Fascinating Dirichlet Beta Function Relationships

One important L-function - closely related to the Riemann zeta function - is known as Dirichlet’s beta function (also Catalan’s beta function) with its L-series (i.e. L1) sum over the integers and product over the primes expressions as follows,

1/1^s – 1/3^s + 1/5^s – 1/7^s + 1/9^s –  …   = 1/(1 + 1/3^s) *  1/(1 – 1/5^s) * 1/(1 + 1/7^s) * …

Then in the simplest case when s = 1, we have,

1 – 1/3 + 1/5 – 1/7 + 1/9 –  …  =  3/4 * 5/4 * 7/8 * …        = π/4.

So the well-known Leibniz formula for π represents a special case of the Dirichlet beta function i.e. β(s), for s = 1.

And when s = 1, 3, 5, …  β(s), results in a value of the form kπ^s, where k is a rational number.

So for example β(3) = 1 – 1/3³ + 1/5³ – 1/7³ + 1/9³ –  …  = π³/32 and

β(5) = 1 – 1/3⁵ + 1/5⁵ – 1/7⁵ + 1/9⁵ –  …  = 5π⁵/1536.

Now this function can be connected with the Riemann zeta function in a surprising manner.

For using a similar type general connection as illustrated in the last blog entry “More on Riemann Generalisation” we have

(1^s – 1/3^s + 1/5^s – 1/7^s + 1/9^s –  …)²/(1^2s + 1/3^2s + 1/5^2s + 1/7^2s + 1/9^2s +  …)

= 1/1^s – 2/3^s + 2/5^s – 2/7^s + 2/9^s –  2/11^s + 2/13^s – 4/15^s  + …)  = k (where k is a rational number). 

And this is another L1 function with corresponding product over the primes expression, i.e.

1/{1 + 2/(3^s + 1)} *  1/{1 – 2/(1/5^s + 1)}* 1/{1 + 2/(7^s + 1)} * …

Thus again the value of the numerator is determined by the number of distinct prime factors in the corresponding denominator value. So with just one distinct prime factor in the denominator, the value of the numerator is 2; however with the denominator 15 for example, we now have two distinct prime factors, so the value of the corresponding numerator is 4 (i.e. 2²).

So for example when s = 1, we obtain,

(1¹ – 1/3¹ + 1/5¹ – 1/7¹ + 1/9¹ –  …)²/(1² + 1/3² + 1/5² + 1/7² + 1/9² +  …)

= 1 – 2/3 + 2/5 – 2/7 + 2/9 – 2/11 + 2/13 – 4/15  + …  = (π/4)²/(π²/8) = 1/2.

And when s = 3, we obtain

(1³ – 1/3³ + 1/5³ – 1/7³ + 1/9³ –  …)²/(1⁶ + 1/3⁶ + 1/5⁶ + 1/7⁶ + 1/9⁶ +  …)

= 1 – 2/3³ + 2/5³ – 2/7³ + 2/9³ – 2/11³ + 2/13³ – 4/15³  + … = (π³/32)²/(π⁶/960) = 15/16. 

In fact interestingly just as we have seen above,

(1¹ – 1/3¹ + 1/5¹ – 1/7¹ + 1/9¹ –  …)²/(1² + 1/3² + 1/5² + 1/7² + 1/9² +  …) = 1/2,

likewise,

(1¹ – 1/3¹ + 1/5¹ – 1/7¹ + 1/9¹ –  …)³/(1³ – 1/3³ + 1/5³ – 1/7³ + 1/9³ –  …) = 1/2.

Here is another very interesting connection!

As we know,

1² + 1/3² + 1/5² + 1/7² + 1/9² +  … = 1/(1 – 1/3²) * 1/(1 – 1/5²) * 1/(1 – 1/7²) * … = π²/8.

So, if we eliminate from the product over primes expression the term involving 3 as a prime, we must then eliminate in turn all terms with 3 as a factor in the in sum over integers expression.

Thus, 1² + 1/5² + 1/7² + 1/11² +  …  =  1/(1 – 1/5²) * 1/(1 – 1/7²) * 1/(1 – 1/11²) * …  = π²/9.

And if we eliminate with respect to the corresponding Dirichlet beta function the term involving 3 as a prime, we must then eliminate in turn all terms with 3 as a factor in the in sum over integers expression.

Thus, 1 + 1/5 – 1/7 – 1/11 + …  = (1 + 1/5) *  1/(1 – 1/7) * 1/(1 + 1/11) * … = π/3.

and (1 + 1/5 – 1/7 – 1/11 +  …)²   = π²/9, where we consistently have two positive terms followed by two negative terms in the series and vice versa.

Therefore,

1² + 1/5² + 1/7² + 1/11² + 1/13² + …    = (1 + 1/5 – 1/7 – 1/11 + 1/13 + …)².      

And as (1 + 1/5^s – 1/7^s – 1/11^s + 1/13^s + …)²/(1 + 1/5^2s + 1/7^2s + 1/11^2s + 1/13^2s + …)

=  1 + 2/5^s – 2/7^s – 2/11^s + 2/13^s + …, this implies that when s = 1,

(1 + 1/5 – 1/7 – 1/11 + 1/13 +  …)²/(1² + 1/5² + 1/7² + 1/11² + 1/13² + …)

=  1 + 2/5 – 2/7 – 2/11 + 2/13 + …    = 1.

So we have the interesting case here of where the value of this L-function (for s = 1) = 1.