<s>[INST] Showing that $2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\left( \frac{x^{a-1}}{\sinh x} - x^{a-2} \right) \mathrm dx$ I want to show that $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\Big( \frac{x^{a-1}}{\sinh x} - x^{a-2}\Big)  \, \mathrm dx \, , \quad 0 < \Re (a) <1. \tag{1}$$
I can show that $$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty} \frac{x^{a-1}}{\sinh x} \, \mathrm dx \,  , \quad \Re (a) >1, \tag{2}$$
but I don't know if $(2)$ can be used to prove $(1)$.
EDIT:
Following the approach used in this paper to show that $$\Gamma(s) \zeta(s) = \int_{0}^{\infty} \left(\frac{1}{e^{x}-1}-\frac{1}{x} \right)x^{s-1} \, \mathrm dx \, , \quad0 < \Re(s) < 1,  $$
subtract $$\frac{\Gamma(a)}{1-a} b^{1-a} = \int_{0}^{\infty} e^{-b x}x^{a-2} \, \mathrm dx \, , \quad \left( \Re(a) >1, \, b>0 \right),$$ from $(2)$ to get
$$\Gamma(a) \left(2 \zeta(a) \left(1-\frac{1}{2^{a}} \right) - \frac{b^{1-a}}{1-a} \right) = \int_{0}^{\infty} \left(\frac{1}{\sinh x} - \frac{e^{-bx}}{x}  \right)x^{a-1} \, \mathrm dx \, , \quad \Re(a) > 0.$$
What this did is compensate for the fact that $\frac{1}{\sinh z}$ has a simple pole at $z=0$ with residue $1$.
Now restrict $a$ to the vertical strip $0 < \Re (a) <1$, and let $b$ tend to zero to get $$ 2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\left( \frac{1}{\sinh x} - \frac{1}{x}\right)x^{a-1}  \, \mathrm dx \ , \quad 0 < \Re(a)  <1.$$
You could then appeal to the identity theorem to argue that $(1)$ actually holds for $-1 < \Re(a) <1$.
 [/INST] 
$$ \color{blue}{\Gamma(s-N)\zeta(s-N)=\int_{0}^{\infty}x^{s-N-2}\left[\frac{x}{e^x-1}-\left(\sum_{n=0}^{N}B_{n}\frac{x^n}{n!}\right)\right]\,dx} $$ 
  $$ {\small \,0\lt\,Re\{s\}\,\lt1 ,\quad N\in\{\,0,\,1,\,2,\,\cdots\,\} ,\quad B_{n}\,\,{Bernoulli\,Number} ,\quad B_{1}=-1/2} $$ 

$$ 
\begin{align} 
I\,& =\,\int_{0}^{\infty}\left(\frac{x^{s-1}}{\sinh{x}}-x^{s-2}\right)\,dx =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{e^x}{e^{2x}-1}-\frac{1}{2x}\right)\,dx \\[2mm] 
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{e^x\color{red}{+1-1}}{e^{2x}-1}-\frac{1}{2x}\color{red}{-\frac{1}{2x}+\frac{1}{2x}}\right)\,dx \\[2mm] 
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}-\frac{1}{e^{2x}-1}+\frac{1}{2x}\right)\,dx 
\end{align} 
$$ 

For ${\small\color{red}{\,0\le\,Re\{s\}\,\le+1\,}}\colon\quad\Gamma(s)\zeta(s)=\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)\,dx\,$; Thus: 
$$ 
\begin{align} 
I\,& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}-\frac{1}{e^{2x}-1}+\frac{1}{2x}\right)\,dx \qquad\color{blue}{\small\{\text{both converge}\}} \\[2mm] 
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx\,-\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{2x}-1}-\frac{1}{2x}\right)\,dx \\[2mm] 
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx\,-\,\frac{2}{2^{s}}\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\right)\,dx \\[2mm] 
& =\,\color{red}{\left(2-2^{1-s}\right)\,\Gamma(s)\zeta(s)} 
\end{align} 
$$ 

For ${\small\color{red}{\,-1\le\,Re\{s\}\,\le0\,}}\colon\quad\Gamma(s)\zeta(s)=\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx\,$; Thus: 
$$ 
\begin{align} 
I\,& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}\color{red}{+\frac{1}{2}}-\frac{1}{e^{2x}-1}+\frac{1}{2x}\color{red}{-\frac{1}{2}}\right)\,dx \qquad\color{blue}{\small\{\text{both converge}\}} \\[2mm] 
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx\,-\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{2x}-1}-\frac{1}{2x}+\frac{1}{2}\right)\,dx \\[2mm] 
& =\,2\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx\,-\,\frac{2}{2^{s}}\int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^{x}-1}-\frac{1}{x}+\frac{1}{2}\right)\,dx \\[2mm] 
& =\,\color{red}{\left(2-2^{1-s}\right)\,\Gamma(s)\zeta(s)} 
\end{align} 
$$ 
</s>