Talk:Dirichlet beta function

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Reciprocal of this function - is this worth adding to the main section?

It may be worth mentioniong that:

   1/beta(s) = sum(n=1..infinity((-1)^n * mu(n+1)/(2*n+1)^s));

which is valid for Re(s)>1. Hair Commodore (talk) 20:05, 28 July 2008 (UTC). Note that mu(*) is the Mobius mu function. (It certainly converges to 4/Pi when s=1)[reply]

This formula is wrong! I tried checking it for 's'=1, 's'=2, and 's'=3, and it works at none of them! 81.102.15.200 (talk) 13:36, 18 August 2008 (UTC)[reply]

Apologies - I should have said that:

   1/beta(s) = sum(n=1..infinity,(((-1)^n * mu(2*n+1))/(2*n+1)^s))

in Maple notation - rather than what I typed previously.The anaonymous user who questioned it was quite correct. I made a wally error, so I have now corrected it. (It is still correct for 'Re'(s)>1) Hair Commodore (talk) 19:24, 18 August 2008 (UTC)[reply]

yes, from the Euler product I just added :

{\frac {1}{\beta (s)}}=\sum _{n=0}^{\infty }\mu _{2n+1}(2n+1)^{-s}(-1)^{n}

I'd still like to know if the Riemann hypothesis for

\beta (s)

would imply (or be implied by) the one for

\zeta (s)

, or if the two are completely independent. 78.227.78.135 (talk) 17:10, 4 January 2016 (UTC)[reply]

Definition

The formula for beta(s) in terms of the polygamma function only seems to be valid at even positive integers. I believe multiplying this formula by (-1)^s makes it valid at odd positive integers as well as even positive integers. The conditions for the validity of the polygamma formula for beta(s) need to be specified or the reader may assume the formula is valid for all complex values of s similar to the two preceding formulas which express beta(s) in terms of the Hurwitz zeta function and the Lerch transcendent.StvC (talk) 16:22, 2 April 2021 (UTC)[reply]

I corrected and clarified the polygamma formula.StvC (talk) 17:06, 4 April 2021 (UTC)[reply]