<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Verify the following integral.

$\ds\int_0^{\infty} \ds\frac{dx x^{-p}}{x-1}=\pi \cot p\pi,\
0&lt;p&lt;1$
\medskip

{\bf Answer}

The integral converges for $0&lt;\rho&lt;1$ only if we can treat the
singularity (pole) of the integrand at $x=+1$ in a suitable way.

This is done by considering:

$$J=\ds\int_C \ds\frac{dz z^{-\rho}}{(z-\alpha)}$$

where we take $C$ as PICTURE \vspace{2in}

for $z \in {\bf{C}},\ \alpha \in {\bf{C}}$

Now $J=2\pi i \times (\rm{residue\ at}\ \alpha)=2\pi i
\alpha^{-\rho}$

$-2\pi i
\alpha^{-\rho}=\ds\int_{\epsilon}^R+\ds\int_{\Gamma_2}+\ds\int_{Re^{2\pi
i}}^{\epsilon}+\ds\int_{\Gamma_1}$

Take limits $R \to \infty,\ \epsilon \to 0$ and
$\ds\int_{\Gamma_1},\ \ds\int_{\Gamma_2} \to 0$

Therefore

\begin{eqnarray*} 2\pi i \alpha^{-\rho} &amp; = &amp; \undb{\ds\int_0^{\infty}
\ds\frac{dx\ x^{-\rho}}{(x-\alpha)}}+ \undb{\ds\int_{\infty
e^{2\pi i}}^{0} \ds\frac{dz\ z^{-\rho}}{(z-\alpha)}}\\ &amp; &amp; \ \ \
z=x\ \ \ \ \ \ \ \ \ z=xe^{2\pi i}\\ &amp; = &amp; \ds\int_0^{\infty}
\ds\frac{dx\ x^{-\rho}}{(x-\alpha)}+\ds\int_{\infty}^0
\ds\frac{d(xe^{2\pi i}) (xe^{2\pi i})^{-\rho}}{(xe^{2\pi
i}-\alpha)}\\ 2\pi i \alpha^{-\rho} &amp; = &amp; \ds\int_0^{\infty}
\ds\frac{dx\ x^{-\rho}}{(x-\alpha)}(1-e^{-2\pi i})\end{eqnarray*}

So $\ds\int_0^{\infty}\ds\frac{dx\
x^{-\rho}}{(x-\alpha)}=\ds\frac{2\pi i \alpha^{-\rho}}{1-e^{-2\pi
i}},\ \ \alpha \not\in {\bf{R}}+$

When $\alpha&gt;0$ it can be either above the cut or below. If above
we have

$$\alpha=|\alpha|e^{i0}$$

If below we have

$$\alpha=|\alpha|e^{2\pi i}$$

Let $|\alpha|=1$ now. Then:

$\un{\alpha=e^{i0}}: \ds\int_0^{\infty}\ds\frac{dx\
x^{-\rho}}{(x-e^{i0})}=\ds\frac{2\pi i(e^{i0})^{-\rho}}{1-e^{-2\pi
i\rho}}$

$\un{\alpha=e^{2\pi i}}: \ds\int_0^{\infty}\ds\frac{dx\
x^{-\rho}}{(x-e^{2\pi i})}=\ds\frac{2\pi i(e^{2\pi
i})^{-\rho}}{1-e^{-2\pi i\rho}}$

What's the answer? It's natural to take the average. This is
called Cauchy's Principal Value and is sometimes written as

\begin{eqnarray*} \ds\int_0^{\infty} \ds\frac{dx\ x^{-\rho}}{x-1} &amp;
= &amp; \ds\frac{2\pi i}{1-e^{-2\pi i \rho}}(1+e^{-2\pi i \rho})\times
\ds\frac{1}{2}\ (\rm{average})\\ &amp; &amp; \rm{where}\ x \ne 1\\ &amp; = &amp;
\ds\frac{2\pi}{2} \times \ds\frac{e^{-i\pi \rho}}{e^{-i\pi
\rho}}\ds\frac{(e^{i\pi \rho}+e^{-i\pi \rho})}{(e^{i\pi
\rho}-e^{-i\pi \rho})}\ds\frac{2i}{2}\\ &amp; = &amp; \pi \ds\frac{\cos
\pi \rho}{\sin \pi \rho}\end{eqnarray*}

$\Rightarrow \ds\int_0^{\infty}\ds\frac{dx\ x^{-\rho}}{x-1}=\pi
\cot \pi \rho,\ \ \ x \ne 1$

\end{document}
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