odd indices >1 are zeros.Using the representation (3.4) in (4.1) and integrating termwise one obtains an equivalent  form of (4.1):
(4.2)∑k=0n∑j=0m(-1)k+j{nk}{mj}(k+j-1)!2k+j=(-1)n-12n+m-1n+mBn+m.
This (double sum) identity extends the known identity [38, page 317, Problem 6.76]
(4.3)∑j=0m(-1)j+1{mj}j!2j+1=2m+1-1m+1Bm+1.
Namely, (4.3) results from (4.2) for n=1. The presence of (-1)n-1 at the right-hand side in (4.1) is not a “break of symmetry,’’ because when n+m is even, then n and m are both even or both odd.Proof.
Starting from(4.4)Γ(z)=∫0∞xz-1e-xdx,
we set x=eλ,  z=a+it, to obtain the representation
(4.5)Γ(a+it)=∫-∞+∞eiλteaλe-eλdλ,
which is a Fourier transform integral. The inverse transform is
(4.6)eaλe-eλ=12π∫ℝe-iλtΓ(a+it)dt.
When a=1, this is
(4.7)-eλe-eλ=ddλe-eλ=-12π∫ℝe-iλtΓ(1+it)dt.
Differentiating (4.7) n-1 times for λ we find
(4.8)(ddλ)ne-eλ=ϕn(-eλ)e-eλ=-12π∫ℝe-iλt(-it)n-1Γ(1+it)dt,
and Parceval's formula yields the equation
(4.9)∫ℝϕn(-eλ)ϕm(-eλ)e-2eλdλ=12π∫ℝ(-it)n-1(it)m-1|Γ(1+it)|2dt,
or, with x=eλ(4.10)∫0∞ϕn(-x)ϕm(-x)e-2xdxx=(-1)nin+m2π∫ℝtn+m-2πtsinh(πt)it.
The right-hand side is 0 when n+m is odd. When n+m is even, we use the integral [1, page 351]
(4.11)∫0∞t2p-1sinh(πt)dt=22p-12p(-1)p=1B2p
to finish the proof.Property (4.1) resembles the semiorthogonal property of the Bernoulli polynomials
(4.12)∫01Bn(x)Bm(x)dx=(-1)n-1n!m!(n+m)!Bn+m,
see, for instance, [25, page 530].


## 5. Gamma Integrals

We use the technique in the previous section to compute certain Fourier integrals and evaluate the moments ofΓ(a+it) and Γ(a+it)Γ(b-it).Proposition 5.1.
For everyn=0,1,… and a,b>0 one has
(5.1)∫ℝe-iμttnΓ(a+it)Γ(b-it)dt=in2πe-bμ∑k=0n∑m=0k(nk){km}(-1)man-kΓ(a+b+m)(1+e-μ)a+b+m,(5.2)∫ℝe-iλttnΓ(a+it)dt=in2πeaλe-eλ∑k=0n(nk)an-k∑m=0k{km}(-1)meλm.
In particular, when λ=μ=0, one obtains the moments
(5.3)Gn(a,b)≡∫ℝtnΓ(a+it)Γ(b-it)dt=inπ∑k=0n∑m=0k(nk){km}(-1)man-kΓ(a+b+m)2a+b+m-1,(5.4)Gn(a)≡∫ℝtnΓ(a+it)dt=2πine∑k=0n∑m=0k(nk){km}(-1)man-k.
Whenn=0 in (5.1) one has the known integral
(5.5)∫ℝe-iμtΓ(a+it)Γ(b-it)dt=2πΓ(a+b)e-bμ(1+e-μ)-a-b,
which can be found in the form of an inverse Mellin transform in [44].Proof.
Using again (4.6)
(5.6)eaλe-eλ=12π∫ℝe-iλtΓ(a+it)dt,
we differentiate both side n times
(5.7)(ddλ)n[eaλe-eλ]=12π∫ℝe-iλt(-it)nΓ(a+it)dt,
and then, according to the Leibniz rule and (1.2) the left-hand side becomes
(5.8)(ddλ)n[eaλe-eλ]=eaλe-eλ∑k=0n(nk)ϕk(-eλ)an-k.
Therefore,
(5.9)eaλe-eλ∑k=0n(nk)ϕk(-eλ)an-k=12π∫ℝe-iλt(-it)nΓ(a+it)dt,
and (5.2) follows from here.
Replacingλ by λ-μ we write (5.6) in the form
(5.10)ebλe-bμe-eλe-μ=12π∫ℝe-iλteiμtΓ(b+it)dt,
and then Parceval's formula for Fourier integrals applied to (5.9) and (5.10) yields
(5.11)e-bμ∑k=0n(nk)an-k∫ℝe(a+b)λe-eλ(1+e-μ)ϕk(-eλ)dλ=(-i)n2π∫ℝe-iμttnΓ(a+it)Γ(b-it)dt.
Returning to the variable x=eλ we write this in the form
(5.12)12π∫ℝe-iμttnΓ(a+it)Γ(b-it)dt=ine-bμ∑k=0n(nk)an-k∫0∞ϕk(-x)xa+b-1e-x(1+e-μ)dx=ine-bμ∑k=0n∑j=0k(nk){kj}an-k(-1)j∫0∞xa+b+j-1e-x(1+e-μ)dx=ine-bμ∑k=0n∑j=0k(nk){kj}an-k(-1)jΓ(a+b+j)(1+e-μ)a+b+j,
which is (5.1). The proof is complete.Next, we observe that for any polynomial(5.13)p(t)=∑n=0mantn
one can use (5.4) to write the following evaluation:(5.14)∫ℝp(t)Γ(a+it)dt=∑n=0manGn(a).
In particular, when a=1 we have(5.15)Gn(1)=2πine-1ϕn+1(-1),
and therefore,(5.16)∫ℝp(t)Γ(1+it)dt=2πe∑n=0maninϕn+1(-1).More applications can be found in the recent papers [9, 20, 21].



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