Another Simple L-Function

This is another of the simpler L-functions (degree 1) on the LMFDB site, which I examined with a view to assessing common features with respect to all functions of this class.

So, L(χ, s) where χ is the Dirichlet character with label 7, 6 is given as,

1 + 1/2^s – 1/3^s + 1/4^s – 1/5^s – 1/6^s + 1/8^s + …

So with respect to the Label, 7 refers to the number by which we divide each successive natural number. Then when the remainder is 1, 2, or 4, then χ(n) = 1; when the remainder is 3, 5 or 6 then χ(n) = – 1; and when the remainder is zero, χ(n) = 0.

Therefore for each cycle of 7 successive natural numbers, we have 6 terms (with numbers exactly divisible by 7 omitted).

The corresponding product over primes expression (for this Dirichlet sum) is given as

(1 – 1/2^s) * (1 + 1/3^s) * (1 + 1/5^s) * (1 – 1/11^s) * …

In this case, when s is a positive odd integer, the value of both expressions = kπ^s/√7 (where k is a rational number).

So for example when s = 1,

1 + 1/2 – 1/3 + 1/4 – 1/5 – 1/6 + 1/8 + …

 = (1 – 1/2) * (1 + 1/3) * (1 + 1/5) * (1 – 1/11) * …

= π/√7

Then when s = 3,

1 + 1/2³ – 1/3³ + 1/4³ – 1/5³ – 1/6³ + 1/8³ + …

= (1 – 1/2³) * (1 + 1/3³) * (1 + 1/5³) * (1 – 1/11³) * …

= √7π³/75

And when s = 5,

1 + 1/2⁵ – 1/3⁵ + 1/4⁵ – 1/5⁵ – 1/6⁵ + 1/8⁵ + …

= (1 – 1/2⁵) * (1 + 1/3⁵) * (1 + 1/5⁵) * (1 – 1/11⁵) * …

= 64π⁵/(7203√7).

Again as always is the case, the above L-function, where s is a positive integer, it can be related to the Riemann zeta function (excluding those terms where 7 is a factor).

(1 + 1/2^s – 1/3^s + 1/4^s – 1/5^s – 1/6^s + 1/8^s + …)²/(1 + 1/2^2s + 1/3^2s + 1/4^2s + 1/5^2s – 1/6^2s + 1/8^2s + …)

= 1 + 2/2^s  – 2/3^s  + 2/4^s – 2/5^s – 4/6^s + 2/8^s + …

with a product over primes expression,

1/{1 – 2/(2^s + 1)} * 1/{1 + 2/(3^s + 1)} * 1/{1 + 2/(5^s + 1)} * 1/{1 – 2/(11^s + 1)} * …

And where s is a positive odd integer, the result is always a rational number.

It is important here to stress that the general relationship holds in all cases, where is a positive odd integer.

However in this case, where s is even, the result is an irrational number.

So for example in the simplest case where s = 1,

(1 + 1/2 – 1/3 + 1/4 – 1/5 – 1/6 + 1/8 + …)²/(1 + 1/2² + 1/3² + 1/4² + 1/5² – 1/6² + 1/8² + …)

= 1 + 2/2  – 2/3  + 2/4 – 2/5 – 4/6 + 2/8 + …,

With a product over primes expression,

1/{1 – 2/(2 + 1)} * 1/{1 + 2/(3 + 1)} * 1/{1 + 2/(5 + 1)} * 1/{1 – 2/(11 + 1)} * …

= {(π/√7)²}/²{(8π²/49)  = 7/8.