= Solution
Use the <Hankel contour> about the negative real axis: it starts at $-\infty$ below the cut, circles the origin counterclockwise, and returns to $-\infty$ above the cut. Take $-\pi<\arg t<\pi$. For $\operatorname{Re}s>0$, the small circle vanishes and the two banks give
$$
\int_{-\infty}^{(0+)}\frac{t^{s-1}}{e^t+e^{-t}}dt
=2i\sin(\pi s)\int_0^\infty\frac{x^{s-1}}{e^x+e^{-x}}dx.
$$
The <gamma reflection formula> then gives
$$
\frac{\Gamma(1-s)}{2\pi i}(2i\sin\pi s)\int_0^\infty\frac{x^{s-1}}{e^x+e^{-x}}dx
=\frac1{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x+e^{-x}}dx=\beta(s).
$$
Near zero,
$$
\frac1{e^t+e^{-t}}=\frac12-\frac14t^2+\frac5{48}t^4-\cdots.
$$
Subtracting any desired number of these terms makes the local contour <integral> converge in successively larger left half-planes. The resulting apparent singularities are removable after multiplication by $\Gamma(1-s)$, so the Hankel formula analytically continues $\beta$ to all $s\in\mathbb C$.
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