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 ?



Binomial transform of Dirichlet series
Mellin transform interpolation is related formally at least to Newton series interpolation. One example is its use to [interpolate the Bernoulli polynomials](https://www.google.com/amp/s/tcjpn.wordpress.com/2015/01/12/umbral-composition-of-the-bernoulli-polynomials-and-the-riemann-zeta-function/amp/) 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.


