Showing that $2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\left( \frac{x^{a-1}}{\sinh x} - x^{a-2} \right) \, dx$

First step, write the integral as $$ I = \int_{0}^{\infty}\left(-{\frac {{x}^{a-2}{{\rm e}^{2\,x}}}{{{\rm e}^{2\,x}}-1}}+2\,{\frac {{x }^{a-1}{{\rm e}^{x}}}{{{\rm e}^{2\,x}}-1}}+{\frac {{x}^{a-2}}{{{\rm e} ^{2\,x}}-1}}\right) dx. $$

Second step, use the change of variables $ 2x=u $. Third step, use the identity (the Hurwitz zeta function)

$$ \zeta(s,q)=\frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}e^{-qt}}{1-e^{-t}}dt.$$

Here are related problems: I, II. By the way, I tried Maple and WolframAlpha, but no answer.