AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation16867210.1155/2009/168672168672Review ArticleExponential Polynomials, Stirling Numbers, and Evaluation of Some Gamma IntegralsBoyadzhievKhristo N.LittlejohnLanceDepartment of MathematicsOhio Northern UniversityAda, OH 45810USAonu.edu200926082009200915052009040820092009Copyright © 2009This 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.

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 involving Γ(a+it) and Γ(a+it)Γ(bit) are evaluated in terms of Stirling numbers.

1. Introduction

We review the exponential polynomials ϕn(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:

(xddx)nf(x),(ddxx)nf(x). Namely, we have (ddt)neaet=ϕn(aet)eaet, or, after the substitution x=et,

(xddx)neax=ϕn(ax)eax.

We also include in this review two properties relating exponential polynomials to Bernoulli numbers, Bk. One is the semiorthogonality

-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

e-itλtnΓ(a+it)dt,e-itλtnΓ(a+it)Γ(b-it)dt for n=0,1,, in terms of Stirling numbers.

2. Exponential Polynomials

The evaluation of the series

Sn=k=0knk!,n=0,1,2, has a long and interesting history. Clearly, S0=e, with the agreement that 00=1. Several reference books (e.g., ) provide the following numbers:

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  evaluated the first eight series S1,,S8 by regrouping S1=1kk!=1+22!+33!+=1+11!+12!+=e,S2=1k2k!=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  suggested a better method, providing a generating function for the numbers Sn:

eez=k=0ekzk!=k=0n=0knk!znn!=n=0Snznn!. Further, an effective iteration formula was found

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  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 series Sn hints at the exponential function. Differentiating the expansion ex=k=0xkk! and multiplying both sides by x we get

xex=k=0kxkk!, which, for x=1, gives S1=e. Repeating the procedure, we find S2=2e from

x(xex)=(x+x2)ex=k=0k2xkk!, and continuing like that, for every n=0,1,2,, we find the relation

(xddx)nex=ϕn(x)ex=k=0knxkk!, where ϕn are polynomials of degree n. Thus,

Sn=ϕn(1)e,n0. The polynomials ϕn deserve a closer look. From the defining relation (2.9) we obtain

x(ϕnex)=x(ϕn+ϕn)ex=ϕn+1ex, that is, ϕn+1=x(ϕn+ϕn), which helps to find ϕn explicitly starting from ϕ0:

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 ϕn+1(x)=xk=0n(nk)ϕk(x). From (2.12) and (2.14) one finds immediately

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 .

The assertion is true for n=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

fn(x)=ϕn(x)ex. Moreover, fn is zero at - and by Rolle's theorem its derivative

ddxfn=ddx(ϕn(x)ex) has n distinct real negative zeros. It follows that the function

ϕn+1(x)ex=xddx(ϕn(x)ex) has n+1 distinct real nonpositive zeros (adding here x=0).

The polynomials ϕn can be defined also by the exponential generating function (extending Ligowski's formula)

ex(ez-1)=n=0ϕn(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 polynomials ϕn(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.

Setting z=2kπi, k=±1,±2,, in the generating function (2.19) one finds

e2kπi=1,ex(ez-1)=e0=1, which shows that the exponential polynomials are linearly dependent:

1=n=0ϕn(x)(2kπi)nn!or0=n=1ϕn(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 for z in the equation

e(x+y)ez=exezeyez yields the binomial identity

ϕn(x+y)=k=0n(nk)ϕk(x)ϕn-k(y). With y=-x this implies the interesting “orthogonality’’ relation for n1:

k=0n(nk)ϕk(x)ϕn-k(-x)=0.

Next, let Bn, n=0,1,, be the Bernoulli numbers. Then for p=0,1,, we have 0xϕp(t)dt=1p+1k=1p+1(p+1k)Bp+1-kϕk(x). For proof see Example  4 in [7, page 51], or .

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  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 . As a matter of fact, Bell introduced in  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 . These polynomials are also a special case of the actuarial polynomials introduced by Toscano  which, on their part, belong to the more general class of Sheffer polynomials . The exponential polynomials appear in a number of papers and in different applications—see [9, 13, 2024] and the references therein. In  they appear on page 524 as the horizontal generating functions of the Stirling numbers of the second kind (see (3.4)).

The numbers

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, 2632]. An extensive list of 202 references for Bell numbers is given in .

We note that (2.9) can be used to extend ϕn to ϕz for any complex number z by the formula

ϕz(x)=e-xk=0kzxkk! (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

es(x,λ)=n=0xnn!(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 ).

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, 3639]. Given a set of n elements, {nk} represents the number of ways by which this set can be partitioned into k nonempty subsets (0kn). 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 {n+1k}=k{nk}+{nk-1} (see [38, page 259]) which helps to compute all {nk} by iteration. For instance,

{n3}=3n-1-2n+12. A general formula for the Stirling numbers of the second kind is

{nk}=1k!j=1k(-1)k-j(kj)jn.

Proposition 3.1.

For every n=0,1,2,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 here x=1 we come to the well-known representation for the numbers Sn

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

{nk}k! appear in the computations of Euler—see .

Next we turn to some special differentiation formulas. Let D=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, (xD)neax=ϕn(ax)eax by the substitution xax. Even further, this extends to (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 (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 of xD on exponentials, formula (3.9) can be “discovered’’ by using Fourier transform. Let F[f](t)=e-ixtf(x)dx be the Fourier transform of some function f. Then f(x)=12πeixtF[f](t)dx,(xD)nf(x)=12πeixtϕn(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

ddtf=(ddxf)dxdt=(ddxf)et=(xD)f, and we see that (1.2) and (1.3) are equivalent: (ddt)neaet=(xD)neax=ϕn(ax)eax=ϕn(aet)eaet.

Proof of Proposition <xref ref-type="statement" rid="prop1">2.1</xref>.

We apply (1.2) to the function fx(z)=ex(ez-1)=exeze-x: (ddz)nfx(z)=ϕn(xez)fx(z). From here, with z=0(ddz)nfx(z)|z=0=ϕn(x) as needed.

Now we list some simple operational formulas. Starting from the obvious relation

(xD)nxk=knxk,n=0,1,,k for any function of the form

f(t)=n=0antn, we define the differential operator

f(xD)=n=0an(xD)n with action on functions g(x):

f(xD)g(x)=n=0an(xD)ng(x). When g(x)=xk, (3.16) and (3.19) show that

f(xD)xk=n=0anknxk=f(k)xk. If now g(x)=k=0ckxk is a function analytical in a neighborhood of zero, the action of f(xD) on this function is given by f(xD)g(x)=k=0ckf(k)xk, provided that the series on the right side converges. When f is a polynomial, formula (3.22) helps to evaluate series like k=0ckf(k)xk in a closed form. This idea was exploited by Schwatt  and more recently by the present author in . For instance, when g(x)=ex, (3.22) becomes

k=0f(k)xkk!=exn=0anϕn(x). As shown in  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 (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 all n,m=0,1,2,ϕn+m(x)=k=0nj=0m(nk){mj}jn-kxjϕk(x).

Proof.

One has ϕn+m(x)=(xD)n+mex=(xD)n(xD)mex=(xD)n(ϕm(x)ex), which by the Leibniz rule (3.25) equals k=0n(nk)[(xD)n-kϕm(x)][(xD)kex]. Using (3.3) and (3.16) we write (xD)n-kϕm(x)=j=0m{mj}jn-kxj, and since also (xD)kex=ϕk(x)ex, we obtain (3.26) from (3.27). The proof is completed.

Setting x=1 in (3.25) yields an identity for the Bell numbers: bn+m=k=0nj=0m(nk){mj}jn-kbk. This identity was recently published by Spivey , who gave a combinatorial proof. After that Gould and Quaintance  obtained the generalization (3.26) together with two equivalent versions. The proof in  is different from the one above.

Using the Leibniz rule for xD we can prove also the following extension of property (2.24).

Proposition 3.3.

For any two integers n,m0(xD)nϕm(x)=k=0m{mk}knxk=k=0n(nk)ϕm+k(x)ϕn-k(-x).

The proof is simple. Just compute

(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 operator Dx. Proofs and details are left to the reader: (Dx)neax=ϕn+1(ax)axeax,(Dx)nf(x)=k=0n{n+1k+1}xkDkf(x),f(Dx)g(x)=k=0ckf(k+1)xk, analogous to (1.3), (3.9), and (3.22) correspondingly.

For a comprehensive study of the Mellin derivative we refer to .

More Stirling Numbers

The polynomials ϕn, 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 1=ϕ0,x=ϕ1,x2=-ϕ1+ϕ2,x3=2ϕ1-3ϕ2+ϕ3,x4=-6ϕ1+11  ϕ2-6ϕ3+ϕ4, and so forth. The coefficients here are also special numbers. If we write xn=k=0n(-1)n-k[nk]ϕk, then [nk] are the (absolute) Stirling numbers of first kind, as defined in . (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, [nk]=(n-1)[n-1k]+[n-1k-1],n1.

4. Semiorthogonality of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M192"><mml:mrow><mml:msub><mml:mrow><mml:mi>ϕ</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula>Proposition 4.1.

For every n,m=1,2,, one has 0ϕn(-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): k=0nj=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] 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 Γ(z)=0xz-1e-xdx, we set x=eλ, z=a+it, to obtain the representation Γ(a+it)=-+eiλteaλe-eλdλ, which is a Fourier transform integral. The inverse transform is eaλe-eλ=12πe-iλtΓ(a+it)dt. When a=1, this is -eλe-eλ=ddλe-eλ=-12πe-iλtΓ(1+it)dt. Differentiating (4.7) n-1 times for λ we find (ddλ)ne-eλ=ϕn(-eλ)e-eλ=-12πe-iλt(-it)n-1Γ(1+it)dt, and Parceval's formula yields the equation ϕn(-eλ)ϕm(-eλ)e-2eλdλ=12π(-it)n-1(it)m-1|Γ(1+it)|2dt, or, with x=eλ0ϕn(-x)ϕm(-x)e-2xdxx=(-1)nin+m2πtn+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] 0t2p-1sinh(πt)dt=22p-12p(-1)p=1B2p to finish the proof.

Property (4.1) resembles the semiorthogonal property of the Bernoulli polynomials 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 of Γ(a+it) and Γ(a+it)Γ(b-it).

Proposition 5.1.

For every n=0,1, and a,b>0 one has e-iμttnΓ(a+it)Γ(b-it)dt=in2πe-bμk=0nm=0k(nk){km}(-1)man-kΓ(a+b+m)(1+e-μ)a+b+m,e-iλttnΓ(a+it)dt=in2πeaλe-eλk=0n(nk)an-km=0k{km}(-1)meλm. In particular, when λ=μ=0, one obtains the moments Gn(a,b)tnΓ(a+it)Γ(b-it)dt=inπk=0nm=0k(nk){km}(-1)man-kΓ(a+b+m)2a+b+m-1,Gn(a)tnΓ(a+it)dt=2πinek=0nm=0k(nk){km}(-1)man-k.

When n=0 in (5.1) one has the known integral e-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 .

Proof.

Using again (4.6) eaλe-eλ=12πe-iλtΓ(a+it)dt, we differentiate both side n times (ddλ)n[eaλe-eλ]=12πe-iλt(-it)nΓ(a+it)dt, and then, according to the Leibniz rule and (1.2) the left-hand side becomes (ddλ)n[eaλe-eλ]=eaλe-eλk=0n(nk)ϕk(-eλ)an-k. Therefore, eaλe-eλk=0n(nk)ϕk(-eλ)an-k=12πe-iλt(-it)nΓ(a+it)dt, and (5.2) follows from here.

Replacing λ by λ-μ we write (5.6) in the form ebλe-bμe-eλe-μ=12πe-iλteiμtΓ(b+it)dt, and then Parceval's formula for Fourier integrals applied to (5.9) and (5.10) yields e-bμk=0n(nk)an-ke(a+b)λe-eλ(1+e-μ)ϕk(-eλ)dλ  =(-i)n2πe-iμttnΓ(a+it)Γ(b-it)dt. Returning to the variable x=eλ we write this in the form 12πe-iμttnΓ(a+it)Γ(b-it)dt=ine-bμk=0n(nk)an-k0ϕk(-x)xa+b-1e-x(1+e-μ)dx=ine-bμk=0nj=0k(nk){kj}an-k(-1)j0xa+b+j-1e-x(1+e-μ)dx=ine-bμk=0nj=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

p(t)=n=0mantn one can use (5.4) to write the following evaluation:

p(t)Γ(a+it)dt=n=0manGn(a). In particular, when a=1 we have

Gn(1)=2πine-1ϕn+1(-1), and therefore,

p(t)Γ(1+it)dt=2πen=0maninϕn+1(-1).

More applications can be found in the recent papers [9, 20, 21].