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Complex Analysis (Lent 2019) keyhole.pdf
In this example, we evaluate, for $0<\alpha<1$, the integral
$$
I(\alpha)=\int_0^{\infty} \frac{x^\alpha}{x(x+1)} d x
$$
We use a 'keyhole' contour $\Gamma$ depending on three fixed parameters $0<\delta<\varepsilon<R$. The contour consists of the straight line segment $\left\{x+i \delta: \sqrt{\varepsilon^2-\delta^2}<x<\sqrt{R^2-\delta^2}\right\}$ in the first quadrant going parallel to the real axis in the positive direction, then going counter-clockwise along the circle $C(R)=\{|z|=R\}$ to $\sqrt{R^2-\delta^2}-i \delta$, then horizontally in the fourth quadrant the straight line segment $\left\{x-i \delta: \sqrt{\varepsilon^2-\delta^2}<x<\sqrt{R^2-\delta^2}\right\}$ in the negative direction and finally closing with arc of the circle $C(\varepsilon)=\{|z|=\varepsilon\}$ clockwise.
Extend $x^\alpha$ to $z^\alpha=\exp (\alpha \log z)$ understood as a holomorphic function on the cut-plane $\mathbb{C} \backslash[0, \infty)$ by choosing the holomorphic branch of $\log z=\ln |z|+i \arg z$ with $0<\arg z<2 \pi$. Consider $\int_{\Gamma} \frac{z^\alpha}{z(z+1)} d z$ and let $R \rightarrow \infty$ and $\varepsilon, \delta \rightarrow 0$.
We find that each of the integrals over arcs of circles $C(R)$ and $C(\varepsilon)$ tends to zero, estimated by
$$
C^{\prime} R \frac{R^\alpha}{R^2} \text { for large } R\quad \text { and }\quad C^{\prime \prime} \varepsilon \frac{\varepsilon^\alpha}{\varepsilon} \text { for small } \varepsilon
$$
for some positive constants $C^{\prime}, C^{\prime \prime}$. The integral along the upper horizontal straight line segment tends to $I(\alpha)$ and the limit of the integral over the lower horizontal straight line segment may be expressed as $-e^{2 \pi i \alpha} I(\alpha)$ because the values of the chosen branch of $z^\alpha$ tend to $e^{2 \pi i \alpha} x^\alpha$ as we approach the cut $[0, \infty)$ from below. Thus the limit of the contour integral is $\left(1-e^{2 \pi i \alpha}\right) I(\alpha)$, taking account of the directions of integration.
On the other hand, by the Residue Theorem (3.4),
$$
\int_{\Gamma} \frac{z^\alpha}{z(z+1)} d z=2 \pi i \operatorname*{Res}_{z=-1} \frac{z^\alpha}{z(z+1)}=-2 \pi i e^{i \pi \alpha}
$$
noting that the only singularity inside $\Gamma$ is a simple pole at $z=-1=e^{i \pi}$. Therefore,
$$
I(\alpha)=-2 \pi i \frac{e^{i \pi \alpha}}{\left(1-e^{2 \pi i \alpha}\right)}=\frac{-\pi i}{e^{-i \pi \alpha}-e^{\pi i \alpha}}=\frac{\pi}{\sin \pi \alpha}
$$
Notice that $I(\alpha) \rightarrow \infty$ as $\alpha \rightarrow 0$ and as $\alpha \rightarrow 1$ |
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