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.

  • $\begingroup$ Note that, $z=0$ is a branch point and you need to use the keyhole contour. $\endgroup$ – Mhenni Benghorbal Apr 26 '13 at 4:59
  • $\begingroup$ 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. $\endgroup$ – Mhenni Benghorbal Apr 26 '13 at 5:06
  • $\begingroup$ @MhenniBenghorbal: According to Ron Gordon's solution below, it should converge I believe. $\endgroup$ – anon271828 Apr 26 '13 at 5:07
  • $\begingroup$ I'll be checking this carefully. $\endgroup$ – Mhenni Benghorbal Apr 26 '13 at 5:09
  • $\begingroup$ It converges. I was making a mistake. $\endgroup$ – Mhenni Benghorbal Apr 26 '13 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 $z=1$:

enter image description here

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

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


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

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


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


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


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

as was to be demonstrated.

  • $\begingroup$ 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? $\endgroup$ – anon271828 Apr 26 '13 at 5:05
  • $\begingroup$ 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. $\endgroup$ – Ron Gordon Apr 26 '13 at 5:11
  • $\begingroup$ 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 $0$. $\endgroup$ – anon271828 Apr 26 '13 at 5:13
  • $\begingroup$ 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. $\endgroup$ – Ron Gordon Apr 26 '13 at 5:14
  • $\begingroup$ 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. $\endgroup$ – anon271828 Apr 26 '13 at 5:16

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