Some Notes on the Dirichlet Lambda Function Johar M. Ashfaque
Let λ(s) = be the Dirichlet lambda function, β(s) =
∞ X
1 (2n + 1)s n=0 ∞ X
(−1)n (2n + 1)s n=0
be its alternating form.
1
Introduction
According to Varadarajan, a long time ago Pietro Mengoli posed the problem of finding the sum of the series ∞ X 1 1 1 = 1 + + + · · ·. 2 n 4 9 n=1 This problem was first solved by Euler in a letter to Bernoulli as 1+ In general, let
1 1 π2 + +···= . 4 9 6 ∞ X 1 ζ(s) = s n n=1
be Riemann’s zeta function, Euler also proved the following formula ζ(2k) =
(−1)k−1 B2k 22k 2k π 2(2k)!
where B2k are the Bernoulli numbers.
2
Relation to Riemann Zeta
In general λ(s) =
λ(2) = ζ(2) −
1 3 π2 ζ(2) = ζ(2) = 22 4 8
λ(4) = ζ(4) −
1 15 π4 ζ(4) = ζ(4) = 24 16 96
∞ X
1 = (1 − 2−s )ζ(s), s (2n + 1) n=0
1
Re(s) > 1.
3
Evaluation of Dirichlet Lambda Function at 2m λ(2m) = (−1)m
π 2m E2m−1 (0), 4(2m − 1)!
m≥1
where the Euler polynomials En (x) are defined by the following generating function ∞ X tn 2 tx e = En (x) . t e +1 n! n=0
The integers En = 2n En (1/2), n ∈ N0 are called Euler numbers. For example, E0 = 1 E2 = −1 E4 = 5 and E6 = −61.
λ(2m) = (−1)
m−1
m−1 X π 2k π 2m m−k (−1) + λ(2m − 2k) . 22m+1 (2m − 1)! 22k (2k)! k=1
At m = 1 λ(2) = At m = 2 1 λ(4) = 5 2 At m = 3 1 λ(6) = 7 2 At m = 4 λ(8) =
1 29
−
π2 8
2 π4 π4 3π − +2 λ(2) = 3! 2! 96
4 2 π6 π6 3π 5π −2 λ(2) + 2 λ(4) = 5! 4! 2! 960
π8 17π 8 π6 π4 π2 + 23 λ(2) − 25 λ(4) + 27 λ(6) = 7! 6! 4! 2! 161280
2