Summary
1 Dirichlet Beta Generating Functions
sech x , sec x and csc x can be expanded to Fourier series and Taylor series. And if the termwise higher order integration of these is carried out, Dirichlet Beta at a natural number are obtained.
Where, these are automorphisms which are expressed by lower betas. However, in this chapter, we stop those so far.
The work that obtain the non-automorphism formulas by removing lower betas from these is done in the next chapter
" 2 Formulas for Dirichlet Beta " .
In this chapter, we obtain the following polynomials from the beta generating functions of each family of sech, sec and csc . Where, Dirichlet Beta and Dirichlet Lambda are as follows.
( ) x = Σ
r=0
( 2 r +1 )
x( -1 )
r, ( ) x = Σ
r=1
( 2 r -1 )
x1
Bernoulli numbers and Euler numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 , E
0=1 , E
2=-1 , E
4=5 , E
6=-61 , E
8=1385 ,
Harmonic number is
H
s= Σ
t=1 s
1/t = ( 1+s ) +
( ) n = Σ
r=0
( 2 r +1 )
n( -1 )
re
-(2r+1)x- 2 ( -1 )
nΣ
r=0
( 2 r + n ) ! E
2rx
2r+n- Σ
s=1 n-1
s !
( -1 )
sx
s ( n - s )
Σ
r=0
( 2 r +1 )
2n+1( -1 )
rsin {( 2 r +1 ) x }
- 2 ( -1 )
nΣ
r=0
( 2 n +1+2 r ) !
E
2rx
2n+1+2r= Σ
s=0 n-1
( 2 s +1 ! )
( -1 )
sx
2s+1 ( 2 n -2 s )
Σ
r=0
( 2 r +1 )
2n( -1 )
rcos {( 2 r +1 ) x }
- 2 ( -1 )
nΣ
r=0
( 2 n +2 r ) !
E
2rx
2n+2r= Σ
s=0 n-1
( 2 s ) !
( -1 )
sx
2s ( 2 n -2 s )
Σ
r=0
( 2 r +1 )
2n+1( -1 )
rcos {( 2 r +1 ) x }
= Σ
s=0 n
( 2 s ) !
( -1 )
sx
2s ( 2 n +1-2 s )
Σ
r=0
( 2 r +1 )
2n( -1 )
rsin {( 2 r +1 ) x }
= Σ
s=0 n-1
( 2 s +1 ! )
( -1 )
sx
2s+1 ( 2 n -1-2 s )
( 2 n ) =
2( 2 n -1 ! ) ( -1 )
n 2
2n-1 log 4 - H
2n-1
+ 2 ( -1 )
nΣ
r=1
2 r ( 2 r +2 n -1 ! )
2
2r-2 B
2r 2
2n-1+2r- Σ
s=1 n-1
( 2 s -1 ! ) ( -1 )
s 2
2s-1 ( 2 n +1-2 s )
Furthermore, if the termwise higher order differentiation of the Fourier series of each family of sech and sec are carried out, the following expressions are obtained.
( - n ) = 2
n+11 Σ
r=0 n
( -1 )
rnK
rn =1 , 2 , 3 ,
( -2 n ) = 2
2n+11 Σ
r=0 2n
( -1 )
r2nK
rn =1 , 2 , 3 ,
= 2 n E
2nn =1 , 2 , 3 ,
Where, n
K
r is a kind of Eulderian Number and is defined as follows.n
K
r= Σ
k=0 r
( -1 )
k n +1 k ( 2 r +1-2 k )
nn =1 , 2 , 3 ,
2 Formulas for Dirichlet Beta
Here, removing the lower betas from the the automorphism formulas in the previous chapter, we obtain the following non-automorphism formulas. Where, Bernoulli numbers and Euler numbers are as follows.
B
0=1 , B
2=1/6 , B
4=-1/30 , B
6=1/42 , B
8=-1/30 , E
0=1 , E
2=-1 , E
4=5 , E
6=-61 , E
8=1385 ,
And, gamma function and incomplete gamma function were as follows.
( ) p =
0t
p-1e
-tdt , ( p,x ) =
xt
p-1e
-tdt
2.1 Formulas for Beta at natural number For 0 < x /2
, ( ) n = Σ
r=0
Σ
s=0 n-1
s ! ( 2 r +1 )
sx
s( 2 r +1 )
n( -1 )
re
-(2r+1)x+ 2 x
nΣ
r=0
- 2 n r ( n +2 r ) ! E
2rx
2rEspecially,
( ) n = Σ
r=0
Σ
s=0 n-1
s !2
s( 2 r +1 )
s( 2 r +1 )
n( -1 )
re
-(r+1/2)+ 2
n+11 Σ
r=0
- 2 n r ( n +2 r ) !2
2rE
2r ( ) n = Σ
r=0
Σ
s=0 n-1
s ! ( 2 r +1 )
s( 2 r +1 )
n( -1 )
re
-(2r+1)+ 2 1 Σ
r=0
- 2 n r ( n +2 r ) ! E
2rExample
( ) 4 = Σ
r=0
1+ 1!2
1
2r+1 +
2!2
2( 2r+1 )
2+
3!2
3( 2r+1 )
3( 2r +1 )
4( -1 )
re
-r-2 1
+ 2
51 Σ
r=0
-4 2r ( 4+2r ) !2
2rE
2r
( ) 4 = Σ
r=0
1+ 1!
2r+1 + 2!
(2r+1)
2+ 3!
(2r+1)
3( 2r +1 )
4( -1 )
re
-2r-1+ 2 1 Σ
r=0
-4 2r ( 4+2r ) ! E
2r2.2 Formulas for Beta at even number For 0 < x /2
, ( 2 n ) = Σ
r=0
Σ
s=0 n-1
( 2 s ) !
E
2s{( 2 r +1 ) x }
2s( 2 r +1 )
2n( -1 )
rcos {( 2 r +1 ) x }
- 2 ( -1 )
nΣ
r=0
Σ
s=0
n-1
2 n 2 +2 s r E
2s( 2 n +2 r ) !
E
2rx
2n+2r ( 2 n ) = - x 1 Σ
r=1
Σ
s=0 n-1
( 2 s ) !
( -1 )
sB
2s 2
2s-2 {( 2 r +1 ) x }
2s( 2 r +1 )
2n+1( -1 )
rsin {( 2 r +1 ) x }
+ ( -1 )
n2 x
2nΣ
r=1
Σ
s=0
n-1
( 2 s ) !( 2 n +1+2 r -2 s ) !
2
2s-2 B
2s2 r
E
2rx
2rEspecially,
( 2 n ) = 2 ( -1 )
n-1Σ
r=0
Σ
s=0
n-1
2 n 2 +2 s r E
2s( 2 n +2 r ) !
E
2r 2
2n+2rExample
( ) 4 = - 2 1 Σ
r=0
4+2 0 r E
0+ 4+2 2 r E
2 ( 4+2 r ) !
E
2r 2
4+2r
( ) 6 = 2 1 Σ
r=0
6+2r 0 E
0+ 6+2r 2 E
2+ 6+2r 4 E
4
( 6+2 r ) !
E
2r 2
6+2r2.3 Formulas for Beta at odd number For 0 < x /2
, ( 2 n -1 ) = Σ
r=0
Σ
s=0 n-1
( 2 s ) !
E
2s{( 2 r +1 ) x }
2s( 2 r +1 )
2n-1( -1 )
rcos {( 2 r +1 ) x }
( 2 n -1 ) = - x 1 Σ
r=0
Σ
s=0 n-1
( 2 s ) !
( -1 )
sB
2s 2
2s-2 {( 2 r +1 ) x }
2s( 2 r +1 )
2n( -1 )
rsin {( 2 r +1 ) x }
Especially,
( 2 n -1 ) = 4
( 2 n -2 ! )
E
2n-2
2
2n-22.4 Formulas for Beta at complex number
When
p
is a complex number such thatp 1 , 0 , -1 , -2 ,
,For x = u+ v i s . t . 0 < | | x 2 , u0
, ( ) p = Σ
r=0
( ) p
p , ( 2 r +1 ) x
( 2 r +1 )
p( -1 )
r+ 2 x
pΣ
r=0
- 2 p r ( p +1+2 r ) E
2rx
2rEspecially,
( ) p = Σ
r=0
( ) p
( p , 2 r +1 )
( 2 r +1 )
p( -1 )
r+ 2 1 Σ
r=0
- 2 p r ( p +1+2 r ) E
2r3 Global definition of Dirichlet Beta and Generalized Euler Number
Diriclet beta function is defined on the whole complex plane with patches as follows.
( ) p = Σ
r=1 2 ( ( 2 -1
1-pr -1 ) cos
r-1)
pp 2 ( 1- p ) Σ
r=1( 2 r -1 )
1-pRe ( ) p 0 ( -1 )
r-1Re ( ) p < 0
This is inconvenient. so, we focus on the following sequence.
n
B
r= Σ
s=0 r
( -1 )
r-srC
s s - 2 1
nr =0 , 1 , 2 , ,n
Using this sequence, we can define Diriclet beta function on the whole complex plane as follows.
Definition 3.2.1
( ) p = Σ
r=1
2
r+11 Σ
s=1 r
( -1 )
s-1 r s ( 2 s -1 )
-pFurthermore, by using this sequence, Euler Number can be defined on the whole complex plane.
Definition 3.3.1
E
p= Σ
r=1
2
r1 Σ
s=1 r
( -1 )
s-1 r s ( 2 s -1 )
p4 Completed Dirichlet Beta
In 4.1, symmetric functional equations are derived from functional equations.
Formula 4.1.1
2
1+z 2 1 + 2 z ( ) z = 2
2-z 2 1 + 1- 2 z ( 1- z )
2
23+z
4 3 + 2 z 2 1 + z = 2
23-z
4 3 - 2 z 2 1 - z
In 4.2, we define the completed Dirichlet beta functions
( ) z , ( ) z
as follows, respectively. ( ) z = 2
1+z 1+ 2 z ( ) z
( ) z = 2
23+z
2
1 2
3 +z 2
1 + z
Then, Formula 4.1.1 is expressed as follows.
( ) z = ( 1- z )
( ) z = ( - z )
From the latter, we can see that
( ) z
is an even function. Therefore, ( ) z
has the same properties as completed Riemann zeta function ( ) z
. ( See " 07 Completed Riemann Zeta ". ) And, as in ( ) z
, the following theorem holds.Theorem 4.2.1
If Dirichlet beta function
( ) z
has a non-trivial zero whose real part is not1/2
, the one set consists of the following four.1/2+
1 i
1, 1/2-
1 i
1( 0 <
1< 1/2 )
05 Factorization of Completed Dirichlet Beta
In 5.1, the following Hadamard product is derived.Formula 5.1.1
Let completed beta function be as follows.
( ) z = 2
1+z 1+ 2 z ( ) z
When non-trivial zeros of
( ) z
arez
k= x
k i y
kk =1, 2,3,
and
is Euler-Mascheroni constant, ( ) z
is expressed by the Hadamard product as follows. ( ) z = e
2
3log -2
- log2 - 4log
4 3 zΠ
k=1
1- z
k
z e
zkz
( ) z = e
2
3log -2
- log2 - 4log
4 3 zΠ
n=1
1- x
n
2
+ y
n22x
nz
+
x
n2+ y
n2z
2e
xn2+yn2 2xn z
And, the following special values are obtained.
Π
n=1
1- x
n2+ y
n2
2 x
n- 1 e
xn2+yn2 2xn
= e
4log
4 3 +2
+ log2 - 2 3log
= 1.08088915
Π
n=1
1- x
n iy
n
2
1 1- x
n i y
n
2
1 = ( -1 ) = 1.16624361
In 5.2, we consider how the formulas in the previous section are expressed when non-trivial zeros whose real part is 1/2 and non-trivial zeros whose real part is not 1/2 are mixed. Then, we obtain the following theorems.
Theorem 5.2.2
Let
be Euler-Mascheroni constant, non-trivial zeros of Dirichlet beta function arex
n+i y
nn =1, 2, 3,
. Among them, zeros whose real part is 1/2 are1/2 i y
rr =1, 2,3,
and zeros whose real parts is not 1/2 are1/2
si
s
0 <
s< 1/2
s =1, 2,3,
. Then the following expressions hold.Π
n=1
1- x
n2+ y
n2
2x
n-1
= 1
Σ
n=1
x
n2+ y
n22x
n= Σ
r=1
1/4 + y
r21 + Σ
s=1
1/2
s2
s2
1 2
s+
1/2
s2
s21 2
sΣ
n=1
x
n2+ y
n22x
n= 4log 4 3 +
2
+ log 2 - 2 3 log
= 0.07778398
Formula 5.2.3 ( Special values )
When non-trivial zeros of Dirichlet beta function are
x
k i y
kk =1, 2,3,
, the following expressions hold.Π
n=1
1- x
n i y
n
1 1- x
n iy
n
1 = 1
Π
n=1
1+ x
n i y
n
1 1+ x
n iy
n
1 = ( -1 ) = 1.1662436
Theorem 5.2.4
Let non-trivial zeros of Dirichlet beta function are
x
n+ i y
nn =1, 2, 3,
and
be Euler-Mascheroni constant.If the following expression holds, non-trivial zeros whose real parts is not
1/2
do not exist.Σ
n=1
1/4 + y
n21 = 4 log 4 3 + 2 + log 2 - 3 log 2 = 0.07778398
Incidentally, when this was calculated using 10000
y
r, both sides coincided with the decimal point 3 digits.In 5.3, we show that
( ) z
is factored completely.Theorem 5.3.1 ( Factorization of
( )z )Let Dirichlet beta function be
( ) z
, the non-trivial zeros arez
n= x
n i y
nn =1, 2,3,
and completed beta function be as follows. ( ) z = 2
1+z 1+ 2 z ( ) z
Then,
( ) z
is factorized as follows. ( ) z = Π
n=1
1- x
n2+ y
n2
2 x
nz +
x
n2+ y
n2z
2In 5.4, we first derive the factorization of
( ) z
.Theorem 5.4.1 ( Factorization of
( )z )Let Dirichlet beta function be
( ) z
, the non-trivial zeros arez
n= x
n i y
nn =1, 2,3,
and completed beta function be as follows. ( ) z = 2
23+z
2
1 2
3 +z 2
1 + z
Then,
( ) z
is factorized as follows. ( ) z = ( ) 0 Π
n=1
1-
x
n-1/2
2+ y
n2
2
x
n-1/2
z
+
x
n-1/2
2+ y
n2z
2Where,
( ) 0 = Π
n=1
x
n2+ y
n2
x
n-1/2
2+ y
n2=
2
3/2 4
3 2
1 = 0.98071361
And, using this theorem and Theorem 4.2.1 in the previous section, we obtaine the following theorem.
Theorem 5.4.4
When Dirichlet beta function is
( ) z
and the non-trivial zeros arez
n= x
n iy
nn =1 , 2 , 3 ,
, If the following expression holds, non-trivial zeros whose real parts is not1/2
do not exist.Π
r=1
1/4 + y
r2y
r2= 2
3/2 4 3 2 1 = 0.98071361
(4.40) Incidentally, when this was calculated using 10000y
r, both sides coincided with the decimal point 4 digits.06 Zeros on the Critical Line of Dirichlet Beta
In 6.1, substituting
z = 0+iy
for the completed Dirichlet beta ( ) z
,
h( ) z = 2
23+iy
2 1 2 3 + iy 2 1 + iy
We use this to calculate the zeros on the critical line. However, this function is too small in absolute value and can only find the zeros up to
y =917
.So we normalize
h( ) y
and define the following sign function.sgn ( ) y = -
h( ) y
h( ) y
= - 2
i y
2
1 2
3 + iy 2
1 + iy
2
1 2
3 + iy 2
1 + iy
Using this sign function
sgn ( ) y
, we can find the zeros at largey
.In 6.2, multiplying this sign function
sgn ( ) y
by the absolute value of the Dirichlet beta ( 1/2+i y )
, we obtain a smooth functionB ( ) y
.B ( ) y = sgn ( ) y 2 1 + iy = - 2
i y
2
1 2
3 + i y
2
1 2
3 + i y
2
1 + i y
Using this
B ( ) y
function, we can find the zeros on the critical line of ( ) z
by the intersection of the curve and they
-axis In 6.3, first, a lemma is prepared.Lemma
When
f ( ) z
is a complex function defined on the domainD
, the following expression holds.e
i Im log f ( )z=
f ( ) z
f ( ) z
Applying this lemma to the gamma function in the 6.2 ,
B ( ) y = - 2
i y
2
1 2
3 + i y
2
1 2
3 + i y
2
1 + i y
= - 2
i ye
i Im log
2
1
2
3+
i y
2
1 + i y
= - e
i
Im log
2
1
2
3+
i y
- 2 ylog4
2 1 + i y
From this, we obtain
B ( ) y = - e
i ( )y 2 1 + i y
where, ( ) y = Im log 2 1 2 3 + i y - 2 y log 4
This is Riemann-Siegel style
B
function .2012.04.13
2019.03.27 Updated
Kano Kono Alien's Mathematics