= Solution
For a positively oriented simple closed <contour> $C$, with no <pole> on $C$ and with $f$ <holomorphic> inside except at finitely many poles, the <residue theorem> states
$$
\oint_Cf(z)\,dz=2\pi i\sum_{z_j\text{ inside }C}\operatorname{Res}(f,z_j).
$$
For a more general closed contour, each <residue> is multiplied by its <winding number> about the pole.
The powers must be defined using the same <branch of the complex logarithm> as the numerator. A branch suitable for both requested calculations is
$$
\operatorname{Log}z=\log|z|+i\arg z,\qquad -\delta<\arg z<2\pi-\delta,
\quad 0<\delta<2\pi-\max\left(\pi,\frac{2\pi}{\beta}\right).
$$
The <branch cut> is a ray just below the positive real axis, and $z^\beta:=e^{\beta\operatorname{Log}z}$. Since $\beta>1$, the displayed choice of $\delta$ is possible. It includes the required upper half-plane and the sector used below. The origin remains an excluded <branch point>.
In $0\leq\arg z<\pi$, all zeros of $1+z^\beta$ are
$$
\zeta_j=\exp\frac{(2j+1)\pi i}{\beta},\qquad j=0,1,\ldots,\quad 2j+1<\beta.
$$
Each is a <simple pole>, since $\beta\zeta_j^{\beta-1}=-\beta/\zeta_j\ne0$. Its <residue> is
$$
\boxed{\operatorname{Res}(f,\zeta_j)=-\frac{\zeta_j\operatorname{Log}\zeta_j}{\beta}=-\frac{i(2j+1)\pi}{\beta^2}\exp\frac{(2j+1)\pi i}{\beta}.}
$$
The strict inequality matters if a root lies on the negative real axis, which is outside the specified region.
To evaluate the two <improper integrals>, write them as $I$ for the logarithmic numerator and $J$ for numerator one, and set $\theta=\pi/\beta$, $\alpha=2\theta$. Use a counterclockwise annular sector with $\epsilon<|z|<R$ and $0<\arg z<\alpha$. It encloses precisely $\zeta_0=e^{i\theta}$. The outward positive-axis segment contributes $I$ in the limit. On the inward segment $z=re^{i\alpha}$, the logarithm is $\log r+i\alpha$ and $z^\beta=r^\beta$, so that segment contributes $-e^{i\alpha}(I+i\alpha J)$.
The small arc is $O(\epsilon|\log\epsilon|)$, and the large arc is $O(R^{1-\beta}\log R)$; both vanish. These estimates follow from the arc lengths and $|1+z^\beta|\geq\big||z|^\beta-1\big|$. The real integrals also converge absolutely at both endpoints for $\beta>1$. The <residue theorem> now gives
$$
(1-e^{2i\theta})I-2i\theta e^{2i\theta}J=2\theta^2e^{i\theta}.
$$
After division by $e^{i\theta}$, the real and imaginary parts are
$$
2\theta\sin\theta\,J=2\theta^2,\qquad -2\sin\theta\,I-2\theta\cos\theta\,J=0.
$$
Since $0<\theta<\pi$, $\sin\theta>0$. Hence the <logarithmic beta integral from a sector contour> yields
$$
\boxed{J=\frac{\pi}{\beta\sin(\pi/\beta)},\qquad I=-\frac{\pi^2\cos(\pi/\beta)}{\beta^2\sin^2(\pi/\beta)}.}
$$
For $1<\beta<2$ the sector angle exceeds $\pi$. This is why choosing the <principal complex logarithm> on the whole sector would be invalid; the displaced cut above handles that range as well as $\beta\geq2$.
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