Maybe I'm jumping the gun, but it doesn't look like anyone is going to attempt this one either.Again I'm going to add the restriction that $\text{Re}(s) >1$ which can be removed at the end.$ \displaystyle \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt = \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s} (e^{2 \pi t}-1)} \ dt $I'm leaving the imaginary sign inside of the integral since bringing it outside would result in a divergent integral.\displaystyle = \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s}} \frac{e^{-\pi t}}{e^{\pi t}-e^{- \pi t}} \ dt = \frac{1}{2} \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s}} \Big( \frac{e^{\pi t} + e^{\pi t}}{e^{\pi t} - e^{- \pi t}} -1 \Big) \ dt
\displaystyle = \frac{1}{2} \int_{0}^{\infty} \text{Im} \ \frac{1}{(1-it)^{s}} \Big( \coth(\pi t) -1 \Big) \ dt = \frac{1}{2} \int_{0}^{\infty} \text{Im} \ \frac{\coth( \pi t) }{(1-it)^{s}} \ dt - \frac{1}{2} \text{Im} \int_{0}^{\infty} \frac{1}{(1-it)^{s}} \ dt
\displaystyle \int_{0}^{\infty} \frac{1}{(1-it)^{s}} \ dt = \frac{i}{1-s}\frac{1}{ (1-it)^{s-1}} \Big|^{\infty}_{0} = \frac{i}{s-1}
\int_{0}^{\infty} \text{Im} \ \frac{\coth( \pi t) }{(1-it)^{s}} \ dt = \frac{1}{2} \int_{-\infty}^{\infty} \ \text{Im} \ \frac{\coth(\pi t)}{(1-it)^{s}} \ dt = \frac{1}{2} \text{Im} \ \text{PV} \int_{-\infty}^{\infty} \frac{\coth(\pi t)}{(1-it)^{s}} \ dtLet z = 1-it= \frac{1}{2} \text{Im} \ \text{PV}\int^{1+ i \infty}_{1- i \infty} i \coth \Big( \pi i (z-1) \Big) \frac{dz}{z^{s}} \ dz = \frac{1}{2} \text{Im} \ \text{PV}\int^{1+ i \infty}_{1- i \infty} \cot \Big( \pi(z-1) \Big) \frac{dz}{z^{s}} = \frac{1}{2} \text{Im} \ \text{PV}\int^{1+ i \infty}_{1- i \infty} \frac{\cot (\pi z)}{z^{s}} \ dzwhere the contour is a vertical lineSince there is a branch point at the origin, close the contour to the right.Then \int_{0}^{\infty} \text{Im} \ \frac{\coth( \pi t) }{(1-it)^{s}} \ dt = \frac{1}{2} \text{Im} \ \Big( \pi i \text{Res} \Big[ \frac{\cot(\pi z)}{z^{s}} , 1 \Big] + 2 \pi i \sum_{n=2}^{\infty} \text{Res} \Big[\frac{\cot (\pi z)}{z^{s}}, n \Big] \Big)
=\frac{1}{2} \text{Im} \ \Big( \pi i \Big( \frac{1}{\pi} \Big) + 2 \pi i \sum_{n=2}^{\infty} \frac{1}{\pi n^{s}} \Bigg) = \frac{1}{2} \text{Im} \ \Big( i + 2 i \zeta(s) - 2i \Big) = \zeta(s) - \frac{1}{2}
Putting things together we get\displaystyle \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt = \frac{1}{2} \Big( \zeta(s) - \frac{1}{2} \Big) - \frac{1}{2} \text{Im} \ \frac{i}{s-1} = \frac{\zeta(s)}{2} - \frac{1}{4} + \frac{1}{2(1-s)}\displaystyle \implies \zeta(s)= 2 \int_{0}^{\infty} \frac{\sin (s \arctan t)}{(1+t^{2})^{s/2} (e^{2 \pi t} - 1)} \ dt+ \frac{1}{2} + \frac{1}{s-1}