On the integral $\int\limits_{0}^{\infty}x^{s-1}\left(\frac{e^{(1-a)x}}{e^{x}-1}-\frac{1}{x}\right)\,dx$ On the integral representation of Hurwitz Zeta Function inside the critical strip,

Show that:
$$ \Gamma(s)\zeta(s,a)=\int_{0}^{\infty}x^{s-1}\left(\frac{e^{(1-a)x}}{e^{x}-1}-\frac{1}{x}\right)\,dx \tag{1} $$
for $\, 0\lt \Re\{s\}\lt1 ,\,\, a\in\mathbb{R}^{+} \,$.

 
  
I am also seeking for a proof (or disproof) of the more general representation (claim): 
$$ \Gamma(s-m)\zeta(s-m,a)=\int_{0}^{\infty}x^{s-m-2}\left[\frac{xe^{(1-a)x}}{e^x-1}-\left(\sum_{n=0}^{m}B_{n}\frac{x^n}{n!}\right)\right]\,dx \tag{2} $$ 
for $\,0\lt\,\Re\{s\}\,\lt1 ,\,\, a\in\mathbb{R}^{+} ,\,\, m\in\mathbb{N} ,\,\, B_{n}={\operatorname{Bernoulli}}_{\#} ,\,\, B_{1}=-1/2 \,$.  
Thanks.
 A: The formula $(2)$ needs some correction: for small $x$ (actually for $|x|<2\pi$) we have $$\frac{xe^{(1-a)x}}{e^x-1}=\frac{xe^{-ax}}{1-e^{-x}}=\sum_{n=0}^\infty(-1)^nB_n(a)\frac{x^n}{n!},$$ where $B_n(a)$ are the Bernoulli polynomials. Both $(1)$ and $(2)$ [fixed] can be deduced from $$\zeta(s,a)=\frac{\Gamma(1-s)}{2\pi i}\int_\lambda\frac{z^{s-1}e^{az}}{1-e^z}\,dz\qquad(s\notin\mathbb{Z}_{>0},\Re a>0)$$ where the contour $\lambda$ encircles the negative real axis (but not the poles $z=2n\pi i$, $n\in\mathbb{Z}_{\neq0}$ of the integrand). In turn, one proves this for $\Re s>1$ by "squeezing" $\lambda$ closely to the negative real axis: $$\int_\lambda\frac{z^{s-1}e^{az}}{1-e^z}\,dz=\int_\infty^0(xe^{-\pi i})^{s-1}\frac{e^{-ax}\,d(-x)}{1-e^{-x}}+\int_0^\infty(xe^{\pi i})^{s-1}\frac{e^{-ax}\,d(-x)}{1-e^{-x}}\\=2i\sin s\pi\int_0^\infty\frac{x^{s-1}e^{-ax}}{1-e^{-x}}\,dx=2i\sin s\pi\sum_{n=0}^\infty\int_0^\infty x^{s-1}e^{-(n+a)x}\,dx\\=2i\sin s\pi\,\Gamma(s)\zeta(s,a)=\frac{2\pi i}{\Gamma(1-s)}\zeta(s,a),$$ and by analytic continuation elsewhere, since the integral is an entire function of $s$.
Now, for $0<\Re s<1$ and $m\in\mathbb{Z}_{\geqslant 0}$, we have $$\int_\lambda\frac{z^{s-m-1}e^{az}}{1-e^z}\,dz=\int_\lambda z^{s-m-2}\left(\frac{ze^{az}}{1-e^z}+\sum_{n=0}^m B_n(a)\frac{z^n}{n!}\right)dz$$ since $\int_\lambda z^\alpha\,dz=0$ for $\Re\alpha<-1$; "squeezing" $\lambda$ the same way, we get a fixed version of $(2)$: $$\Gamma(s-m)\zeta(s-m,a)=\int_0^\infty x^{s-m-2}\left(\frac{xe^{-ax}}{1-e^{-x}}-\sum_{n=0}^m(-1)^n B_n(a)\frac{x^n}{n!}\right)dx.$$
