26 Higher and Super Calculus of Zeta Function etc
26.1 Higher and Super Calculus of Riemann Zeta Function
26.1.1 Higher and Super Integral of Riemann Zeta Function
Formula 26.1.1h ( Higher Integral ) n <>n When z is Riemann zeta function, z is the lineal n -th order primitive, Hn ( =Σ1/k ) is k=1 a harmonic number ( where H0 =0 ) , the following expression holds on whole complex plane.
n-1 r+n n z -1 r z -1 ()z = logz -1 -Hn-1 +Σ()-1 r n =1,2,3, n -1 ! r=0 r +n ! (1.1h)
Where, r is Stieltjes constant defined by the following expression. r n ()log k r ()log n +1 r = lim Σ - n k=1 k r+1
Proof It is known that the Riemann zeta function z is expanded to Laurent series around 1 as follows.
r 1 r ()z-1 ()z = + Σ()-1 r z-1 r=0 r! Then, integrating the both sides with respect to z without considering the constant of the integration,
r+1 <>1 r z -1 ()z = logz -1 + Σ()-1 r r=0 r +1 ! Integrating this once more without considering the constant of the integration,
1 r+2 <>2 z -1 r z -1 ()z = logz -1 -1 + Σ()-1 r 1! r=0 r +2 ! Integrating this once more without considering the constant of the integration,
2 r+3 <>3 z -1 1 r z -1 ()z = logz -1 - 1+ +Σ()-1 r 2! 2 r=0 r +3 ! Hereafter, by induction,
n-1 r+n n z -1 1 1 r z -1 ()z = logz -1 - 1+ ++ +Σ()-1 r n -1 ! 2 n -1 r=0 r +n !
Rewriting the harmonic number as Hn-1 , we obtain the desired expression.
Example The 1st order integral <>1 When the real part and the imaginary part of x +iy are illustrated, it is as follows. The left figure is the real part and the right figure is the imaginary part.
- 1 -
Note If the left side of (1.1h) is represented by integral symbols, it is as follows.
z z n-1 r+n z -1 r z -1 ()zdz = logz -1 -Hn +Σ()-1 r n -1 an a1 -1 ! r=0 r +n ! Here, the lower limits of the integral are as follows.
a1 = 1.669008 , a2 = 2.641300 , a3 = 3.610288 , That is, this is a higher integral with variable lower limits. And this is a lineal higher integral . cf.
The higher integral with variable lower limits a1 = a2 = = an = 0 is as folows. z z z -1 n-1 z -1 r+n n r ()zdz = logz -1 -Hn-1 +Σ()-1 r 0 0 n -1 ! r=0 r +n ! n-1 r n-1- r s n-s z -1 n -1 ()-1 Hr z n ()-1 r z - i + Σ -ΣΣ n -1 ! r=1 r!n -1-r ! r=0s=1 r +s !n -s ! As seen from the existence of constant-of-integration polynomials, this is a collateral higher integral.
Formula 26.1.1s ( Super Integral ) <>p When p is a complex number, z is Riemann zeta function, z is the lineal p -th order primitive,
p is gamma function, p is digamma function and r is Stieltjes constant, the following expression holds on whole complex plane.
r+p p log()z-1 -()p -0 p-1 r ()z-1 ()z = ()z-1 +Σ()-1 r (1.1s) ()p r=0 ()1+r+p
Proof From (1.1h) ,
- 2 - n-1 r+n n z -1 r z -1 ()z = logz -1 -Hn-1 +Σ()-1 r n =1,2,3, n -1 ! r=0 r +n ! At first,
()n -1 ! = ()n,Hn-1 = ()n +0 , ()r+n ! = ()1+r+n Using these,
n-1 r+n n z -1 r z -1 ()z = logz -1 -n +0 +Σ()-1 r n r=0 1+r +n Then, this expression also holds at n =0 . Because,
logz -1 -0 logz -1 -0 -()0 = = 0 , = 1 ()0 ()0 So, replacing the natural number n with a complex number p , we obtain the desired expression.
Example The 0.1th order integral <>0.1 When z = x +iy , z and z are illustrated as follows. The left figure is a real part and the right <>0.1 figure is an imaginary part. In both figures, the orange is z and the blue is z . Since p is neer 0 , both curved surfaces look double.
26.1.2 Higher and Super Derivative of Riemann Zeta Function
Formula 26.1.2h ( Higher Derivative ) ()n When z is Riemann zeta function, z is the lineal n -th order derivative and r is Stieltjes constant, the following expression holds on whole complex plane.
-n r-n n ()-1 n ! r ()z-1 (1.2h) ()z = n + Σ()-1 r ()z-1 +1 r=0 1+r-n
Proof It is known that the Riemann zeta function z is expanded to Laurent series around 1 as follows.
- 3 - r 1 r ()z-1 ()z = + Σ()-1 r z-1 r=0 r! Differentiating the both sides n times with respect to z , n 1 r ()n n ()z = + Σ()-1 r ()z-1 r z-1 r=0 r! According to Formula 9.2.1 in " 09 Higher Derivative ",
n 1+ x = x-n 0 1+ -n
-n - +n -n = ()-1 x < 0 - Applying this,
()n ()1+r r n r! r n ()z-1 r = ()z-1 - = ()z-1 - 1+r-n 1+r-n 1 n ()1+n n ! = ()-1 -n ()z-1 -1-n = ()-1 -n z-1 ()1 ()z-1 n+1 Substituting these for the above,
-n r-n n ()-1 n ! r ()z-1 (1.2h) ()z = n + Σ()-1 r ()z-1 +1 r=0 1+r-n
()1 ()2 Example ()0.3 , - 1.1 + 2.3 i When these are calculated by formula manipulation soft Mathematica , it is as follows. We can see that this formula is numerically right.
Formula 26.1.2s ( Super Derivative ) ()p When p is a complex number, z is Riemann zeta function, z is the lineal p -th order derivative,
p is gamma function, p is digamma function and r is Stieltjes constant, the following expression holds on whole complex plane.
logz p r-p p -1 - - - 0 -p-1 r z -1 ()z = z -1 + Σ()-1 r (1.2s) -p r=0 1+r -p
- 4 - Proof From Formula 26.1.2h ,
-n r-n n ()-1 n ! r ()z-1 (1.2h) ()z = n + Σ()-1 r ()z-1 +1 r=0 r-n ! According to Formula 1.3.1 in " 01 Gamma Function & Digamma Function " , ()-n ()-1 -n n ! = - ,n=0,1,2,3, ()-n Using this, (1.2h) is rewritten as follows.
logz n r-n n -1 - - - 0 -n-1 r z -1 ()z = z -1 + Σ()-1 r -n r=0 1+r -p Because,
logz -1 -0 logz -1 -0 = = 0 for n =0,1,2,3, -n So, replacing the natural number n with a complex number p , we obtain the desired expression.
logz p r-p p -1 - - - 0 -p-1 r z -1 ()z = z -1 + Σ()-1 r (1.2s) -p r=0 1+r -p
Note If the sign of p is inverted in (1.2s), it becomes as follows.
logz p r+p -p -1 - - 0 p-1 r z -1 ()z = z -1 + Σ()-1 r p r=0 1+r +p This results in Formula 26.1.1s . That is, the lineal super calculus of Riemann zeta function is seamless.
Example The 1.9th order derivative ()2 ()1.9 When z = x +iy , z and z are illustrated as follows. The left figure is a real part and the ()2 ()1.9 right figure is an imaginary part. In both figures, the orange is z and the blue is z . Since p is near 2 , both curved surfaces look double.
- 5 - 26.2 Higher and Super Calculus of Dirichlet Lambda Function
26.2.1 Higher and Super Integral of Dirichlet Lambda Function
Formula 26.2.1h ( Higher Integral ) n <>n When z is Dirichlet lambda function, z is the lineal n -th order primitive, Hn ( =Σ1/k ) k=1 is a harmonic number ( where H0 =0 ) , the following expression holds on whole complex plane. n-1 <>n z -1 z = logz -1 -H 2n -1 ! n-1
r+1 r-1 r r+n 1 r log 2 r-s z -1 + Σ()-1 r+ -Σ s log2 n =1,2,3, 2 r=0 r +1 s=0s r +n ! (2.1h)
Where, r is Stieltjes constant defined by the following expression. r n ()log k r ()log n +1 r = lim Σ - n k=1 k r+1
Proof According to Formula 3.1.3 in " 03 Complementary Series of Dirichlet Series " ( Dirichlet Series ), Dirichlet lambda function is expanded to Laurent series around 1 as follows.
r+1 r-1 r r 1 1 r log 2 r-s z -1 ()z = + Σ()-1 r + -Σ s log2 2z -1 2 r=0 r +1 s=0s r ! Where, in { } is absent for r =0 . Hereafter, in a way similar to the proof of Formula 26.1.1h , the desired expression is obtained.
Example The 2nd order integral <>1 When the real part and the imaginary part of x +iy are illustrated, it is as follows. The left figure is the real part and the right figure is the imaginary part.
Note This is also lineal higher integral with variable lower limits.The first few of the integral lower limits are as follows.
- 6 - a1 = 1.509052 , a2 = 2.203125 , a3 = 2.891846 ,
Formula 26.2.1s ( Super Integral ) <>p When p is a complex number, z is Dirichlet lambda function, z is the lineal p -th order primitive,
p is gamma function, p is digamma function and r is Stieltjes constant, the following expression holds on whole complex plane. <>p logz -1 - p -0 z = z -1 p-1 2p
r+1 r-1 r r+p 1 r log 2 r-s z -1 + Σ()-1 r+ -Σ s log2 (2.1s) 2 r=0 r +1 s=0s 1+r +p
Proof In a way similar to the proof of Formula 26.1.1s , the desired expression is obtained.
Example The 0.05th order integral <>0.05 z and z are illustrated as follows.
26.2.2 Higher and Super Derivative of Dirichlet Lambda Function
Formula 26.2.2h ( Higher Derivative ) ()n When z is Dirichlet lambda function, z is the lineal n -th order derivative and r is Stieltjes constant, the following expression holds on whole complex plane. -n ()n ()-1 n ! z = 2z -1 n+1
r+1 r-1 r r-n 1 r log 2 r-s z -1 + Σ()-1 r+ -Σ s log2 n =0,1,2, 2 r=0 r +1 s=0s 1+r -n (2.2h)
Proof As seen in the proof of Formula 26.2.1h , the Dirichlet lambda function z is expanded to Laurent series
- 7 - around 1 as follows.
r+1 r-1 r r 1 1 r log 2 r-s z -1 ()z = + Σ()-1 r + -Σ s log2 2z -1 2 r=0 r +1 s=0s r ! Where, in { } is absent for r =0 . Hereafter, in a way similar to the proof of Formula 26.1.2h , the desired expression is obtained.
()2 ()3 Example ()- 2.1 , - 0.8 - 1.9 i When these are calculated by formula manipulation soft Mathematica , it is as follows. We can see that this formula is numerically right.
Formula 26.2.2s ( Super Derivative ) ()p When p is a complex number, z is Dirichlet lambda function, z is the lineal p -th order derivative,
p is gamma function, p is digamma function and r is Stieltjes constant, the following expression holds on whole complex plane.
logz -1 --p -0 p ()z = ()z-1 -p-1 2-p r p r+1 r-1 r - 1 r log 2 r-s ()z-1 + Σ()-1 r+ -Σ s log2 (2.2s) 2 r=0 r +1 s=0s 1+r -p
Proof In a way similar to the proof of Formula 26.1.2s , the desired expression is obtained.
Note If the sign of p is inverted in (2.2s), it becomes as follows.
()-p logz -1 -p -0 z = z -1 p-1 2p
r+1 r-1 r r+p 1 r log 2 r-s z -1 + Σ()-1 r+ -Σ s log2 2 r=0 r +1 s=0s 1+r +p This results in Formula 26.2.1s . That is, the lineal super calculus of Dirichlet lambda function is seamless.
- 8 - Example The 0.9th order derivative ()1 ()0.9 When z = x +iy , z and z are illustrated as follows. The left figure is a real part and the ()1 ()0.9 right figure is an imaginary part. In both figures, the orange is z and the blue is z . Since p is near 1 , both curved surfaces look double.
- 9 - 26.3 Higher and Super Calculus of Dirichlet Eta Function
26.3.1 Higher and Super Integral of Dirichlet Eta Function
Formula 26.3.1h ( Higher Integral ) <>n When z is Dirichlet eta function, z is the lineal n -th order primitive, (1) The following expression holds for z s.t. Rez > 0 .
n -n n z n r log r z -1 n , , , (3.1h) ()= + ()-1 Σ()-1 z =0 1 2 n ! r=2 r (2) The following expression holds on whole complex plane.
n k r-1 -n n z n ()-1 k log r (3.1h') ()z = + ()-1 ΣΣ k z n =0,1,2, n ! k=2 r=2 2 +1 rr
Proof At Rez >0, Dirichlet eta function z is expressed with the following series which is called DIrichlet eta series. 1 1 1 r-1 -zlogr ()z = 1 + Σ()-1 e = 1 - z + z - z +- r=2 2 3 4 So, integrating the both sides n times with respect to z without considering the constant of the integration,
n -n n z n r log r z -1 n , , , (3.1h) ()= + ()-1 Σ()-1 z =0 1 2 n ! r=2 r Applying Euler transformation to this second term, we obtain (3.1h') . By this transformation, (3.1h) is analytically continued from Rez >0 to the whole complex plane. In addition, about Euler transformation, see " 10 Convergence Acceleration & Summation Method by Double Series of Functions " ( A la carte ).
Example The 1st order integral <>1 The imaginary part of x +iy are illustrated as follows. The left is (3.1h) and the right is (3.1h') . In the left figure, the line of convergence is visible at x =0 .
- 10 - Note If the left side of (3.1h) is represented by integral symbols, it is as follows. z z zn log -nr ()zdz = + ()-1 n ()-1 r-1 Σ z an a1 n ! r=2 r Here, the lower limits of the integral are as follows.
a1 = -1.809613 + i 1.766080 , a2 = 1.216967 ,
a3 = 1.337211 + i 1.289222 , a4 = 2.163768 , That is, this is a higher integral with variable lower limits. And this is a lineal higher integral .
Formula 26.3.1s ( Super Integral ) <>p When p is a complex number, z is Dirichlet eta function, z is the lineal p -th order primitive and p is gamma function, (1) The following expression holds for z s.t. Rez > 0 .
p -p p z p i r log r z e -1 (3.1s) ()= + Σ()-1 z ()1+p r=2 r (2) The following expression holds on whole complex plane. r p k -1 k -p p z p i ()-1 log r (3.1s') ()z = + e ΣΣ k z ()1+p k=2 r=2 2 +1 rr
Proof n -n i In Formula 26.3.1h , replacing n! with 1+n , replacing ()-1 - with e and replacing the natural number n with a complex number p , we obtain the desired expressions.
Example The 0.2th order integral ()0.2 When z = x +iy , the real parts of z and z are illustrated as follows. The left figure is (3.1s) ()0.2 and the right figure is (3.1s'). In both figures, the orange is z and the blue is z . Since p is near 0 , both curved surfaces look double. In the left figure, the line of convergence is visible at x =0 .
- 11 - 26.3.2 Higher and Super Derivative of Dirichlet Eta Function
Formula 26.3.2h ( Higher Derivative ) ()n When z is Dirichlet eta function, z is the lineal n -th order drivative and n is gamma function, (1) The following expression holds for z s.t. Rez > 0 .
-n n n z n r log r z - -1 n , , , (3.2h) ()= +()-1 Σ()-1 z =0 1 2 ()1-n r=2 r (2) The following expression holds on whole complex plane.
-n k r-1 n n z -n ()-1 k log r (3.2h') ()z = +()-1 ΣΣ k z n =0,1,2, ()1-n k=2 r=2 2 +1 rr
Proof As seen in the proof of Formula 26.3.1h , 1 1 1 r-1 -zlogr ()z = 1 + Σ()-1 e = 1 - z + z - z +- r=2 2 3 4 0 So, assuming 1=z /0! and differentiating the both sides n times with respect to z , n z- log nr n z -n r-1 n , , , ()= + ()-1 Σ()-1 z =0 1 2 ()-n ! r=2 r Replacing -n ! with 1-n , we obtain (3.2h) . In addition, z-n 1 for n =0 = ()1-n 0 for n =1,2,3, And applying Euler transformation to this second term, we obtain (3.2h') . By this transformation, (3.2h) is analytically continued from Rez >0 to the whole complex plane.
()1 ()3 Example ()0.2 , 0.5 + 1 4.1 i If these are calculated according to (3,2h') by formula manipulation soft Mathematica , it is as follows. We can see that this formula is numerically right.
Formula 26.3.2s ( Super Derivative ) ()p When p is a complex number, z is Dirichlet eta function, z is the lineal p -th order derivative
- 12 - and p is gamma function,
(1) The following expression holds for z s.t. Rez > 0 .
-p p p z -pi r log r z e -1 (3.2s) ()= + Σ()-1 z ()1-p r=2 r (2) The following expression holds on whole complex plane. r -p k -1 k p p z -pi ()-1 log r (3.2s') ()z = + e ΣΣ k z ()1-p k=2 r=2 2 +1 rr
Proof n -n i In Formula 26.3.2h , replacing ()-1 - with e and replacing the natural number n with a complex number p , we obtain the desired expressions.
Note If the sign of p is inverted in (3.2s), it becomes as follows. zp log -p r -p z epi r-1 ()= + Σ()-1 z ()1+p r=2 r This results in Formula 26.3.1s . That is, the lineal super calculus of Dirichlet eta function is seamless.
Example The 2.2th order derivative ()2 ()2.2 When z = x +iy , the real parts of z and z are illustrated as follows. The left figure is ()2 ()2.2 (3.2s) and the right figure is (3.2s'). In both figures, the orange is z and the blue is z . In the left figure, the line of convergence is visible at x =0 .
- 13 - 26.4 Higher and Super Calculus of Dirichlet Beta Function
26.4.1 Higher and Super Integral of Dirichlet Beta Function
Formula 26.4.1h ( Higher Integral ) <>n When z is Dirichlet beta function, z is the lineal n -th order primitive, (1) The following expression holds for z s.t. Rez > 0 .
n -n n z n r log 2r-1 z -1 n , , , (4.1h) ()= + ()-1 Σ()-1 z =0 1 2 n ! r=2 2r-1 (2) The following expression holds on whole complex plane.
n k r-1 -n <>n z n ()-1 k log 2r -1 z = + ()-1 - n =0,1,2, (4.1h') () ΣΣ k z n ! k=2 r=2 2 +1 r 2r -1
Proof At Rez >0, Dirichlet beta function z is expressed with the following series which is called DIrichlet beta series. 1 1 1 r-1 -zlog2r-1 ()z = 1 + Σ()-1 e = 1 - z + z - z +- r=2 3 5 7 So, integrating the both sides n times with respect to z without considering the constant of the integration,
n -n n z n r log 2r-1 z -1 n , , , (4.1h) ()= + ()-1 Σ()-1 z =0 1 2 n ! r=2 2r-1 Applying Euler transformation to this second term, we obtain (4.1h') . By this transformation, (4.1h) is analytically continued from Rez >0 to the whole complex plane.
Example The 3rd order integral <>3 The imaginary part of x +iy are illustrated as follows. The left is (4.1h) and the right is (4.1h') . In the left figure, the line of convergence is visible at x =0 .
- 14 - Note If the left side of (4.1h) is represented by integral symbols, it is as follows. z z zn log -n2r-1 ()zdz = + ()-1 nΣ()-1 r-1 a z an 1 n ! r=2 2r-1 Here, the lower limits of the integral are as follows.
a1 = -1.027077 + i 0.978760 , a2 = 0.754520 ,
a3 = 0.831711 + i 0.807718 , a4 = 1.357089 , That is, this is a higher integral with variable lower limits. And this is a lineal higher integral .
Formula 26.4.1s ( Super Integral ) <>p When p is a complex number, z is Dirichlet beta function, z is the lineal p -th order primitive and p is gamma function, (1) The following expression holds for z s.t. Rez > 0 .
p -p <>p z p i r log 2r-1 z e -1 (4.1s) ()= + Σ()-1 z ()1+p r=2 2r-1 (2) The following expression holds on whole complex plane.
p k r-1 -p <>p z p i ()-1 k log 2r -1 z = + e (4.1s') () ΣΣ k z ()1+p k=2 r=2 2 +1 r 2r -1
Proof n -n i In Formula 26.4.1h , replacing n! with 1+n , replacing ()-1 - with e and replacing the natural number n with a complex number p , we obtain the desired expressions.
Example The 0.3th order integral ()0.3 When z = x +iy , the real parts of z and z are illustrated as follows. The left figure is ()0.3 (4.1s) and the right figure is (4.1s'). In both figures, the orange is z and the blue is z . In the left figure, the line of convergence is visible at x =0 .
- 15 - 26.4.2 Higher and Super Derivative of Dirichlet Beta Function
Formula 26.4.2h ( Higher Derivative ) ()n When z is Dirichlet beta function, z is the lineal n -th order drivative and n is gamma function, (1) The following expression holds for z s.t. Rez > 0 .
-n n ()n z n r log 2r -1 z = +()-1 - ()-1 -1 n =0,1,2, (4.2h) Σ z 1-n r=2 2r -1 (2) The following expression holds on whole complex plane.
-n k r-1 n ()n z ()-1 k log 2r -1 z = +()-1 -n n =0,1,2, ΣΣ k z 1-n k=2 r=2 2 +1 r 2r -1 (4.2h')
Proof As seen in the proof of Formula 26.4.1h , 1 1 1 r-1 -zlog2r-1 ()z = 1 + Σ()-1 e = 1 - z + z - z +- r=2 3 5 7 Hereafter, in a way similar to the proof of Formula 26.3.2h , (4.2h) is obtained. And applying Euler transformation to this second term, we obtain (4.2h') . By this transformation, (4.2h) is analytically continued from Rez >0 to the whole complex plane.
()2 ()3 Example 0.9 + 8.7 i , ()1.3 If these are calculated according to (4,2h) by formula manipulation soft Mathematica , it is as follows. Though the convergence is slow, we can see that this formula is numerically right.
Formula 26.4.2s ( Super Derivative ) ()p When p is a complex number, z is Dirichlet beta function, z is the lineal p -th order derivative and p is gamma function,
(1) The following expression holds for z s.t. Rez > 0 .
-p p p z -pi r-1 log ()2r-1 (4.2s) ()z = + e Σ()-1 z ()1-p r=2 ()2r-1
- 16 - (2) The following expression holds on whole complex plane. r -p k -1 k p p z -pi ()-1 log ()2r-1 (4.2s') ()z = + e ΣΣ k z ()1-p k=2 r=2 2 +1 r ()2r-1
Proof n -n i In Formula 26.4.2h , replacing ()-1 - with e and replacing the natural number n with a complex number p , we obtain the desired expressions.
Note If the sign of p is inverted in (4.2s), it becomes as follows. zp log -p ()2r-1 -p pi r-1 ()z = + e Σ()-1 z ()1+p r=2 ()2r-1 This results in Formula 26.4.1s . That is, the lineal super calculus of Dirichlet beta function is seamless.
Example The 1.1th order derivative ()1 ()1.1 When z = x +iy , the imaginary parts of z and z are illustrated as follows. The left figure ()1 ()1.1 is (4.2s) and the right figure is (4.2s'). In both figures, the orange is and the blue is z . In the left figure, the line of convergence is visible at x =0 .
2018.01.13 Kano Kono Alien's Mathematics
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