# On the Fock Kernel for the Generalized Fock Space and Generalized Hypergeometric Series

Authors: Jong-Do Park
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we compute the reproducing kernelBm,αz,w for the generalized Fock space Fm,α2ℂ. The usual Fock space is the case when m=2. We express the reproducing kernel in terms of a suitable hypergeometric series 1Fq. In particular, we show that there is a close connection between B4,αz,w and the error function. We also obtain the closed forms of Bm,αz,w when m=1,2/3,1/2. Finally, we also prove that Bm,αz,z~eαzmzm−2 as ∣z∣⟶∞.


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## Body

## 1. Introduction

For any fixed parameterα>0, we consider
(1)dλmz≔dλm,αz=cm,αe−αzmdAz,where dAz is the Euclidean area measure on the complex plane ℂ. Here, cm,α is a normalizing constant so that dλm,α is a probability measure on ℂ.We call thegeneralized Fock spaceFm2ℂ≔Fm,α2ℂ the set of all entire functions f in L2ℂ,dλmz. It is easy to see that Fm2ℂ is a Hilbert space with the inner product:
(2)f,g≔∫Cfzgz¯dλmz.Letϕj·:j∈ℕ be the countable orthonormal basis for Fm2ℂ. Then, the generalized Fock kernelBmz,w≔Bm,αz,w for Fm2ℂ is defined by
(3)Bmz,w≔∑j∈ℕϕjzϕjw¯.Ifm=2, then F22ℂ is the usual Fock space. In fact, it is well known that B2z,w=eαzw¯ for z,w∈ℂ. See the detailed properties on the usual Fock space in the book [1] written by Zhu. In fact, the explicit form of B2z,w is very useful for studying the properties of the Fock space in [2].In this paper, we focus on the following natural question.Question: compute the Fock kernelBmz,w for any positive rational number m.In the theory of the Bergman kernel, it is difficult to find the closed form of the Bergman kernel for a general domain. Instead, in the case of a complex ellipsoid or similar domains, one can see the expression of the Bergman kernel in terms of the hypergeometric series in [3, 4].Thegeneralized hypergeometric seriespFqa1,⋯,ap;b1,⋯,bq;x is defined by
(4)pFqa1,⋯,ap;b1,⋯,bq;x=∑k=0∞a1k⋯apkb1k⋯bqkxkk!,where ak is the Pochhammer symbol defined by
(5)ak=Γa+kΓa=aa+1⋯a+k−1,k≥1,1,k=0.Ifp=q+1, then the series converges for ∣x∣<1 and diverges for ∣x∣>1. If p<q+1, then the series converges for all x. If p>q+1, then the series converges only at x=0.It is well known that the Bergman kernel for the complex ellipsoid(6)Dp1,⋯,pn≔z1,⋯,zn∈ℂn:z12p1+⋯+zn2pn<1is closely connected with 2F1 and its higher dimensional hypergeometric series (Appell hypergeometric series or Lauricella hypergeometric series). Using the theory of the hypergeometric series, new formulas of the Bergman kernel have been computed in [3, 5–7].Recently, new interesting generalized Fock spaces have been studied. In [8], Gonessa investigated the duality on the generalized Fock space with respect to the minimal norm. In [9], one can see the boundedness of the Bergman projection on the generalized Fock-Sobolev space with respect to dλmz. But they did not obtain the explicit forms of the integral kernel. In [10], Cho et al. computed the Fock kernel for the space with respect to dμαz=cαz2αe−z2dVz. For α>0, the kernel Kαz,w is represented by Kαz,w=1F1n,n+α;z,w for z,w∈ℂn.The main theorem of this paper is the following. At first, we consider the case whenm is a positive integer.Theorem 1.
Letm be any positive integer and let ζ≔α2/mzw¯.
(i)
Ifm is even, then(7)Bmz,w=mα2/m2π∑r=0m/2−1ζrΓ2r+2/m1F11;2r+2m;ζm/2.(ii)
Ifm is odd, then(8)Bmz,w=mα2/m2π∑r=0m−1ζrΓ2r+2/m1F21;r+1m,r+1m+12;ζm4.Now, we generalize to the case whenm is a positive rational number.Theorem 2.
Letm be any positive rational number and let ζ≔α2/mzw¯.
(i)
Ifm=2p/q, where 2p and q are relatively prime, then(9)Bmz,w=mα2/m2π∑r=0p−1ζrΓqr+1/p1Fq1;r+1p+j⟶q;ζpqq,where
(10)r+1p+j⟶q=r+1p,r+1p+1q,r+1p+2q,⋯,r+1p+q−1q.(ii)
Ifm=2p+1/q, where 2p+1 and q are relatively prime, then(11)Bmz,w=mα2/m2π∑r=02pζrΓ2qr+1/2p+11F2q1;r+12p+1+j⟶2q;ζ2p+12q2q,where
(12)r+12p+1+j⟶2q=r+12p+1,r+12p+1+12q,⋯,r+12p+1+2q−12q.In particular, ifm=4, then there is a close connection between B4z,w and the error function.Theorem 3.
Letα>0. Then,
(13)B4z,w=2απzw¯eαzw¯2erfαzw¯+1+2αππ,where erfx is the error function denoted by
(14)erfx=2π∫0xe−t2dt.In general, it is difficult to find the closed form of the generalized hypergeometric seriespFq. Using the hypergeometric series in Theorems 1 and 2, we obtain the following closed forms for m=1,2/3,1/2.Theorem 4.
Letα>0. Then,
(i)
B1z,w=α/2πsinhαzw¯1/2/zw¯1/2,(ii)
B2/3z,w=αe−α/2zw¯1/3/9πzw¯2/3e3/2αzw¯1/3−2sin3/2αzw¯1/3+π/6,(iii)
B1/2z,w=α/8πzw¯3/4sinhαzw¯1/4−sinαzw¯1/4.Finally, we discuss the asymptotic behavior of the Fock kernel. Now, we writeAx~Bx if Ax/Bx converges to nonzero constant as x goes to some number or infinity. Denote KDz,w by the Bergman kernel for the bounded domain D⊂ℂn. It is a well-known fact that KDz,z diverges to infinity under some condition. More precisely, if dz is the distance to the boundary bD, then
(15)KDz,z~dzn+1,as z approaches the strongly pseudoconvex boundary point p∈bD.Using the properties of the incomplete gamma function, we can obtain the similar result also for the generalized Fock space.Theorem 5.
Letm be any positive even integer. Then,
(16)Bmz,z~eαzmzm−2as∣z∣⟶∞.Remark 6.
The usual Fock kernelB2z,w=eαzw¯ is very simple but plays an important role in the research of the function theoretic properties of the Fock space F2,α2ℂ. Theorems 1 and 2 in this paper are the first result on the generalized Fock space Fm,α2ℂ for any m≠2. Also, we hope that the explicit formulas in Theorems 3 and 4 can give a clue on studying optimal pointwise estimates for Bmz,w for some m.


## 2. Computation ofBmz,w

Considerdλmz=cm,αe−αzmdAz, where cm,α is a normalizing constant so that dλmz is a probability measure on ℂ. In fact, we can obtain cm,α from the following lemma.Lemma 7.
For any nonnegative integersk, we have
(17)∥zk∥2=2πmα2k+2/mΓ2k+2m,where Γ· is the usual gamma function. In particular, we have
(18)cm,α=mα2/m2πΓ2/m.Proof.
Recall that the usual gamma functionΓ is defined by
(19)Γz=∫0∞xz−1e−xdx,Rz>0.
Using the polar coordinate change, we have(20)∥zk∥2=∫ℂzk2e−αzmdAz=2π∫0∞r2k+1e−αrmdr.
If we can substitutes=αrm, then by (19)
(21)∥zk∥2=2π∫0∞sα2k+1/me−s1mα1/ms1/m−1ds=2πmα2k+2/m∫0∞s2k+2/m−1e−sds=2πmα2k+2/mΓ2k+2m.
It completes the proof.☐It follows that the reproducing kernelBmz,w is written as
(22)Bmz,w=∑k=0∞zw¯k∥zk∥2=mα2/m2π∑k=0∞α2/mzw¯kΓ2k+2/m.Throughout this paper, we are focusing on computing the function(23)Gmζ≔∑k=0∞ζkΓ2k+2/m.Then, we have(24)Bmz,w=mα2/m2πGmα2/mzw¯.Remark 8.
Ifm=2, then G2ζ=∑k=0∞ζk/k!=eζ. In this case,
(25)B2z,w=απeαzw¯,which is just the usual Fock kernel.Now, we investigate the relation betweenGmζ and generalized hypergeometric series for any positive rational number m.


## 3. Proof of Theorem1

In this section, we express the Fock kernelBmz,w in terms of the suitable hypergeometric series pFq when m is a positive integer. The crucial term for computing the form of Bmz,w is Γ2k+2/m.


### 3.1. Proof of Theorem1 (i)

Assume thatm is an even integer. Let m=2p for some p∈ℕ. Then, we have
(26)Gmζ=∑k=0∞ζkΓk+1/p.Theorem1 (i) can be easily proven by the following proposition using (24).Proposition 9.
Letm be any even positive integer, and let ζ≔α2/mzw¯. Then, we have
(27)Gmζ=∑r=0m/2−1ζrΓ2r+2/mΦ1;2r+2m;ζm/2,where Φa;b;x:=1F1a;b;x is the confluent hypergeometric series.Proof.
Note that there exist unique integersℓ and r such that k=pℓ+r with 0≤r≤p−1. Thus, we have
(28)Gmζ=∑r=0p−1ζr∑ℓ=0∞ζpℓΓℓ+r+1/p=∑r=0∞ζrΓr+1/p∑ℓ=0∞ζpℓr+1/pℓ.
Note that(29)Φ1;b;x=1F11;b;x=∑k=0∞1bℓxℓ.
It follows that(30)Gmζ=∑r=0p−1ζrΓr+1/pΦ1;r+1p;ζp,which completes the proof.☐


### 3.2. Proof of Theorem1 (ii)

Assume thatm is an odd integer. Let m=2p+1 for some p∈ℕ. Then,
(31)Gmζ=∑k=0∞ζkΓ2k+2/2p+1.Theorem1 (ii) can be easily proven by the following proposition using (24).Proposition 10.
Letm be any odd positive integer, and let ζ≔α2/mzw¯. Then,
(32)Gmζ=∑r=0m−1ζrΓ2r+2/m1F21;r+1m,r+1m+12;ζm4.Proof.
Note that there exist unique integersℓ and r such that k=2p+1ℓ+r with 0≤r≤2p. Then,
(33)Gmζ=∑r=02pζr∑ℓ=0∞ζ2p+1ℓΓ2ℓ+2r+2/2p+1.
Now, we will use the identity(34)Γ2ℓ+2t=22ℓtℓt+12ℓΓ2t,for any nonnegative integer ℓ and t∈ℝ. In fact, the identity (34) can be proven by
(35)Γ2ℓ+2tΓ2t=2t2t+1⋯2t+2ℓ−1=22ℓtt+1·t+ℓ−1t+12t+32⋯t+2ℓ−12=22ℓtℓt+12ℓ.
Then, by (34), we have
(36)Gmζ=∑r=02pζrΓ2r+2/2p+1∑ℓ=0∞1r+1/2p+1ℓr+1/2p+1+1/2ℓζ2p+14ℓ=∑r=02pζrΓ2r+2/2p+11F21;r+12p+1,r+12p+1+12;ζ2p+14,since 1F21;b1,b2;x=∑ℓ=0∞xℓ/b1ℓb2ℓ.☐


## 3.1. Proof of Theorem1 (i)

Assume thatm is an even integer. Let m=2p for some p∈ℕ. Then, we have
(26)Gmζ=∑k=0∞ζkΓk+1/p.Theorem1 (i) can be easily proven by the following proposition using (24).Proposition 9.
Letm be any even positive integer, and let ζ≔α2/mzw¯. Then, we have
(27)Gmζ=∑r=0m/2−1ζrΓ2r+2/mΦ1;2r+2m;ζm/2,where Φa;b;x:=1F1a;b;x is the confluent hypergeometric series.Proof.
Note that there exist unique integersℓ and r such that k=pℓ+r with 0≤r≤p−1. Thus, we have
(28)Gmζ=∑r=0p−1ζr∑ℓ=0∞ζpℓΓℓ+r+1/p=∑r=0∞ζrΓr+1/p∑ℓ=0∞ζpℓr+1/pℓ.
Note that(29)Φ1;b;x=1F11;b;x=∑k=0∞1bℓxℓ.
It follows that(30)Gmζ=∑r=0p−1ζrΓr+1/pΦ1;r+1p;ζp,which completes the proof.☐


## 3.2. Proof of Theorem1 (ii)

Assume thatm is an odd integer. Let m=2p+1 for some p∈ℕ. Then,
(31)Gmζ=∑k=0∞ζkΓ2k+2/2p+1.Theorem1 (ii) can be easily proven by the following proposition using (24).Proposition 10.
Letm be any odd positive integer, and let ζ≔α2/mzw¯. Then,
(32)Gmζ=∑r=0m−1ζrΓ2r+2/m1F21;r+1m,r+1m+12;ζm4.Proof.
Note that there exist unique integersℓ and r such that k=2p+1ℓ+r with 0≤r≤2p. Then,
(33)Gmζ=∑r=02pζr∑ℓ=0∞ζ2p+1ℓΓ2ℓ+2r+2/2p+1.
Now, we will use the identity(34)Γ2ℓ+2t=22ℓtℓt+12ℓΓ2t,for any nonnegative integer ℓ and t∈ℝ. In fact, the identity (34) can be proven by
(35)Γ2ℓ+2tΓ2t=2t2t+1⋯2t+2ℓ−1=22ℓtt+1·t+ℓ−1t+12t+32⋯t+2ℓ−12=22ℓtℓt+12ℓ.
Then, by (34), we have
(36)Gmζ=∑r=02pζrΓ2r+2/2p+1∑ℓ=0∞1r+1/2p+1ℓr+1/2p+1+1/2ℓζ2p+14ℓ=∑r=02pζrΓ2r+2/2p+11F21;r+12p+1,r+12p+1+12;ζ2p+14,since 1F21;b1,b2;x=∑ℓ=0∞xℓ/b1ℓb2ℓ.☐


## 4. Proof of Theorem2

In this section, we focus on computingGm when m is a positive rational number.


### 4.1. Proof of Theorem2 (i): Even Numerator

Letm=2p/q, where 2p and q are relatively prime. Then, we have
(37)Gmζ=∑k=0∞ζkΓk+1q/p=∑r=0p−1ζr∑ℓ=0∞ζpℓΓqℓ+qr+1/p,where k=pℓ+r with 0≤r≤p−1.Lemma 11.
The gamma functionΓ satisfies the identity
(38)ΓxΓx+1nΓx+2n⋯Γx+n−1n=2πn−1/2n1/2−nxΓnx.Using the above lemma, we can prove the following.Lemma 12.
(39)Γqℓ+qt=qqℓ∏j=0q−1t+jqℓΓqt.Proof.
We will prove it in two different methods. Using the propertyΓx+1=xΓx, we have
(40)Γqℓ+qtΓqt=∏i=0qℓ−1qt+i=qqℓ∏i=0qℓ−1t+iq.
Then, there existsx,y∈ℤ such that i=qj+y with 0≤j≤ℓ−1 and 0≤y≤q−1. It follows that
(41)∏i=0qℓ−1t+iq=∏j=0q−1∏x=0ℓ−1t+jq+x=∏j=0q−1t+jqℓ.
It can be proven also using Lemma11. Note that
(42)Γqℓ+qt=Γℓ+tΓℓ+t+1/q⋯Γℓ+t+q−1/q2πq−1/2q1/2−qℓ+t=ΓtΓt+1/q⋯Γt+q−1/q2πq−1/2q1/2−qℓ+ttℓt+1qℓ⋯t+q−1qℓ=2πq−1/2q1/2−qtΓqt2πq−1/2q1/2−qℓ+ttℓt+1qℓ⋯t+q−1qℓ=qqℓΓqttℓt+1qℓ⋯t+q−1qℓ.☐Now, we prove Theorem2 (i) using Lemma 12.Theorem 13 (Theorem2 (i) again).
Letm=2p/q, where 2p and q are relatively prime. Then,
(43)Gmζ=∑r=0p−1ζrΓqr+1/p1Fq1;r+1p+j⟶q;ζpqq,where
(44)r+1p+j⟶q=r+1p,r+1p+1q,r+1p+2q,⋯,r+1p+q−1q.
Thus, we have(45)Bmz,w=mα2/m2π∑r=0p−1ζrΓqr+1/p1Fq1;r+1p+j⟶q;ζpqq.Proof.
By Lemma12, we have
(46)Gmζ=∑r=0p−1ζrΓqr+1/p∑ℓ=0∞1∏j=0q−1r+1/p+j/qℓζpqqℓ.
By the definition (4), we see that
(47)1Fq1;b1,⋯,bq;x=∑ℓ=0∞xℓb1ℓ⋯bqℓ.
It follows that(48)Gmζ=∑r=0p−1ζrΓqr+1/p1Fq1;r+1p,r+1p+1q,⋯,r+1p+q−1q;ζpqq.
If we use (24), then it completes the proof.☐


### 4.2. Proof of Theorem2 (ii): Odd Numerator

Letm=2p+1/q, where 2p+1 and q are relatively prime. Then,
(49)Gmζ=∑k=0∞ζkΓ2k+2q/2p+1=∑r=02pζr∑ℓ=0∞ζ2p+1ℓΓ2qℓ+2qr+1/2p+1,where k=2p+1ℓ+r with 0≤r≤2p. By Lemma 12, we have
(50)Γ2qℓ+2qr+12p+1=Γ2qt2q2qℓ∏j=02q−1t+j2qℓ,where t≔r+1/2p+1. It follows that
(51)Gmζ=∑r=02pζrΓ2qt∑ℓ=0∞1∏j=02q−1t+j/2qℓζ2p+12q2qℓ=∑r=02pζrΓ2qt1F2q1;t,t+12q,⋯,t+2q−12q;ζ2p+12q2q.If we use (24), then we obtain the following.Theorem 14 (Theorem2 (ii) again).
Letm=2p+1/q, where 2p+1 and q are relatively prime. Then,
(52)Gmζ=∑r=02pζrΓ2qr+1/2p+11F2q1;r+12p+1+j⟶2q;ζ2p+12q2q,where
(53)r+12p+1+j⟶2q=r+12p+1,r+12p+1+12q,⋯,r+12p+1+2q−12q.
Thus, we have(54)Bmz,w=mα2/m2π∑r=02pζrΓ2qr+1/2p+11F2q1;r+12p+1+j⟶2q;ζ2p+12q2q.


## 4.1. Proof of Theorem2 (i): Even Numerator

Letm=2p/q, where 2p and q are relatively prime. Then, we have
(37)Gmζ=∑k=0∞ζkΓk+1q/p=∑r=0p−1ζr∑ℓ=0∞ζpℓΓqℓ+qr+1/p,where k=pℓ+r with 0≤r≤p−1.Lemma 11.
The gamma functionΓ satisfies the identity
(38)ΓxΓx+1nΓx+2n⋯Γx+n−1n=2πn−1/2n1/2−nxΓnx.Using the above lemma, we can prove the following.Lemma 12.
(39)Γqℓ+qt=qqℓ∏j=0q−1t+jqℓΓqt.Proof.
We will prove it in two different methods. Using the propertyΓx+1=xΓx, we have
(40)Γqℓ+qtΓqt=∏i=0qℓ−1qt+i=qqℓ∏i=0qℓ−1t+iq.
Then, there existsx,y∈ℤ such that i=qj+y with 0≤j≤ℓ−1 and 0≤y≤q−1. It follows that
(41)∏i=0qℓ−1t+iq=∏j=0q−1∏x=0ℓ−1t+jq+x=∏j=0q−1t+jqℓ.
It can be proven also using Lemma11. Note that
(42)Γqℓ+qt=Γℓ+tΓℓ+t+1/q⋯Γℓ+t+q−1/q2πq−1/2q1/2−qℓ+t=ΓtΓt+1/q⋯Γt+q−1/q2πq−1/2q1/2−qℓ+ttℓt+1qℓ⋯t+q−1qℓ=2πq−1/2q1/2−qtΓqt2πq−1/2q1/2−qℓ+ttℓt+1qℓ⋯t+q−1qℓ=qqℓΓqttℓt+1qℓ⋯t+q−1qℓ.☐Now, we prove Theorem2 (i) using Lemma 12.Theorem 13 (Theorem2 (i) again).
Letm=2p/q, where 2p and q are relatively prime. Then,
(43)Gmζ=∑r=0p−1ζrΓqr+1/p1Fq1;r+1p+j⟶q;ζpqq,where
(44)r+1p+j⟶q=r+1p,r+1p+1q,r+1p+2q,⋯,r+1p+q−1q.
Thus, we have(45)Bmz,w=mα2/m2π∑r=0p−1ζrΓqr+1/p1Fq1;r+1p+j⟶q;ζpqq.Proof.
By Lemma12, we have
(46)Gmζ=∑r=0p−1ζrΓqr+1/p∑ℓ=0∞1∏j=0q−1r+1/p+j/qℓζpqqℓ.
By the definition (4), we see that
(47)1Fq1;b1,⋯,bq;x=∑ℓ=0∞xℓb1ℓ⋯bqℓ.
It follows that(48)Gmζ=∑r=0p−1ζrΓqr+1/p1Fq1;r+1p,r+1p+1q,⋯,r+1p+q−1q;ζpqq.
If we use (24), then it completes the proof.☐


## 4.2. Proof of Theorem2 (ii): Odd Numerator

Letm=2p+1/q, where 2p+1 and q are relatively prime. Then,
(49)Gmζ=∑k=0∞ζkΓ2k+2q/2p+1=∑r=02pζr∑ℓ=0∞ζ2p+1ℓΓ2qℓ+2qr+1/2p+1,where k=2p+1ℓ+r with 0≤r≤2p. By Lemma 12, we have
(50)Γ2qℓ+2qr+12p+1=Γ2qt2q2qℓ∏j=02q−1t+j2qℓ,where t≔r+1/2p+1. It follows that
(51)Gmζ=∑r=02pζrΓ2qt∑ℓ=0∞1∏j=02q−1t+j/2qℓζ2p+12q2qℓ=∑r=02pζrΓ2qt1F2q1;t,t+12q,⋯,t+2q−12q;ζ2p+12q2q.If we use (24), then we obtain the following.Theorem 14 (Theorem2 (ii) again).
Letm=2p+1/q, where 2p+1 and q are relatively prime. Then,
(52)Gmζ=∑r=02pζrΓ2qr+1/2p+11F2q1;r+12p+1+j⟶2q;ζ2p+12q2q,where
(53)r+12p+1+j⟶2q=r+12p+1,r+12p+1+12q,⋯,r+12p+1+2q−12q.
Thus, we have(54)Bmz,w=mα2/m2π∑r=02pζrΓ2qr+1/2p+11F2q1;r+12p+1+j⟶2q;ζ2p+12q2q.


## 5. Special Cases

In the last section, we expressBmz,w in terms of the generalized hypergeometric series 1Fq for a suitable q. However, in general, it is difficult to find the closed form of 1Fq1;b1,⋯,bq;x for any b1,⋯,bq.


### 5.1. Proof of Theorem3: The Case When m=4

In this case, we show that1F11,1/2;x is represented in terms of the error function. In fact, we will conclude that
(55)B4z,w=2απzw¯eαzw¯2erfαzw¯+1+2αππ.By Proposition9, we need to study Φ1;b;x for any rational number b with 0<b≤1. It is easy to see that Φ1;1;x=ex. Also, if 0<b<1, then there is a connection between Φ1;b;x and the incomplete gamma function.Proposition 15.
Φ satisfies the following identities.
(i)
Kummer’s transformation:Φa;b;x=exΦb−a;b;−x(ii)
IfRb>Ra, then the confluent hypergeometric series Φa;b;x has the integral representation(56)Φa;b;x=ΓbΓaΓb−a∫01exuua−11−ub−a−1du.
Theupper incomplete gamma functionΓs,x and the lower incomplete gamma functionγs,x are defined by
(57)Γs,x=∫x∞ts−1e−tdt,γs,x=∫0xts−1e−tdt.Now,Φ1;b;x can be written in terms of the lower incomplete gamma function.Proposition 16.
For any0<b<1, we have
(58)Φ1;b;x=b−1exx1−bγb−1,x.Proof.
By Proposition15 (i), we have
(59)Φ1;b;x=exΦb−1;b;−x.
By Proposition15 (ii), we have
(60)Φb−1;b;−x=ΓbΓb−1Γ1∫01e−xuub−2du=b−1x1−b∫0xe−ttb−2dt=b−1x1−bγb−1,x.
It completes the proof.☐In particular, ifm=4, then we can write G4ζ and B4z,w in a simple form using the error function. Recall that the error functionerfx is denoted by
(61)erfx=2π∫0xe−t2dt.It is easy to see thatγ1/2,x=πerfx.The following lemma can be proven easily by the integration parts of the integral.Lemma 17.
For anys, we have
(62)γs+1,x=sγs,x−xse−x.By Lemma17, we have
(63)γ−12,x=−2γ12,x−2e−xx=−2πerfx−2e−xx.By Proposition16, we have
(64)Φ1,12;x=−12exxγ−12,x=πxexerfx+1.Now, we are ready to expressG4ζ and B4z,w in terms of the error function.Theorem 18 (Theorem3 again).
Ifm=4, then
(65)G4ζ=ζeζ2erfζ+1+1π.
Thus, we have(66)B4z,w=2απzw¯eαzw¯2erfαzw¯+1+2αππ.Proof.
By Proposition9, we have
(67)G4ζ=1πΦ1,12;ζ2+ζΦ1,1;ζ2.
If we use (64) and the identity Φ1,1;ζ=eζ, then we obtain (65). Since B4z,w=2α/πG4αzw¯, we obtain the formula of B4z,w.☐


### 5.2. Proof of Theorem4: The Case when m=1,2/3,1/2

It is surprising that we can obtain the explicit forms ofB1z,w, B2/3z,w, and B1/2z,w.Theorem 19 (Theorem4 (i) again).
Ifm=1, then
(68)G1ζ=sinhζζ.
Thus, we have(69)B1z,w=α2πsinhαzw¯1/2zw¯1/2.Proof.
Note thatG1ζ=1F21;1,3/2;ζ/4. Use the identity
(70)1F21;1,32;x=sinh2x2x.
In fact, the identity (70) can be proven as follows. Note that
(71)1F21;1,32;x=∑k=0∞xk3/2kk!,32kk!=32·52⋯k+12k!=3·5⋯2k+12kk!=2k+1!4k.
It follows that(72)1F21;1,32;x=∑k=0∞4k2k+1!xk=sinh2x2x.☐In general, the explicit forms of the most hypergeometric series are unknown. But the very special following the hypergeometric series including (70) can be computed.Proposition 20.
For anyx, we have
(i)
1F31;1,4/3,5/3;x=2e−3/2x1/3/27x2/3e9/2x1/3−2sin33/2x1/3+π/6(ii)
1F41;1,5/4,6/4,7/4;x=3/64x3/4sinh4x1/4−sin4x1/4One can find the closed forms of various hypergeometric series in [11]. In particular, one can find the closed forms of
(73)1F31;1,43,53;x=0F2;43,53;x,1F41;1,54,64,74;x=0F3;54,64,74;x,in [12, 13], respectively.Now, we prove Theorem4 (ii) and (iii) as finding the closed forms of B2/3z,w and B1/2z,w using Proposition 20. Since we have
(74)G2/3ζ=121F31;1,43,53;ζ27,it follows that
(75)B2/3z,w=α33πG2/3α3zw¯=α36π1F31;1,43,53;α3zw¯27.By Proposition20 (i), we have
(76)B2/3z,w=αe−α/2zw¯1/39πzw¯2/3e32αzw¯1/3−2sin32αzw¯1/3+π6.Since we have(77)G1/2ζ=161F41;1,54,64,74;ζ44,it follows that
(78)B1/2z,w=α44πG1/2α4zw¯=α424π1F41;1,54,64,74;α4zw¯44.By Proposition20 (ii), we have
(79)B1/2z,w=α8πzw¯3/4sinhαzw¯1/4−sinαzw¯1/4.It completes the proof of Theorem4 (ii) and (iii).


### 5.3. Proof of Theorem5

In this section,Ax~Bx means that Ax/Bx converges to a nonzero constant as x goes to some number or infinity.Theorem 21 (Theorem5 again).
Letm be any positive even integer. Then,
(80)Bmz,z~eαzmzm−2as∣z∣⟶∞.Proof.
Letm=2p. Then, by Theorem 1 (i),
(81)B2pz,w=pα1/p2π∑r=0p−1ζrΓr+1/pΦ1;r+1p;ζp=pα1/p2π∑r=0p−2ζrΓr+1/pΦ1;r+1p;ζp+ζp−1Φ1;1;ζp.
If0≤r≤p−2, then by Proposition 16,
(82)Φ1;r+1p;ζp=r+1p−1eζpζp−r−1γr+1p−1,ζp,and Φ1;1;ζp=eζp. It follows that
(83)B2pz,w=pα1/p2π∑r=0p−2eζpζp−1γr+1/p−1,ζpΓr+1/p−1+ζp−1eζp.
Sinceγr+1/p−1,x⟶Γr+1/p as x⟶∞, it completes the proof.☐In fact, it is easily checked that (80) holds also when m=1,3/2,1/2 using the explicit forms in Theorem 4. We can conjecture that (80) holds for any m>0.


## 5.1. Proof of Theorem3: The Case When m=4

In this case, we show that1F11,1/2;x is represented in terms of the error function. In fact, we will conclude that
(55)B4z,w=2απzw¯eαzw¯2erfαzw¯+1+2αππ.By Proposition9, we need to study Φ1;b;x for any rational number b with 0<b≤1. It is easy to see that Φ1;1;x=ex. Also, if 0<b<1, then there is a connection between Φ1;b;x and the incomplete gamma function.Proposition 15.
Φ satisfies the following identities.
(i)
Kummer’s transformation:Φa;b;x=exΦb−a;b;−x(ii)
IfRb>Ra, then the confluent hypergeometric series Φa;b;x has the integral representation(56)Φa;b;x=ΓbΓaΓb−a∫01exuua−11−ub−a−1du.
Theupper incomplete gamma functionΓs,x and the lower incomplete gamma functionγs,x are defined by
(57)Γs,x=∫x∞ts−1e−tdt,γs,x=∫0xts−1e−tdt.Now,Φ1;b;x can be written in terms of the lower incomplete gamma function.Proposition 16.
For any0<b<1, we have
(58)Φ1;b;x=b−1exx1−bγb−1,x.Proof.
By Proposition15 (i), we have
(59)Φ1;b;x=exΦb−1;b;−x.
By Proposition15 (ii), we have
(60)Φb−1;b;−x=ΓbΓb−1Γ1∫01e−xuub−2du=b−1x1−b∫0xe−ttb−2dt=b−1x1−bγb−1,x.
It completes the proof.☐In particular, ifm=4, then we can write G4ζ and B4z,w in a simple form using the error function. Recall that the error functionerfx is denoted by
(61)erfx=2π∫0xe−t2dt.It is easy to see thatγ1/2,x=πerfx.The following lemma can be proven easily by the integration parts of the integral.Lemma 17.
For anys, we have
(62)γs+1,x=sγs,x−xse−x.By Lemma17, we have
(63)γ−12,x=−2γ12,x−2e−xx=−2πerfx−2e−xx.By Proposition16, we have
(64)Φ1,12;x=−12exxγ−12,x=πxexerfx+1.Now, we are ready to expressG4ζ and B4z,w in terms of the error function.Theorem 18 (Theorem3 again).
Ifm=4, then
(65)G4ζ=ζeζ2erfζ+1+1π.
Thus, we have(66)B4z,w=2απzw¯eαzw¯2erfαzw¯+1+2αππ.Proof.
By Proposition9, we have
(67)G4ζ=1πΦ1,12;ζ2+ζΦ1,1;ζ2.
If we use (64) and the identity Φ1,1;ζ=eζ, then we obtain (65). Since B4z,w=2α/πG4αzw¯, we obtain the formula of B4z,w.☐


## 5.2. Proof of Theorem4: The Case when m=1,2/3,1/2

It is surprising that we can obtain the explicit forms ofB1z,w, B2/3z,w, and B1/2z,w.Theorem 19 (Theorem4 (i) again).
Ifm=1, then
(68)G1ζ=sinhζζ.
Thus, we have(69)B1z,w=α2πsinhαzw¯1/2zw¯1/2.Proof.
Note thatG1ζ=1F21;1,3/2;ζ/4. Use the identity
(70)1F21;1,32;x=sinh2x2x.
In fact, the identity (70) can be proven as follows. Note that
(71)1F21;1,32;x=∑k=0∞xk3/2kk!,32kk!=32·52⋯k+12k!=3·5⋯2k+12kk!=2k+1!4k.
It follows that(72)1F21;1,32;x=∑k=0∞4k2k+1!xk=sinh2x2x.☐In general, the explicit forms of the most hypergeometric series are unknown. But the very special following the hypergeometric series including (70) can be computed.Proposition 20.
For anyx, we have
(i)
1F31;1,4/3,5/3;x=2e−3/2x1/3/27x2/3e9/2x1/3−2sin33/2x1/3+π/6(ii)
1F41;1,5/4,6/4,7/4;x=3/64x3/4sinh4x1/4−sin4x1/4One can find the closed forms of various hypergeometric series in [11]. In particular, one can find the closed forms of
(73)1F31;1,43,53;x=0F2;43,53;x,1F41;1,54,64,74;x=0F3;54,64,74;x,in [12, 13], respectively.Now, we prove Theorem4 (ii) and (iii) as finding the closed forms of B2/3z,w and B1/2z,w using Proposition 20. Since we have
(74)G2/3ζ=121F31;1,43,53;ζ27,it follows that
(75)B2/3z,w=α33πG2/3α3zw¯=α36π1F31;1,43,53;α3zw¯27.By Proposition20 (i), we have
(76)B2/3z,w=αe−α/2zw¯1/39πzw¯2/3e32αzw¯1/3−2sin32αzw¯1/3+π6.Since we have(77)G1/2ζ=161F41;1,54,64,74;ζ44,it follows that
(78)B1/2z,w=α44πG1/2α4zw¯=α424π1F41;1,54,64,74;α4zw¯44.By Proposition20 (ii), we have
(79)B1/2z,w=α8πzw¯3/4sinhαzw¯1/4−sinαzw¯1/4.It completes the proof of Theorem4 (ii) and (iii).


## 5.3. Proof of Theorem5

In this section,Ax~Bx means that Ax/Bx converges to a nonzero constant as x goes to some number or infinity.Theorem 21 (Theorem5 again).
Letm be any positive even integer. Then,
(80)Bmz,z~eαzmzm−2as∣z∣⟶∞.Proof.
Letm=2p. Then, by Theorem 1 (i),
(81)B2pz,w=pα1/p2π∑r=0p−1ζrΓr+1/pΦ1;r+1p;ζp=pα1/p2π∑r=0p−2ζrΓr+1/pΦ1;r+1p;ζp+ζp−1Φ1;1;ζp.
If0≤r≤p−2, then by Proposition 16,
(82)Φ1;r+1p;ζp=r+1p−1eζpζp−r−1γr+1p−1,ζp,and Φ1;1;ζp=eζp. It follows that
(83)B2pz,w=pα1/p2π∑r=0p−2eζpζp−1γr+1/p−1,ζpΓr+1/p−1+ζp−1eζp.
Sinceγr+1/p−1,x⟶Γr+1/p as x⟶∞, it completes the proof.☐In fact, it is easily checked that (80) holds also when m=1,3/2,1/2 using the explicit forms in Theorem 4. We can conjecture that (80) holds for any m>0.


## 6. Concluding Remarks

In fact, we can consider the more generalized Fock space. Letdλϕz=cϕe−ϕzdAz, where dAz is the Euclidean area measure on the complex plane ℂ. We assume that ϕr is radial and increasing on 0,∞ with limr⟶∞ϕr=∞. We call the (generalized) Fock space Fϕ2ℂ as the set of all entire functions f in L2ℂ,dλϕ. Another simple example is ϕr=lnr. In this case, we can show that the Fock kernel can be written in terms of the Meijer-G function. It will be interesting that one finds the relation between the other hypergeometric series and the new Fock kernel with respect to ϕ.



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*Source: 1365674-2021-08-23.xml*