Сalculate the integral PV $\int_{0}^{\infty} \frac{dx}{x^\alpha(x-a)}dx.$

Question

Сalculate the integral $$ \mathrm{PV}\hspace{-0.5ex}\int_{0}^{\infty} \frac{dx}{x^\alpha(x-a)}, $$ where $0<\alpha <1$ and $a>0$.

So we have simple poles $z = 0$ and $z = a$; We can build contour like in this question to exclude these points.

But for example, in the general case $\int_{0}^{\infty} \frac{R(x)}{x^\alpha} \, dx$, where $R(x)$ is a rational function without poles in $\mathbb{R}$ we can apply something like that:^src)

\begin{align*} &\lim_{\substack{\varepsilon \to 0 \\ R \to \infty}} \int_{\partial\mathcal{D}_{R,\varepsilon}} \frac{R(z)}{z^{\alpha}} \, dz = 2\pi i \cdot \sum_k \mathop{\underset{\substack{z=z_k \\ z_k \neq 0}}{\mathrm{Res}}} \frac{R(z)}{z^{\alpha}} \\ &\hspace{3ex}= \lim_{\substack{\varepsilon \to 0 \\ R \to \infty}} \Biggl[ \, \underbrace{\int_{\partial\mathcal{C}_{\varepsilon}} \frac{R(z)}{z^{\alpha}} \, dz}_{\to 0 \text{ as } \varepsilon \to 0} + \int_{[\varepsilon, R]^+} \frac{R(z)}{z^{\alpha}} \, dz + \underbrace{\int_{\partial\mathcal{C}_{R}} \frac{R(z)}{z^{\alpha}} \, dz}_{\to 0 \text{ as } R \to \infty} + \int_{[R, \varepsilon]^-} \frac{R(z)}{z^{\alpha}} \, dz \,\Biggr] \\ &\hspace{3ex}= \int_{0}^{\infty} \frac{R(x)}{x^{\alpha}} \, dx + \int_{\infty}^{0} \frac{R(x)}{x^{\alpha}e^{2\pi \alpha i}} \, dx \\ &\hspace{3ex}= \int_{0}^{\infty} \frac{R(x)}{x^{\alpha}} \, dx - \biggl( \int_{0}^{\infty} \frac{R(x)}{x^{\alpha}e^{2\pi \alpha i}} \, dx \biggr) e^{-2\pi \alpha i} \\[1ex] &\hspace{3ex}= I - I e^{-2\pi \alpha i} = \bigl(1 - e^{-2\pi \alpha i}\bigr) I \\[2ex] &\hspace{3ex}\implies \ \bbox[border:1px black solid;padding:5px;]{I = \frac{2\pi i}{1 - e^{-2\pi \alpha i}} \sum_k \mathop{\underset{\substack{z=z_k \\ z_k \neq 0}}{\mathrm{Res}}} \frac{R(z)}{z^{\alpha}}} \end{align*}

So, my question: How can I avoid long calculations from first answer.

Сan I apply this theorem for my case? (For example use this^src) instead of $2\pi i \mathop{\underset{z=z_k}{\mathrm{Res}}} \frac{R(z)}{z^\alpha}$ from my calculation above.)

Theorem 9.13. Suppose $f(z)$ has a simple pole at $z_0$. Let $C_r$ be the semicircle $\gamma(\theta) = z_0 + r\mathrm{e}^{i\theta}$, with $0 \leq \theta \leq \pi$. Then $$\lim_{r \to 0} \int_{C_r} f(z) \, \mathrm{d}z = \pi i \operatorname{Res} (f, z_0). $$

Answer 1

The following result may be useful:

Lemma. Let $f$ be a holomorphic function defined on an open set $U$ containing the interval $(0, \infty)$ such that $\int_{0}^{\infty} \frac{|f(x)|}{1+x} \, \mathrm{d}x < \infty$. Then for any $a > 0$, we have \begin{align*} \mathrm{PV}\hspace{-0.5ex}\int_{0}^{\infty} \frac{f(x)}{x-a} \, \mathrm{d}x &= \lim_{\varepsilon \to 0^+} \int_{0}^{\infty} \frac{f(x)}{x-a-i\varepsilon} \, \mathrm{d}x - \pi i f(a) \\ &= \lim_{\varepsilon \to 0^+} \int_{0}^{\infty} \frac{f(x)}{x-a+i\varepsilon} \, \mathrm{d}x + \pi i f(a). \end{align*}

Using this lemma and OP's computation, we have

\begin{align*} \mathrm{PV}\hspace{-0.5ex}\int_{0}^{\infty} \frac{1}{x^{\alpha}(x-a)} \, \mathrm{d}x &= \lim_{\varepsilon \to 0^+} \int_{0}^{\infty} \frac{1}{x^{\alpha}(x-a-i\varepsilon)} \, \mathrm{d}x - \frac{\pi i}{a^{\alpha}} \tag{by lemma} \\ &= \lim_{\varepsilon \to 0^+} \frac{2\pi i}{1-e^{-2\pi\alpha i}} \cdot \frac{1}{(a+i\varepsilon)^{\alpha}} - \frac{\pi i}{a^{\alpha}} \tag{by OP} \\ &= \frac{2\pi i}{1-e^{-2\pi\alpha i}} \cdot \frac{1}{a^{\alpha}} - \frac{\pi i}{a^{\alpha}} \\ &= \frac{\pi \cot(\pi \alpha)}{a^{\alpha}}. \end{align*}

---

Proof of Lemma. Let $\delta > 0$ be sufficiently small so that the closed disk $\{z\in\mathbb{C} : |z-a|\leq\delta\}$ is contained in $U$, and consider the contour $(0, a-\delta) \cup \gamma_{\delta} \cup (a+\delta, \infty)$ as below:

Then by the Cauchy integration theorem, we can deform the integral path $(0, \infty)$ into the above contour without changing the value of the integral:

\begin{align*} \int_{0}^{\infty} \frac{f(x)}{x-a-i\varepsilon} \, \mathrm{d}x &= \int_{0}^{a-\delta} \frac{f(x)}{x-a-i\varepsilon} \, \mathrm{d}x + \int_{\gamma_{\delta}} \frac{f(z)}{z-a-i\varepsilon} \, \mathrm{d}z + \int_{a+\delta}^{\infty} \frac{f(x)}{x-a-i\varepsilon} \, \mathrm{d}x \end{align*}

Letting $\varepsilon \to 0^+$ and then $\delta \to 0^+$, and then applying the dominated convergence theorem and Theorem 9.13 in OP, we get

\begin{align*} &\lim_{\varepsilon \to 0^+} \int_{0}^{\infty} \frac{f(x)}{x-a-i\varepsilon} \, \mathrm{d}x \\ &\quad= \int_{0}^{a-\delta} \frac{f(x)}{x-a} \, \mathrm{d}x + \int_{\gamma_{\delta}} \frac{f(z)}{z-a} \, \mathrm{d}z + \int_{a+\delta}^{\infty} \frac{f(x)}{x-a} \, \mathrm{d}x \\ &\quad\to \mathrm{PV}\hspace{-0.5ex}\int_{0}^{\infty} \frac{f(x)}{x-a} \, \mathrm{d}x + \pi i f(a) \qquad \text{as } \delta \to 0^+. \end{align*}

This proves the first part of the desired equality. The second part can be proved in a similar manner.

Answer 2

This can be done without any contour integrals with some clever manipulations. It's not too hard to show that $$ \mathrm{PV}\int_0^\infty \frac{dx}{x^\alpha (x-a)} = \frac{1}{a^\alpha}\mathrm{PV}\int_0^\infty \frac{du}{u^\alpha (u-1)} =\frac{1}{\alpha^a}\int_0^{1}\frac{u^{\alpha-1}-u^{-\alpha}}{1-u}du. $$ We can then use this integral representation of the digamma function, $$ \psi(z) = \frac{d}{dz}\ln\Gamma(z) = \int_0^1\left(\frac{u^{z-1}}{u-1}-\log u\right)du, $$ to compute the integral as $$ \int_0^{1}\frac{u^{\alpha-1}-u^{-\alpha}}{1-u}du= \psi(1-\alpha)-\psi(\alpha) = -\frac{d}{d\alpha}\ln\left[\Gamma(1-\alpha)\Gamma(\alpha)\right] = -\frac{d}{d\alpha}\ln[\pi\csc(\pi \alpha)] = \pi\cot(\pi\alpha) $$ which gives $$ \mathrm{PV}\int_0^\infty \frac{dx}{x^\alpha (x-a)} =\frac{\pi\cot(\pi\alpha)}{\alpha^a} $$