Prove $\int_{1}^{\infty}\widetilde{B}_{2n+1}(x)x^{-(s+2n+1)}dx$ is analytic

Question

Given $B_{n}(x)$ satisfies \begin{equation} \frac{ze^{xz}}{e^z-1}=\sum_{k=0}^{\infty}\frac{B_k(x)}{k!}z^k \end{equation} which is call Bernoulli polynomial. And $\widetilde{B}_{n}$ is 1-periodic extension of $B_n(x)\vert[0,1)$

Prove that for each $n\in \mathbb{N}$, map $F_n:\{z\in \mathbb{C}: Rez>-2n\}, s \mapsto \int_{1}^{\infty}\widetilde{B}_{2n+1}(x)x^{-(s+2n+1)}dx$ is analytic.

This is an exericse from H.Amann's Analysis.

I have already known that

1. $B_n(x)=\sum_{k=0}^{k=n}\binom{n}{k}B_kx^{n-k}$
2. $\widetilde{B}_{2n+1}(x)\in C^{n-2}(\mathbb{R})$
3. $\int_{k}^{k+1}\widetilde{B}_n(x)dx=0$

By using the above formulas, I obtained which as follows, but I don't know how to continue. \begin{eqnarray*} % \nonumber % Remove numbering (before each equation) RHS &=& \sum_{n=0}^{\infty}\int_{n}^{n+1}\sum_{k=0}^{2n+1}\binom{2n+1}{k}B_{k}(x-n)^{n-k}x^{-(s+2n+1)}dx \\ &=&\sum_{n=0}^{\infty}\int_{n}^{n+1}\sum_{k=0}^{2n+1}\binom{2n+1}{k}B_{k}\sum_{i=0}^{n-k}\binom{n-k}{i}x^i(-n)^{n-k-i}x^{-(s+2n+1)}dx \\ &=&\sum_{n=0}^{\infty}\sum_{k=0}^{2n+1}\sum_{i=0}^{n-k}\binom{2n+1}{k}\binom{n-k}{i}(-1)^{n-k-i}B_{k}n^{n-k-i}\frac{(n+1)^{-s-2n+i}-n^{-s-2n+i}}{-s-2n+i} \end{eqnarray*}

Answer 1

This amounts to proving analyticity of $z\mapsto\int_1^\infty\widetilde{B}_{2n+1}(x)x^{-z-1}dx$ in $\{z\in\mathbb{C} : \Re z>0\}$. For this (i.e. to prove the ability to differentiate under the integral sign), just apply DCT. No need to use any special formulae for $\widetilde{B}_n$ (it's enough to know it is bounded).

Answer 2

Let $G(s)=\int_1^\infty \widetilde B_{2n+1}(x) x^{-(s+2n+1)}dx$ and $G_m(s)=\int_1^m \widetilde B_{2n+1}(x) x^{-(s+2n+1)}dx$.

The outline is the following:

(a) $G_m(s)$ is convergent pointwise to $G(s)$,

(b) $G'_m(s)=-(s+2n+1)\int_1^m \widetilde B_{2n+1}(x) x^{-(s+2n+2)}dx$,

(c) $G'_m(s)$ is convernget locally uniformly to $-(s+2n+1)\int_1^\infty \widetilde B_{2n+1}(x) x^{-(s+2n+2)}dx$, which is $G'(s)$,

(d) $G(s)$ is holomorphic, and hence analytic.