# Contour Integration: $\int_0^\infty\frac{1}{x^a(1-x)}\,dx$ for $0<a<1$.

I've been trying to calculate $$\int_0^\infty\frac{1}{x^a(1-x)}\,dx\quad\text{with }0<a<1.$$I haven't had much luck. I tried taking the branch cut with of the positive reals and estimating that way, but I wasn't sure how to handle the poles at $z=0$ and $z=1$ when dealing with branch cuts. I also looked at the Wikipedia page to try to get some insight, but I didn't see any similar examples with the contour closing in on poles that are on a branch cut.

• Note that, $z=0$ is a branch point and you need to use the keyhole contour. Apr 26, 2013 at 4:59
• By the way, this integral does not converge for this range of $a$. Just try to evaluate it for $a=1/2$. Make sure you got the right integral. Apr 26, 2013 at 5:06
• @MhenniBenghorbal: According to Ron Gordon's solution below, it should converge I believe. Apr 26, 2013 at 5:07
• I'll be checking this carefully. Apr 26, 2013 at 5:09
• It converges. I was making a mistake. Apr 26, 2013 at 5:14

You use an almost-keyhole contour, except that you indent both paths above and below the real axis with a small semicircle to avoid the pole at $z=1$: In doing this, you end up with not $4$, but $8$ contour segments. I will avoid writing them all out by noting that the integrals over the outer circular arc and inner circular arc at the origin vanish in the limits of their radii going to $\infty$ and $0$, respectively. We are left with

$$\oint_C dz \frac{z^{-a}}{1-z} = \int_{\epsilon}^{1-\epsilon} dx \frac{x^{-a}}{1-x} + i \epsilon \int_{\pi}^0 d\phi\, e^{i \phi} \frac{(1+\epsilon e^{i \phi})^{-a}}{-\epsilon e^{i \phi}} + \int_{1+\epsilon}^{\infty} dx \frac{x^{-a}}{1-x} \\+e^{-i 2 \pi a} \int_{\infty}^{1+\epsilon} dx \frac{x^{-a}}{1-x} +e^{-i 2 \pi a} i \epsilon \int_{2 \pi}^{\pi} d\phi\, e^{i \phi} \frac{(1+\epsilon e^{i \phi})^{-a}}{-\epsilon e^{i \phi}} +e^{-i 2 \pi a} \int_{1-\epsilon}^{\epsilon} dx \frac{x^{-a}}{1-x}$$

Combining like terms, we get

$$\oint_C dz \frac{z^{-a}}{1-z} = \left ( 1-e^{-i 2 \pi a}\right ) PV\int_{0}^{\infty} dx \frac{x^{-a}}{1-x} + \left ( 1+e^{-i 2 \pi a}\right ) i \pi = 0$$

because of Cauchy's Theorem. $PV$ denotes the Cauchy principal value. After a little algebra, the result is

$$PV\int_{0}^{\infty} dx \frac{x^{-a}}{1-x} = -i \pi \frac{1+e^{-i 2 \pi a}}{1-e^{-i 2 \pi a}}=-\pi \cot{\pi a}$$

EXAMPLE

Let's check the result for $a=1/2$. This would imply that

$$PV \int _{0}^{\infty} dx \frac{1}{\sqrt{x} (1-x)} = 0$$

Consider

\begin{align}\underbrace{\int_0^{1-\epsilon} dx \frac{1}{\sqrt{x} (1-x)}}_{x=1/u} &= \int_{1/(1-\epsilon)}^{\infty} \frac{du}{u^2} \frac{\sqrt{u}}{1-(1/u)} \\ &= -\int_{1+\epsilon}^{\infty} du \frac{1}{\sqrt{u} (1-u)}\end{align}

Thus

$$\int_0^{1-\epsilon} dx \frac{1}{\sqrt{x} (1-x)} + \int_{1+\epsilon}^{\infty} du \frac{1}{\sqrt{u} (1-u)} = 0$$

or

$$PV \int _{0}^{\infty} dx \frac{1}{\sqrt{x} (1-x)} = 0$$

as was to be demonstrated.

• Do we ever calculate the residue? Or is that more of a by-product for using the almost-keyhole contour, as when doing integration before learning about the residue theorem? Apr 26, 2013 at 5:05
• No residue here, because we excluded the pole at $z=1$ from the interior of the contour. I would say that we just get contributions from the segments we use to avoid integrating over a pole. It is these contributions that give rise to the value of the integral we seek. This happens more often than you think, and mastery of this technique will provide you with a powerful tool. BTW the integral we sought is the Cauchy principal value because it avoids the pole on the real axis. Apr 26, 2013 at 5:11
• Is there ever convergence of the Cauchy principal value and not of the ordinary integral (assuming they're different)? According to Mhenni's comments above, the integral should diverge, say, for $a=\frac{1}{2}$. According to this solution, it is $0$. Apr 26, 2013 at 5:13
• Mhenni speaks of the integration right through the pole, which obviously diverges. My result is the principal value, which converges to the value I posted. Apr 26, 2013 at 5:14
• I see; thank you very much for your help! As an update, Mhenni retracted his statement about divergence. I'll appreciate this answer more and more as I work through the details. Apr 26, 2013 at 5:16

$$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\on}{\operatorname{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\bbox[5px,#ffd]{\large An\ Alternative:}}$$ \begin{align} &\bbox[5px,#ffd]{% \left.{\rm P.V.}\int_{0}^{\infty}{\dd x \over x^{a}\pars{1 - x}} \,\right\vert_{\,a\ \in\ \pars{0,1}}} = -\,\Re\int_{0}^{\infty}{x^{-a} \over x - 1 + \ic 0^{+}}\,\dd x \\[5mm] = &\ \Re\int_{\infty}^{0}{\ic^{-a}\,\,y^{-a} \over \ic y - 1} \,\ic\,\dd y = -\,\Im\bracks{\ic^{-a}\int_{0}^{\infty} {\,y^{\pars{\color{red}{1 - a}} - 1} \over 1 - \ic y}\,\dd y} \end{align} I'll evaluate the last integral with the Ramanujan's Master Theorem. Note that $$\ds{{1 \over 1 - \ic y} = \sum_{k = 0}^{\infty}\pars{\ic y}^{k} = \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{1 + k}\expo{-k\pi\ic/2}}\,\,{\pars{-y}^{k} \over k!}}$$.

Then, \begin{align} &\bbox[5px,#ffd]{% \left.{\rm P.V.}\int_{0}^{\infty}{\dd x \over x^{a}\pars{1 - x}} \,\right\vert_{\,a\ \in\ \pars{0,1}}} \\[5mm] = &\ -\,\Im\pars{\ic^{-a}\,\,\Gamma\pars{1 - a}\braces{\Gamma\pars{1 -\bracks{1 - a}}\expo{\pars{1 - a}\pi\ic/2}\,}} \\[5mm] = &\ -\,\Re\bracks{\expo{-\pi a\ic}\,\, \Gamma\pars{1 - a}\Gamma\pars{a}} = -\cos\pars{\pi a}\,{\pi \over \sin\pars{\pi a}} \\[5mm] = &\ \bbx{-\pi\cot\pars{\pi a}} \\ & \end{align}