I am trying to compute the following integral through the use contour integration.

$$ \int_0^1 \frac{dx}{\sqrt{x^2-1}} $$

So, I am considering the same integrand but from $-1$ to $1$, then doing the usual switch to a complex variable $z$. The contour I am considering is a dumbbell type (please let me known if I need to elaborate on its shape) where the "bells" are centered around $-1$ and $1$, and the two horizontal contours are just above and below the real axis. I run into trouble when picking the branch cuts. What would be a good way to choose the branch in this case? A reasonable choice for me seems to be the cut $[-1,1]$, but when doing this I have trouble defining the argument on $z$. If someone could enlighten me on this I would be very thankful. The computation does not interest me as much as the intuition and method for defining the branch. Any help is appreciated.


The main question is how do we want to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$? There are two possibilities: $\frac1{\sqrt{x^2-1}}=\frac{\pm i}{\sqrt{1-x^2}}$. Once we decide that, things are pretty simple.

We can define $$ \log\left(\frac{z+1}{z-1}\right) =\log(3)+\int_2^z\left(\frac1{w+1}-\frac1{w-1}\right)\mathrm{d}w\tag{1} $$ where the integral is evaluated along any path which does not intersect $[-1,1]$. A closed path avoiding $[-1,1]$ will circle both poles an equal number of times and the residues will cancel.

Therefore, $(1)$ defines $\log\left(\frac{z+1}{z-1}\right)$ with a branch cut along $[-1,1]$. Using $(1)$, we can define $$ \frac1{\sqrt{z^2-1}}=\frac1{z+1}e^{\frac12\log\left(\frac{z+1}{z-1}\right)}\tag{2} $$ According $(2)$, the integrand along the top of $[-1,1]$ is $\frac{-i}{\sqrt{1-z^2}}$ and along the bottom of $[-1,1]$ is $\frac{i}{\sqrt{1-z^2}}$. The integral around the two dumbbell ends vanish as their size gets smaller. Thus, the integral counter-clockwise along the whole dumbbell is $$ 4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}\tag{3} $$ The integral of $\frac1{\sqrt{z^2-1} }$, as defined in $(2)$, counter-clockwise around a circle of essentially infinite radius is $2\pi i$.

Cauchy's Integral Theorem says that the integral around the dumbbell and a circle of essentially infinite radius are the same. Thus, $(3)$ says $$ 4i\int_0^1\frac{\mathrm{d}x}{\sqrt{1-x^2}}=2\pi i\tag{4} $$ We are back to the question I raised at the beginning: how to define $\frac1{\sqrt{x^2-1}}$ along $[-1,1]$.

If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the top of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{-i}{\sqrt{1-x^2}}$ and we get $$ \int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=-i\frac\pi2\tag{5} $$ If we look at $\frac1{\sqrt{x^2-1}}$ as $\frac1{\sqrt{z^2-1}}$ along the bottom of $[-1,1]$, then $\frac1{\sqrt{x^2-1}}=\frac{i}{\sqrt{1-x^2}}$ and we get $$ \int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=i\frac\pi2\tag{6} $$

  • $\begingroup$ Thank you for this lovely exposition! $\endgroup$ Apr 2 '15 at 16:57
  • $\begingroup$ I think I'm missing something simple. Why is it true that the integral, as defined in (2), around a circle of essentially infinite radius is $2\pi i$? The function is not meromorphic in the interior (it's not defined on $[-1,1]$), so we cannot apply the residue theorem. $\endgroup$
    – Potato
    Jul 3 '15 at 12:46
  • $\begingroup$ Or are we just using the fact that the exponential term is going to get very close to $1$ as the radius of the circle becomes large? $\endgroup$
    – Potato
    Jul 3 '15 at 12:49
  • 1
    $\begingroup$ @Potato: consider $e^{\frac12\log\left(\frac{z+1}{z-1}\right)}-1$ when $|z|\to\infty$ then look at $(2)$ again while considering the integral of $\frac1{z+1}$ around a circle of near infinite radius. $\endgroup$
    – robjohn
    Jul 3 '15 at 13:02

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