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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.

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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 $$ \int_0^1\frac{\mathrm{d}x}{\sqrt{x^2-1}}=i\frac\pi2\tag{6} $$

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  • $\begingroup$ 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
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    $\begingroup$ @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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