= Solution
The <residue theorem> states that if a meromorphic <function> has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour <integral> is $2\pi i$ times the sum of the enclosed residues.
Apply it to
$$
F(z)=\frac{e^{\alpha L(z)}}{(1+z)^2}
$$
on a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha<1$ and $\alpha>-1$, respectively. On the upper bank the numerator tends to $x^\alpha$, while on the lower bank it tends to $e^{2\pi i\alpha}x^\alpha$ and the direction is reversed. Therefore the limiting contour <integral> is
$$
(1-e^{2\pi i\alpha})I.
$$
The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,
$$
\operatorname{Res}_{z=-1}F
=\left.\frac d{dz}e^{\alpha L(z)}\right|_{z=-1}
=-\alpha e^{i\pi\alpha}.
$$
The residue theorem now gives
$$
(1-e^{2\pi i\alpha})I=-2\pi i\alpha e^{i\pi\alpha}.
$$
For $\alpha\ne0$, division and
$$
1-e^{2\pi i\alpha}=-2ie^{i\pi\alpha}\sin(\pi\alpha)
$$
yield
$$
\boxed{I=\frac{\pi\alpha}{\sin(\pi\alpha)}}.
$$
At $\alpha=0$, the <integral> is one, agreeing with the continuous <limit>. This is the <positive-axis keyhole beta integral>.
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