2
$\begingroup$

Let $\Theta(s)$ be a Dirichlet series , and let $\beta$ be its abscissa of convergence: $$\Theta(s)=\sum_{n=1}^{\infty}\frac{\theta(n)}{n^{s}}\;\;\;\;\;\;\Re(s)>\beta$$ And let $\left\{a_{n}\right\}_{n\in\mathbb{N}}$ be a sequence, whose ordinary generating function is $g(x)$: $$g(x)=\sum_{n=0}^{\infty}a_{n}x^{n}\;\;\;\;\;\ |x|<x_{0}\leq 1$$ We want to write the related Dirichlet series: $$\Phi (z)=\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}$$ in terms of $\Theta(s)$ and $g(x)$. A practical example of such a construction is the everywhere-convergent series for the Riemann zeta function: $$\zeta(z)-\frac{1}{z-1}=\sum_{n=0}^{\infty}\left | G_{n+1} \right |\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{1}{(k+1)^{z}}$$ $\left | G_{n+1} \right |$ being the absolute Gregory coefficients.

My attempt :

We use the formula (due to Apostol) : $$\theta(k+1)=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\Theta(\sigma+it)(k+1)^{\sigma+it}dt\;\;\;\;\;\;\sigma>\beta$$ Ignoring issues of convergence, we have: $$\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}$$$$=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\Theta(\sigma+it)\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(k+1)^{\sigma+it-z}dt$$ Now, we have that : $$\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}(k+1)^{-\mu}=\frac{1}{\Gamma(\mu)}\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\int_{0}^{\infty}x^{\mu-1}e^{-(k+1)x}dx$$ $$=\frac{1}{\Gamma(\mu)}\sum_{n=0}^{\infty}a_{n}\int_{0}^{\infty}x^{\mu-1}e^{-x}\left(1-e^{-x}\right)^{n}dx=\frac{1}{\Gamma(\mu)}\int_{0}^{\infty}x^{\mu-1}e^{-x}g(1-e^{-x})dx\;\;\;\Re(\mu)>0$$ Assuming this last integral admits analytic continuation, and denoting it by $\Omega (\mu)$, we have: $$\sum_{n=0}^{\infty}a_{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\frac{\theta(k+1)}{(k+1)^{z}}=\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\Theta(\sigma+it)\Omega(\sigma+it-z)dt$$ But i doubt that this kind of integrals could ever be evaluated, even using the residue theorem, due to the limit, and the $T^{-1}$ factor. Is there an alternative to my approach ?

$\endgroup$
2
$\begingroup$

Mellin transform interpolation is related formally at least to Newton series interpolation. One example is its use to interpolate the Bernoulli polynomials from their relation to the Mellin kernel for the Riemann zeta function (times $(s-1)!$) to obtain the Hurwitz zeta function. Maybe you could modify the arguments.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.