= Solution
Use a keyhole contour about the negative real axis, indented symmetrically around the pole $s=t$. The jump of $s^{z-1}$ across the cut is determined by the chosen branch, and the small and large circular contributions vanish for $0<\operatorname{Re}z<1$. The <residue theorem>, with the symmetric indentation interpreted as a <Cauchy principal value>, gives
$$
\boxed{\operatorname{PV}\int_{-\infty}^{\infty}
\frac{s^{z-1}}{s-t}\,ds=\pi i\,t^{z-1}}.
$$
Splitting the real axis at zero and substituting $s\mapsto-s$ in the negative part gives two linear relations. Solving them yields, for real $0<z<1$,
$$
\boxed{\int_0^\infty\frac{s^{z-1}}{s+t}\,ds
=\pi t^{z-1}\csc(\pi z)},
$$
$$
\boxed{\operatorname{PV}\int_0^\infty\frac{s^{z-1}}{s-t}\,ds
=-\pi t^{z-1}\cot(\pi z)}.
$$
Both sides are holomorphic functions of $z$ throughout the vertical strip $0<\operatorname{Re}z<1$: convergence is locally uniform there, and the trigonometric expressions are holomorphic away from integer poles. The <identity theorem> therefore extends the identities from real $z$ to the whole strip.
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