# New Recursive Representations for the Favard Constants with Application to Multiple Singular Integrals and Summation of Series

Authors: Snezhana Georgieva Gocheva-Ilieva; Ivan Hristov Feschiev
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

There are obtained integral form and recurrence representations for some Fourier series and connected with them Favard constants. The method is based on preliminary integration of Fourier series which permits to establish general recursion formulas for Favard constants. This gives the opportunity for effective summation of infinite series and calculation of some classes of multiple singular integrals by the Favard constants.


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## Body

## 1. Introduction

The Fourier series and related with them Achieser-Krein-Favard constants, often simply called Favard constants, have significant theoretical and practical roles in many areas [1, 2]. These remarkable mathematical constants are introduced firstly in the theory of Fourier series and approximations of functions by trigonometric polynomials [3].The classical definitions of Favard constants are given by the infinite series [1, 4, 5]
(1)Kr=4π∑ν=0∞(-1)ν(r+1)(2ν+1)r+1,r=0,1,2,…,(2)K~r=4π∑ν=0∞(-1)νr(2ν+1)r+1,r=1,2,….These constants find wide applications in the approximation theory for exact and asymptotic results on the approximation of functions, and especially for the best approximations of trigonometric and other classes of functions in different spaces and related inequalities [1, 5–14]. In particular, many important applications are concerned with the approximation of Euler, cardinal, periodic, and other type of splines [15–17]. It can be noted that Favard constants are connected to approximations that are best in a pointwise sense in comparison, for instance, with the Lebesgue constants which are connected to approximations that are best in a least-squares sense (Fourier series) [5]. The Favard constants play also an important role in estimating optimal quadrature and cubature formulas, calculation of singular integrals, some classes of differential, integrodifferential and integral equations [18–24], and in other areas.Nevertheless widely used, as a whole, the properties of the Favard constants have not been investigated well enough [14], except for some particular cases.Different methods for their calculation are given, for instance, in [2, Ch. 5.2]. In general these methods are based on the properties of the well-known constants and special functions as gamma function Γ(z), generalized Riemann zeta function ζ(z,a), the Bernoulli polynomials Bn(x) and the Bernoulli numbers Bn, and the Euler polynomials En(x) and the Euler numbers En, given by the following expressions [2]:
(3)Γ(z)=∫0∞tz-1e-tdt(Rez>0),Ψ(z)=Γ′(z)Γ(z),ζ(z,a)=∑k=0∞1(k+a)z=1Γ(z)∫0∞tz-1e-at1-e-tdt(Rez>1),Bn(x):textet-1=∑k=0∞Bk(x)tkk!(|t|<2π),Bn=Bn(0),Bn(x)=∑k=0n(nk)Bkxn-k,En(x):2extet+1=∑k=0∞Ek(x)tkk!,(1cosh(t)=∑k=0∞Ektkk!),(|t|<π).From now on we will use the following notations:(4)A~n(x)=∫0xdx⋯∫0xlntanθ2dθ(0≤x≤π),B~n(x)=∫0xdx⋯∫0xln(2sinθ2)dθ(0≤x≤2π),C~n(x)=∫0xdx⋯∫0xln(2cosθ2)dθ(-π≤x≤π)(n=1,2,3,…)(5)Dn(x)=xnn!,(6)Tr=∑n=1∞1nr(r=2,3,…),Qr=∑n=1∞(-1)nnr(r=1,2,3,…),
where the three multiple singular integrals (4) contain exactly n integral operations.The main purpose of this paper is to establish new recursive formulas for the Favard constants (1), including only finite number of terms and recursive formulas for (2). They will be further used to obtain new integral representations for the previously stated objects, in particular for calculation of the multiple singular integrals (4) and summation of series.


## 2. Recursive Representations forKr and Some Fourier Series

We will prove the following.Theorem 1.
For the constantsKr (r=1,2,3,…), the following recursive representations hold:(7a)
K
2
s
+
1
=
1
2
[
(
-
1
)
s
D
2
s
+
1
(
π
)
+
∑
p
=
1
s
K
2
s
+
1
-
2
p
(
-
1
)
p
-
1
π
2
p
(
2
p
)
!
]
(
K
0
=
1
,
K
1
=
π
2
)
,
(7b)
K
2
s
=
(
-
1
)
s
D
2
s
(
π
2
)
+
∑
p
=
1
s
K
2
s
+
1
-
2
p
(
-
1
)
p
-
1
(
2
p
-
1
)
!
(
π
2
)
2
p
-
1
(
s
=
1,2
,
3
,
…
)
.Proof.
We will use the method of induction and preliminary integration of appropriate Fourier series. Let us start by the well-known expansion [2]
(8)4π∑ν=0∞sin[(2ν+1)x]2ν+1=1(0<x<π),
where for x=π/2 we will have K0=1. By integration of both sides of (8) in [0,x], we get
(9)-4π∑ν=0∞cos[(2ν+1)x](2ν+1)2+4π∑ν=0∞1(2ν+1)2=x=D1(x)(0≤x≤π).
Forx=π, (9) gives us 2K1=π=D1(π), and consequently K1=D1(π)/2=π/2. The same results for K0 and K1 can be achieved starting from the equality [2]
(10)∑n=1∞sinnxn=π2-x2=π2-12D1(x)(0<x<2π).
Forx=π/2 we find again that K0=1. After integration of the both sides of (10), we have
(11)-∑n=1∞cosnxn2+T2=-x24+πx2=-12D2(x)+π2D1(x)(0≤x≤2π).
Forx=π we find that K1=π/2. Next after integration of the both sides of (11), we obtain
(12)-∑n=1∞sinnxn3+T2x=-12D3(x)+π2D2(x)(0≤x≤2π),
which for x=π gives us
(13)T2=-12πD3(π)+12D2(π)=π26.
If we putx=π/2 in the same equality (12) and make a little processing, we will arrive at the value K2=π2/8.
On the other hand, after integration of both sides of (9), we obtain
(14)-4π∑v=0∞sin[(2v+1)x](2v+1)3+K1x=D2(x)(0≤x≤π).
Here forx=π/2, we get -K2+(π/2)K1=D2(π/2), and consequently
(15)K2=π2K1-D2(π2)=π28.
For the constantK3, we must now integrate both sides of (12) as
(16)∑n=1∞cosnxn4-T4+T2D2(x)=-12D4(x)+π2D3(x)(0≤x≤2π),
and put x=π here, or integrate both sides of (14) as
(17)4π∑v=1∞cos[(2v+1)x](2v+1)4-4π∑v=1∞1(2v+1)4+K1D2(x)=D3(x)(0≤x≤π),
and put again x=π. Then we will have
(18)K3=1πD4(π)-32D3(π)+π2D2(π)=π24K1-12D3(π)=π324.
Going on the indicated procedure on the base of induction, we easily arrive at the recursive representations (7a) and (7b) and complete the proof.Remark 2.
The scheme of this proof is valid for the most of the other statements in this paper.In connection with Theorem1, we would like to note another representation of Kr (see, e.g., [25]). It can be written in terms of Lerch transcendent [25], or as it is shown in [2, Section 5.1.4]
(19)K2s-1=2π(2s)!(22s-1)π2s|B2s|,K2s=2π(2s)!(π2)2s+1|E2s|(s=1,2,3,…),
where the Bernoulli and Euler numbers are specified in (3).Data for values of magnitudes ofKr using (7a) and (7b) are shown in Table 1.Table 1
Exact and approximate values of the Favard constantsKr, calculated by the recursive formulas (7a) and (7b) using Mathematica software package.
r
Exact values ofKr
Approximate values ofKr
1
π
/
2
1.5707963267948966192313216916
2
π
2
/
8
1.2337005501361698273543113749
3
π
3
/
24
1.2919281950124925073115131277
4
5
π
4
/
384
1.2683475395052400681828168318
5
π
5
/
240
1.2750820199386727219280887918
6
61
π
6
/
46080
1.2726723265645306132561498711
7
17
π
7
/
40320
1.2734371248066831633864461900
8
277
π
8
/
2064384
1.2731754806526058136347769671
9
31
π
9
/
725760
1.2732612424724875463814366656
10
50521
π
10
/
3715891200
1.2732323827293948495082797108
11
691
π
11
/
159667200
1.2732419458721540967715077901
12
540553
π
12
/
392398110720
1.2732387471572495304117396905The equalities (10)–(13) outline a procedure for summing up the numerical series T2s, (s=1,2,3,…) in (6). It leads to the assertion.Corollary 3.
The following recursive representation holds:(20)T2s=(-1)s[12πD2s+1(π)-12D2s(π)]+∑p=1s-1(-1)p+1π2p(2p+1)!T2s-2p(s=1,2,3,…),
where for s=1 by definition T0=0.It can be noted that the numbersT2s can be also represented by the well known formula (see, e.g., [2, 5.1.2])
(21)T2s=22s-1π2s(2s)!|B2s|(s=1,2,3,…).The same procedure applied on the base of the equality(22)∑n=1∞(-1)nsinnxn=-12D1(x)(-π<x<π),
leads to the following.Corollary 4.
The following recursive representation holds:(23)Q2s=(-1)s2πD2s+1(π)+∑p=1s-1(-1)p+1π2p(2p+1)!Q2s-2p(s=1,2,3,…),
where for s=1 by definition  Q0=0.In this connection we will note the explicit formula forQ2s represented by the Bernoulli numbers (see [2, 5.2.1])
(24)Q2s=(1-22s-1)π2s(2s)!|B2s|,(s=1,2,3,…).It is easy to see that the constantsKr satisfy the following inequalities (see also [1]):
(25)1=K0<K2<K4<⋯<4π<⋯<K5<K3<K1=π2
and limr→∞Kr=4/π.The procedure of getting the representations (11), (12), and (16) with the help of (10) gives us an opportunity to lay down the following.Theorem 5.
The following recursive representations hold:(26)∑n=1∞sinnxn2s-1=(-1)s[12D2s-1(x)-π2D2s-2(x)]+∑p=1s-1(-1)p+1T2s-2pD2p-1(x),∑n=1∞cosnxn2s=(-1)s-1[12D2s(x)-π2D2s-1(x)]+∑p=0s-1(-1)pT2s-2pD2p(x)
(s=1,2,3,…, (0≤x≤2π),  for s=1 by definition D0(x)=1 and T0=0, 0<x<2π).At the same time both series in (26) have the well-known representations [2, 5.4.2]
(27)∑n=1∞sinnxn2s+1=(-1)s-12(2s+1)!(2π)2s+1B2s+1(x2π),((0≤x≤2π);(s=1,2,3,…),0<x<2πfors=0),∑n=1∞cosnxn2s=(-1)s-12(2s)!(2π)2sB2s(x2π)(0≤x≤2π;s=1,2,3,…).The previously stated procedure for obtaining (26) can now be applied on the strength of (22). This leads to the assertion.Theorem 6.
The following recursive representations hold:(28)∑n=1∞(-1)nsinnxn2s-1=(-1)s2D2s-1(x)+∑p=1s-1(-1)p+1Q2s-2pD2p-1(x),∑n=1∞(-1)ncosnxn2s=(-1)s-12D2s(x)+∑p=0s-1(-1)pQ2s-2pD2p(x),
(s=1,2,3,…, (-π≤x≤π); for s=1 by definition D0(x)=1 and Q0=0, -π<x<π).At the same time, both series in (28) have the well-known representations [2, 5.4.2]
(29)∑n=1∞(-1)nsinnxn2s+1=(-1)s-12(2s+1)!(2π)2s+1B2s+1(x+π2π)(-π<x≤π;s=0,1,2,…),∑n=1∞(-1)ncosnxn2s=(-1)s-12(2s)!(2π)2sB2s(x+π2π)(-π≤x≤π;s=1,2,3,…).By analogy with the previous, the procedure for obtaining (9), (14), and (17) with the help of (8) leads us to the following.Theorem 7.
The following recursive representations hold:(30)∑v=0∞sin[(2v+1)x](2v+1)2s-1=π(-1)s-14[D2s-2(x)+∑p=1s-1(-1)pK2p-1D2s-2p-1(x)],∑v=0∞cos[(2v+1)x](2v+1)2s=π(-1)s4[D2s-1(x)+∑p=1s(-1)pK2p-1D2s-2p(x)],s=1,2,3,… (0≤x≤π); for s=1  by definition D0(x)=1  (0<x<π).At the same time, both series in (30) have the well-known formulas [2, 5.4.6]
(31)∑v=0∞sin[(2v+1)x](2v+1)2s+1=(-1)sπ2s+14(2s)!E2s(xπ)(0<x<π;s=0,1,2,…),∑v=0∞cos[(2v+1)x](2v+1)2s=(-1)sπ2s4(2s-1)!E2s-1(xπ)(0≤x≤π;s=1,2,…).The same procedure applied on the base of the equality(32)4π∑v=0∞(-1)vcos[(2v+1)x]2v+1=1(-π2<x<π2),
gives us the next theorem.Theorem 8.
The following recursive representations hold:(33)∑v=0∞(-1)vcos[(2v+1)x](2v+1)2s-1=π(-1)s-14[D2s-2(x)+∑p=1s-1(-1)pK2pD2s-2p-2(x)],∑v=0∞(-1)vsin[(2v+1)x](2v+1)2s=π(-1)s-14[D2s-1(x)+∑p=1s-1(-1)pK2pD2s-2p-1(x)]
(s=1,2,3,…, (-π/2≤x≤π/2), for s=1: K0 declines, D0(x)=1,  (-π/2<x<π/2)).At the same time, both series in (33) have the well-known representations ([2, 5.4.6])
(34)∑v=0∞(-1)vcos[(2v+1)x](2v+1)2s+1=(-1)sπ2s+14(2s)!E2s(xπ+12)(-π2<x<π2;s=0,1,…),∑v=0∞(-1)vsin[(2v+1)x](2v+1)2s=(-1)s-1π2s4(2s-1)!E2s-1(xπ+12)(-π2≤x≤π2;s=1,2,…).Meanwhile it is important to note that the number of addends in our recurrence representations (26), (28), (30), and (33) is two times less than the number of the addends in the corresponding cited formulas from [2]. So our method appears to be more economic and effective.Moreover, one can get many other representations of the constantsKr (r=1,2,3,…) and numerical series Q2s (s=1,2,…) from Theorems 5–8 putting, in particular, x=π/2 or x=π. For completeness we will note the main results.From Theorem5 for x=π/2 and  x=π immediately follows the following.Corollary 9.
For the Favard constantsKr, the following recursive representations hold:
(35)K2s-2=4π{∑p=1s-1(-1)p+1T2s-2pD2p-1(π2)(-1)s[12D2s-1(π2)-π2D2s-2(π2)]+∑p=1s-1(-1)p+1T2s-2pD2p-1(π2)},K2s-1=2π{∑p=1s-1(-1)p+1T2s-2pD2p(π)(-1)s[12D2s(π)-π2D2s-1(π)]+∑p=1s-1(-1)p+1T2s-2pD2p(π)},
(s=1,2,…, for s=1: D0(x)=1,T0=0).Forx=π, one can get (20) too by replacing previously s by s+1.Forx=π/2, we obtain the next corollary.Corollary 10.
For numbersQ2s,  the following recursive representations hold:
(36)Q2s=4s{∑p=0s-1(-1)pT2s-2pD2p(π2)(-1)s-1[12D2s(π2)-π2D2s-1(π2)]+∑p=0s-1(-1)pT2s-2pD2p(π2)}
(s=1,2,…, for s=1: D0(x)=1).Similarly, from Theorem6 for x=π/2 and  x=π,  we obtain, respectively, the following.Corollary 11.
For the Favard constantsKr, the following recursive representations hold:
(37)K2s-2=4π{∑p=1s-1(-1)pQ2s-2pD2p-1(π2)(-1)s-12D2s-1(π2)+∑p=1s-1(-1)pQ2s-2pD2p-1(π2)},K2s-1=2π{∑p=1s-1(-1)pQ2s-2pD2p-1(π2)(-1)s-12D2s(π)+∑p=1s-1(-1)pQ2s-2pD2p(π)},
where s=1,2,…, and for s=1: D0(x)=1, Q0=0.Forx=π, one can get (23) too by replacing previously s by s+1.Forx=π/2 from Theorem 6, (the second formula in (28)), we will have also the next analogous corollary.Corollary 12.
For numbersQ2s,  the following recursive representations hold:
(38)Q2s=4s4s-1{∑p=1s-1(-1)p+1Q2s-2pD2p(π2)(-1)s2D2s(π2)+∑p=1s-1(-1)p+1Q2s-2pD2p(π2)},
(s=1,2,…, for s=1: D0(x)=1, Q0=0).By the same manner from Theorem7 for x=π and x=π/2,  we obtain, respectively, the formulas for Kr (r=1,2,3,…) different from these in Theorem 1.Corollary 13.
For the Favard constantsK2s-3 and K2s-1, the following recursive representations hold:
(39)K2s-3=(-1)sπ{D2s-2(π)+∑p=1s-2(-1)pK2p-1D2s-2p-1(π)},
(s=2,3,…, for s=2, K1D1(π)  must be canceled) and
(40)K2s-1=(-1)s-1{D2s-1(π2)+∑p=1s-1(-1)pK2p-1D2s-2p(π2)},
(s=1,2,3,…, for s=1, K1D0(π/2) must be canceled).The remaining cases forx=π/2 and x=π immediately lead to Theorem 1 after replacing s by s+1.From Theorem8 for x=π/2, one can get, respectively, other representations for Kr (r=1,2,3,…), different from these in Theorem 1.Corollary 14.
For the Favard constantsKr, the following recursive representations hold:
(41)K2s-2=(-1)s{D2s-2(π2)+∑p=1s-2(-1)pK2pD2s-2p-2(π2)},K2s-1=(-1)s-1{D2s-1(π2)+∑p=1s-1(-1)pK2pD2s-2p-1(π2)},
where s=1,2,3,…, for s=1: D0(π/2)=-1, and for s=2: K2D0(π/2)  must be canceled.Corollary 15.
From the differenceT2s-Q2s=(π/2)K2s-1 (s=1,2,…) and after replacing s by s+1 in the obtained expression, one gets the following formula:
(42)K2s+1=(-1)sπD2s+2(π)+∑p=1s(-1)p+1π2p(2p+1)!K2s-2p+1(s=0,1,…).This is somewhat better than the corresponding formula in Theorem1, because (2s+1)!>2(2s)! for s=1,2,….


## 3. Recursive Representations forK~r and Some Fourier Series

Here we will get down to the integral representation and recursive formulas for the constantsK~r, defined in (2). As one can see they are closely linked with the approximation of the conjugate classes of functions obtained on the base of the Hilbert transform [1, 26]. First of all let us note their representations easily obtained by means of special functions in (3) as it is shown in [2, 5.1.4] for the Catalan constant as follows:(43)K~r={4π(1-2-(2s+1))ζ(2s+1),r=2s,s=1,2,3,…,22-4sπ(ζ(2s,14)-ζ(2s,34))=2πΓ(2s)∫0∞t2s-1chtdt,r=2s-1,s=1,2,3,….In the beginning we will prove the following assertion.Theorem 16.
For the constantsK~r (r=1,2,…), the following recursive representations hold:
(44)K~2s-1=2(-1)sπA~2s-1(π2)+∑p=1s-1K~2s-2p(-1)p-1(2p-1)!(π2)2p-1,K~2s=2(-1)sπA~2s(π2)+∑p=1s-1K~2s-2p(-1)p-1(2p)!(π2)2p(s=1,2,3,…;K~0=def0).Proof.
The proof of this theorem is based on induction again. However, for completeness we must give somewhat more detailed considerations at first steps, which underline further discussions.
Let us start by the well-known Fourier expansions (see, e.g., [2, 5.4])
(45)∑v=0∞sin[(2v+1)x](2v+1)2=-12A~1(x)(0≤x≤π),∑n=1∞sinnxn2=-B~1(x)(0≤x≤2π).
Forx=π/2, the first equality gives us immediately (π/4)K~1=-(1/2)A~1(π/2). At the same time, the second equality in (45) leads to (π/4)K~1=-B~1(π/2). So we obtain the integral representation of  K~1 in the form
(46)K~1=-4πB~1(π2)=-2πA~1(π2)=1.166243616123275⋯.
We have another integral representation of the same constantK~1 in our paper [8].
Further by integration of both sides of the first equality in (45), we get
(47)∑v=0∞cos[(2v+1)x](2v+1)3=12A~2(x)+π4K~2(0≤x≤π),
from where for x=π/2 and x=π we have simultaneously 0=(1/2)A~2(π/2)+(π/4)K~2 and -(π/4)K~2=(1/2)A~2(π)+(π/4)K~2, and consequently
(48)K~2=-2πA~2(π2)=-1πA~2(π).
Then, after the integration of both sides of the second equality in (45), we obtain
(49)∑n=1∞cosnxn3=B~2(x)+T3(0≤x≤2π),
from where for x=π we have (π/2)K~2=-B~2(π). If we correspond this with (48), we find
(50)K~2=-2πB~2(π)=-2πA~2(π2)=-1πA~2(π)=1.339193086109090⋯.
In order to obtain the integral representation ofK~3, we must, at first, integrate the both sides of (47) as
(51)∑v=0∞sin[(2v+1)x](2v+1)4=12A~3(x)+π4K~2D1(x)(0≤x≤π).
Next it remains to putx=π/2 as
(52)K~3=2πA~3(π2)+π2K~2=2πA~3(π2)-A~2(π2).
As another integral representation ofK~3, we can get after integration of both sides of (49)
(53)∑n=1∞sinnxn4=B~3(x)+T3D1(x)(0≤x≤2π).
On one hand, forx=π, (53) gives
(54)T3=-1πB~3(π)=1.20205690⋯.
On the other hand, the same equality (53) for x=π/2 gives ∑v=0∞sin[(2v+1)π/2]/(2v+1)4=B~3(π/2)-(1/2)B~3(π). Then
(55)K~3=4πB~3(π2)-2πB~3(π)=2πA~3(π2)-A~2(π2)=1.259163310827165⋯.
For the constantK~4 there are also different ways to receive its integral representations. One of them is based on the integration of both sides of (53) as
(56)-∑n=1∞cosnxn5+T5=B~4(x)+T3D2(x)(0≤x≤2π).
Forx=π, (56) gives  (π/2)K~4=B~4(π)+(π2/2)T3. Admitting (54) we will have
(57)K~4=2πB~4(π)-B~3(π)=1.278999378416936⋯.
Another integral representation ofK~4 can be obtained by integration of both sides of (51) as
(58)-∑v=0∞cos[(2v+1)x](2v+1)5+π4K~4=12A~4(x)+π4K~2D2(x)(0≤x≤π).
Forx=π/2 and x=π, we get, respectively,
(59)K~4=2πA~4(π2)-π4A~2(π2)=2πA~4(π2)+π28K~2=1πA~4(π)+π24K~2.
Further after integration of both sides of (56), we get
(60)-∑v=0∞sinnxn6+T5D1(x)=B~5(x)+T3D3(x)(0≤x≤2π).
On one hand, if we put herex=π and admit (54), we will have
(61)T5=1πB~5(π)+π23!T3=1πB~5(π)-π6B~3(π)=1.03692776⋯.
On the other hand, the same formula (60) for x=π/2 gives us -(π/4)K~5+(π/2)T5=B~5(π/2)+(π3/233!)T3. Then admitting (54) and (61), we obtain
(62)K~5=-π4B~5(π2)+2πB~5(π)-π4B~3(π)=1.271565517671139⋯.
Next, in order to receive another integral representation ofK~5, we must integrate both sides of (58). So we will have
(63)-∑v=0∞sin[(2v+1)x](2v+1)6+π4K~4D1(x)=12A~2(x)+π4K~2D3(x)(0≤x≤π).
Puttingx=π/2, here, we get
(64)K~5=-2πA~5(π2)+π2K~4-π3233!K~2=-2πA~5(π2)+A~4(π2)-π222·3A~2(π2).
Now after this preparatory work, we can go on the indicated procedure which leads us to the general recursive representations (43) and so complete the proof.The previously stated procedure gives us the opportunity to obtain recursive formulas for the multiple integralsA~r, B~r, and C~r in (4) for a given x.First of all we will note that on the base of induction and with the help of the first formula in (45), (47), (51), (58), and (63), we can lay down the following.Theorem 17.
For the multiple singular integralsA~r, the following recursive representations hold:
(65)A~2s-1(x)=2(-1)s∑v=0∞sin[(2v+1)x](2v+1)2s+π2∑p=1s-1(-1)pK~2pD2s-2p-1(x),A~2s(x)=-2(-1)s∑v=0∞cos[(2v+1)x](2v+1)2s+1+π2∑p=1s(-1)pK~2pD2s-2p(x),s=1,2,3,… (0≤x≤π), D0=1, for s=0: A~0(x)=ln(tan(x/2)) (0<x<π), and K~0=0, where
(66)∑v=0∞sin[(2v+1)x](2v+1)2s=12{∑n=1∞sinnxn2s-∑n=1∞(-1)nsinnxn2s}=1Γ(2s)∫0∞t2s-1et(e2t+1)sinx(e2t+1)2-4e2tcos2xdt,∑v=0∞cos[(2v+1)x](2v+1)2s+1=12{∑n=1∞cosnxn2s+1-∑n=1∞(-1)ncosnxn2s+1}=1Γ(2s+1)∫0∞t2set(e2t-1)cosx(e2t+1)2-4e2tcos2xdt.By analogy with the previous and by means of the second formula in (45), (49), (53), (56), and (60), one can get the following.Theorem 18.
For the multiple singular integralsB~r, the following recursive representations hold:
(67)B~2s-1(x)=(-1)s{∑n=1∞sinnxn2s+∑p=1s-1(-1)pT2s-2p+1D2p-1(x)},B~2s-2(x)=(-1)s{∑n=1∞cosnxn2s-1+∑p=1s-1(-1)pT2s-2p+1D2p-2(x)},s=1,2,3,… (0≤x≤2π), D0=1, for s=1: T1Dk (k=0,1) declines, by definition B~0(x)=ln(2sin(x/2))  (0<x<2π), where (see [2])
(68)∑n=1∞sinnxn2s=1Γ(2s)∫0∞t2s-1etsinx1-2etcosx+e2tdt(x≠0,s=1,2,3,…),∑n=1∞cosnxn2s-1=1Γ(2s-1)∫0∞t2s-2(etcosx-1)1-2etcosx+e2tdt(x≠0,s=1,2,3,…).As similar representations ofC~r(x), one can get on the base of the expansion (see [2, 5.4.2])
(69)∑n=1∞(-1)ncosnxn=-ln(2cos(x2))(-π<x<π),Q1=-ln2.Theorem 19.
For the multiple singular integralsC~r, the following recursive representations hold:
(70)C~2s-1(x)=(-1)s{∑n=1∞(-1)nsinnxn2s+∑p=1s-1(-1)pQ2s-2p+1D2p-1(x)},C~2s-2(x)=(-1)s{∑n=1∞(-1)ncosnxn2s-1+∑p=1s-1(-1)pQ2s-2p+1D2p-2(x)},s=1,2,3,… (-π≤x≤π), for s=1: C~0(x)=ln(2cos(x/2)) (-π<x<π), Q1Dk (k=0,1) declines, where (see [2])
(71)∑n=1∞(-1)nsinnxn2s=-1Γ(2s)∫0∞t2s-1etsinx1+2etcosx+e2tdt(s=1,2,3,…),∑n=1∞(-1)ncosnxn2s-1=-1Γ(2s-1)∫0∞t2s-2(etcosx+1)1+2etcosx+e2tdt(s=1,2,…).By analogy with the previous, if we start from the expansion [2, 5.4.6]
(72)∑v=0∞(-1)vsin[(2v+1)x]2v+1=-12ln(tan(π4-x2))(-π2<x<π2),
we can obtain the next theorem.Theorem 20.
The following recursive formulas hold:(73)∫0π/2-xA~2s-1(u)du=-2(-1)s∑v=0∞(-1)vsin[(2v+1)x](2v+1)2s+1+π2∑p=1s(-1)pK~2pD2s-2p(π2-x),∫0π/2-xA~2s-2(u)du=2(-1)s∑v=0∞(-1)vcos[(2v+1)x](2v+1)2s-π2∑p=1s-1(-1)pK~2pD2s-2p-1(π2-x),s=1,2,3,… (-π/2≤x≤π/2),  D0(x)=1,  for s=1: A~0(u)=lntan(u/2) (-π/2<u<π/2), where
(74)∑v=0∞(-1)vsin[(2v+1)x](2v+1)2s+1=1Γ(2s+1)∫0∞t2set(e2t-1)sinx(e2t+1)2-4e2tsin2xdt(s=1,2,…),∑v=0∞(-1)vcos[(2v+1)x](2v+1)2s=1Γ(2s)∫0∞t2s-1et(e2t+1)cosx(e2t+1)2-4e2tsin2xdt(s=1,2,…).The proof of this theorem requires to take integration fromπ/2 to x and then to make the substitution  t=π/2-x.From Table2 one can see that A~r(x)=B~r(x)-C~r(x) (0≤x≤π), r=1,2,…. The constants K~r satisfy the following inequalities [1]:
(75)1<K~1<K~3<K~5<⋯<4π<⋯<K~6<K~4<K~2<π2.Table 2
Calculated values of the magnitudesA~r(x), B~r(x), and C~r(x) for x=π/2,π, by using formulas (65), (67), and (70), respectively.
r
A
~
r
(
π
/
2
)
B
~
r
(
π
/
2
)
C
~
r
(
π
/
2
)
0
0.346573590279972654709
0.34657359027997265471
1
−1.831931188354438030
−0.91596559417721901505
0.91596559417721901506
2
−2.103599580529289999
−1.31474973783080624966
0.78884984269848374979
3
−1.326437390660483389
−0.89924201634043412749
0.42719537432004926159
4
−0.586164434174921923
−0.41567176475313624651
0.17049266942178567636
5
−0.200415775434410589
−0.14636851588819525831
0.05404725954621533063
6
−0.055999130148030908
−0.04177075772178286763
0.01422837242624804042
7
−0.013242890305882861
−0.01003832077074076074
0.00320456953514210000
8
−0.002716221784223860
−0.00208540018388810697
0.00063082160033575329
9
−0.000492033289602442
−0.00038173093475665008
0.00011030235484579206
10
−0.000079824623761109
−0.00006247448634651347
0.00001735013741459518
11
−0.000011727851308441
−0.00000924763886476940
0.00000248021244367167
12
−0.000001574609462951
−0.00000124968150024699
0.00000032492796270360
r
A
~
r
(
π
)
B
~
r
(
π
)
C
~
r
(
π
)
0
0.693147180559945309417
1
0
0
0
2
−4.2071991610585799989
−2.10359958052928999945
2.10359958052928999945
3
−6.6086529882853881226
−3.77637313616307892720
2.83227985212230919540
4
−6.3627527878802461560
−3.92286552530160912232
2.43988726257863703368
5
−4.5591894893017946564
−2.95428020294335768590
1.60490928635843697049
6
−2.6255391709472096141
−1.76271265891422894113
0.86282651203298067295
7
−1.2684915655900274347
−0.87472015815662857450
0.39377140743339886016
8
−0.5287656648068309189
−0.37237211738940254175
0.15639354741742837714
9
−0.1940031559506448617
−0.13897143291435930434
0.05503172303628555739
10
−0.0635985732918145806
−0.04620733330628432785
0.01739123998553025277
11
−0.0188489390019113804
−0.01385968931040814707
0.00498924969150323336
12
−0.0050985327885974909
−0.00378779048977775268
0.00131074229881973820Remark 21.
The equalities (49), (53), (56), and (60) outline a procedure for summing up the numerical series T2s+1 (s=1,2,…) in (2). It leads us to the following.Corollary 22.
The following recursive representation holds:(76)T2s+1=(-1)sπB~2s+1(π)+∑p=1s-1(-1)p+1π2p(2p+1)!T2s-2p+1,s=1,2,…fors=1  the term(π2/6)T1  declines.For comparison we will note also the well-known formula [2, 5.1.2]
(77)T2s+1=ζ(2s+1)=1Γ(2s+1)∫0∞t2set-1dt=-1(2s)!Ψ(2s)(1)(s=1,2,…).The same procedure based on induction and preliminary multiple integration of both sides of (69) leads us to the next.Corollary 23.
The following recursive representation holds:(78)Q2s+1=(-1)sπC~2s+1(π)+∑p=1s-1(-1)p+1π2p(2p+1)!Q2s-2p+1,s=1,2,…, fors=1 the term (π2/6)Q1declines.As in previous, let us give the alternate formula [2, 5.1.3]
(79)Q2s+1=(2-2s-1)ζ(2s+1)=-1Γ(2s+1)∫0∞t2set+1dt(s=1,2,…).At the end we would like to note also that one can get many other representations (through multiple integrals) of the constantsK~r (r=1,2,…) from Theorems 17–20 setting, in particular,  x=π/2 or x=π, as it is made for the constants Kr.From the differenceT2s+1-Q2s+1=(π/2)K~2s  (s=1,2,…), one can get the following formula:
(80)K~2s=2(-1)sπ2A~2s+1(π)+∑p=1s-1(-1)p+1π2p(2p+1)!K~2s-2p(s=1,2,…),
which is somewhat inferior to the similar recursion formula in Theorem 16 because (2s-1)!≤22s-2(2s-2)! for  s=1,2,….As an exception we will give the inverse formula of (80) as
(81)A~2s+1(π)=(-1)s∑p=1s(-1)p+1π2p2(2p-1)!K~2s-2p+2(s=1,2,…).If we, by definition, layK~0=0, then the equality (80) is valid for s=0 too.


## 4. Some Notes on Numerical and Computer Implementations of the Derived Formulas

We will consider some aspects of numerical and symbolic calculations of the Favard constantsKr, singular integrals (4), and summation of series.Let us take only the firstm-1, m≤s+1 terms in the final sums in the right-hand side of the formulas (7a) and (7b) and denote the remaining truncation sums by
(82)S1,m=∑j=ms(-1)j-1πjp(2j)!K2s-2j+1,S2,m=∑j=ms(-1)j-1(2j-1)!(π2)2j-1K2s-2j+1,
respectively.Theorem 24.
Form≥2 and any s>m, the following estimates for the truncation errors (82) hold:
(83)|S1,m|=O(π2m(2m)!),|S2,m|=O((π2)2m-11(2m-1)!),
where Landau big O notation is used. For the Landau notation see [27].Proof.
By means of the inequalities (25), it is easy to obtain
(84)|S1,m|=|∑j=ms(-1)j-1πjp(2j)!K2s-2j+1|≤π2∑j=msπ2j(2j)!=π2π2m(2m)!×(1+π2(2m+1)(2m+2)+⋯+π2s-2m(2m+1)⋯(2s))≤π2m+12(2m)!(1+105.6+10252.62+⋯)≤π2m+12(2m)!(1+13+132+⋯)=π2m+12(2m)!32≤3π2m(2m)!.
Consequently,|S1,m|=O(π2m/(2m)!).  In the same way for the second truncation sum, we have
(85)|S2,m|≤2(π2)2m-11(2m-1)!,so|S2,m|=O((π2)2m-11(2m-1)!).By the details of the proof it follows that the obtained representations did not depend ons, for any m<s. The a priori estimates (83) can be used for calculation of Kr with a given numerical precision ε in order to decrease the number of addends in the sums for larger s at the condition m<s.Remark 25.
The error estimates, similar to (83), can be established for other recurrence representations in this paper too.In Algorithm1, we provide the simplest Mathematica code for exact symbolic and numerical calculation with optional 30 digits accuracy of the Favard constants Kr for r=1,2,3,…,m, based on recursive representations (7a) and (7b) for a given arbitrary integer m>0. The obtained results are given in Table 1. We have to note that this code is not the most economic. It can be seen that the thrifty code will take about 2m2+8m or O(m2) arithmetic operations in (7a) and (7b).Algorithm 1:
Mathematica code for exact symbolic and approximate computation with optional 30 digits accuracy of the Favard constants by recursive representations (7a) and (7b).
m
=
12;   Array  [K,m];  K1=π2;  d[n-,x-]≔xnn!
For
[[2s,π2]s=1,s≤m/2,s++,
K2s+1=12(∑p=1sK2s+1-2p(-1)p-1π2p(2p)!(-1)sd[2s+1,π]+∑p=1sK2s+1-2p(-1)p-1π2p(2p)!);
K2s=(-1)sd[2s,π2]+∑p=1sK2s+1-2p(-1)p-1(π2)2p-1(2p-1)!;
[2s,π2]]
For [s = 1,  s ≤ m,  s ++, Print [Ks,  " ",  N[Ks, 30]]];The basic advantage of using formulas (7a) and (7b) is that they contain only finite number of terms (i.e., finite number of arithmetic operations) in comparison with the initial formula Kr=(4/π)∑ν=0∞((-1)ν(r+1)/(2ν+1)r+1) (r=0,1,2,…), which needs the calculation of the slowly convergent infinite sum. It must be also mentioned that in Mathematica, Maple, and other powerful mathematical software packages, the Favard constants are represented by sums of Zeta and related functions, which are calculated by the use of Euler-Maclaurin summation and functional equations. Near the critical strip they also use the Riemann-Siegel formula (see, e.g., [28, A.9.4]).Finally, we will note that the direct numerical integration in (4) is very difficult. This way a big opportunity is given by Theorems 17, 18, and 19 (formulas (65), (67) and (70)) for effective numerical calculation of the classes of multiple singular integrals A~r(x), B~r(x) and C~r(x) for any x. Their computation is reduced to find the finite number of the Favard constants K~v for all v≤r and the calculation of one additional numerical sum. Numerical values for some of the singular integrals are presented in Table 2.


## 5. Concluding Remarks

As it became clear, there are many different ways and well-developed computer programs at present, for calculation of the significant constants in mathematics (in particular Favard constants) and for summation of important numerical series. Most of them are based on using of generalized functions as one can see in [2–5]. Nonetheless, we hope that our previously stated approach through integration of Fourier series appears to be more convenient and has its theoretical and practical meanings in the scope of applications, in particular for computing of the pointed special types of multiple singular integrals. The basic result with respect to multiple integrals reduces their calculation to finite number of numerical sums.



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