Would someone please help me understand how to integrate $$ \ \int_0^1 (x^2-1)^{-1/2}dx\, ? $$

This is a homework problem from Marsden Basic Complex Analysis. The text book suggested using a "dog bone" contour and finding the residue of a branch of $(z^2-1)^{-1/2}$ at infinity. I believe the residue at infinity is 1.

After factoring $$ \ (z^2-1)^{-1/2}\ = (z-1)^{-1/2}\ (z+1)^{-1/2}\ $$ I chose a branch cut of $(-\infty , -1] \;$ for $\;(z+1)^{-1/2}$ and $(-\infty , 1]$ for $(z-1)^{-1/2}$. I pretty sure that means $\: -\pi \: <\arg(z-1)< \:\pi$ and $\: -\pi \: <\arg(z+1)< \:\pi$.

This problem is so confusing. I've working on it for days and it's driving me crazy. Any help would be greatly appreciated.

  • 1
    $\begingroup$ Do you understand about the residue at infinity? There is a good example here en.wikipedia.org/wiki/Methods_of_contour_integration. $\endgroup$
    – Mark Viola
    May 2, 2015 at 4:57
  • 1
    $\begingroup$ That was the best example I've seen so far. In that example, the integral was from 0 to 3 and the branch cut for the integral was from 0 to 3. But for this problem, I'm integrating from 0 to 1 but my branch cut is from -1 to 1. It's tempting to just divide the answer by 2, but that doesn't seem right. $\endgroup$
    – Mmmath
    May 2, 2015 at 5:22

2 Answers 2


Choose the branch cuts as $(-\infty,-1]$ for $(z+1)^{-1/2}$ and $(-\infty,+1]$ for $(z-1)^{-1/2}$.

Then, $f(z) =(z^2-1)^{-1/2}$ is continuous across the negative real axis and the "effective" branch cut is $[-1,+1]$.

We will integrate $f$ on the clockwise contour $C$, which is the "dog-bone" clock-wise contour that encompasses $z=\pm 1$. To that end, we have

$$\begin{align} \oint_C f(z) dz &= \oint_C (z+1)^{-1/2} (z-1)^{-1/2} dz\\\\ &=\int_{-1}^1 \frac{dx}{+\sqrt{x^2-1}}\,dx+\int_{1}^{-1} \frac{dx}{-\sqrt{x^2-1}}\,dx\\\\ &=4\int_{0}^1 \frac{dx}{\sqrt{x^2-1}}\,dx \end{align}$$

Note that we tacitly used the fact that the contributions to the small "circles" (i.e., at the ends of the contour) around $z=\pm 1$ tend to zero as the radii of those circles approach zero.

We now compute the residue at infinity (Note: This is equivalent to evaluating the integral of $f$ on a counter-clockwise spherical contour of radius $R$ in the limit as $R \to \infty$). This is given by

$$\text{Res}_{z=\infty} f(z)=\text{Res}_{z=0} \left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)=-1$$

Putting it together gives

$$4\int_{0}^1 \frac{dx}{\sqrt{x^2-1}}\,dx=-2\pi i(-1)$$

from which we have

$$\int_{0}^1 \frac{dx}{\sqrt{x^2-1}}\,dx=i\pi/2$$

  • $\begingroup$ Oh my goodness! Thank you so much. $\endgroup$
    – Mmmath
    May 2, 2015 at 7:14
  • $\begingroup$ You're welcome. It was my pleasure. $\endgroup$
    – Mark Viola
    May 2, 2015 at 13:43
  • $\begingroup$ Hi @Dr.MV, are you using the principal branch of log and restricting the arguments to satisfy $-\pi < \theta \le \pi$? What arguments do the two lines pick up, when approaching the real axis from above and below? Does the line below pick up $-\pi$ from -1 to 0, and an argument of 0 from 0 to 1, and similarly the line from above picks up an argument of $+\pi$ from -1 to 0, and an argument of 0 from 0 to 1? And also, you have a factor of -4, but shouldn't it be +4? (from inverting limits in your second integral...) thanks, $\endgroup$
    – User001
    Jan 1, 2016 at 1:33
  • $\begingroup$ ...I am thinking that I am incorrect about the line from above, from 0 to 1 -- and that it picks up an argument of $+\pi$, not 0...@Dr.MV, what do you think? Thanks, $\endgroup$
    – User001
    Jan 1, 2016 at 1:36
  • $\begingroup$ (because of the jump discontinuity from 0 to 1) $\endgroup$
    – User001
    Jan 1, 2016 at 1:37

Use the right tool for the right task. In this context, it is much easier to integrate as follows:

$$\int_0^1 \dfrac{dx}{\sqrt{x^2-1}} = \underbrace{i\int_0^1\dfrac{dx}{\sqrt{1-x^2}} = i\int_0^{\pi/2} \dfrac{\cos(t)dt}{\cos(t)}}_{x = \sin(t)} = \dfrac{i \pi}2$$

  • 4
    $\begingroup$ Unfortunately, the exercise is not to use the "right tool," but a tool that is potentially more powerful, but much less efficient here ... to continue your metaphor (or is it a simile?). $\endgroup$
    – Mark Viola
    May 2, 2015 at 4:34

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