# Connection between Bernoulli polynomials and polygamma function

There is an intricate connection between Hurwitz Zeta and the (traditional) polygamma function:

$$\psi_n(z)=(-1)^{n+1}n!\zeta(n+1,z)$$

If to use a generalization for Bernoulli numbers, this can be considered a formula, connecting polygamma and Bernoulli polynomials of negative order:

$$\psi_n(z)=(-1)^{n+1}n!\frac{B_{-n}(x)}n$$ (1)

As much as this equality is impressing, it is limited as it only holds for natural n.

After a couple of unsuccessful attempts to find a more general formula, I encountered a more natural, "balanced" generalization of polygamma function explained in this paper. It turned out that while the old formula still holds for natural n (since the old polygamma and balanced polygamma coincide in integer positive orders), a completely new formula connecting this balanced polygamma with Zeta and Bernoulli numbers can be derived which holds for any z:

$$\zeta(z,q)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,\frac{q}{2}+\frac{1}{2}\right)+\psi \left(z-1,\frac{q}{2}\right)\right)-\psi(z-1,q)\right)}{\ln(2)}$$

$$B_z(q) = -\frac{\Gamma (z+1) \left(2^{z-1} \left(\psi\left(-z,\frac{q}{2}+\frac{1}{2}\right)+\psi\left(-z,\frac{q}{2}\right)\right)-\psi(-z,q)\right)}{\ln (2)}$$

Both of them can be expressed completely in terms of balanced polygamma and elementary functions if to notice that $\Gamma(x)=e^{\psi(-1,x)+\frac 12 \ln(2\pi)}$, which allows to get rid of the Gamma function.

While the target was reached, these expressions still leave a bad impression. I cannot simplify it as no CAS system is capable of operations with the balanced polygamma.

Hence I am asking for help on how to simplify the expressions so they could be easier to manage and use. It is also not evident how the letter formulas become the former ones at positive real z.

• This appears to be a duplicate of mathoverflow.net/questions/42696 – Robin Chapman Oct 19 '10 at 13:08
• Usually people ask first here and then promote it to MO if it is sufficiently advanced that it stands a better chance of getting answered there. So I'm not entirely sure why you posted here if you couldn't get answers there. – J. M. is a poor mathematician Oct 19 '10 at 14:14