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I'm trying to evaluate the integral $$ I =\int_{-1}^{1}\frac{\sqrt{1-x^{2}}}{1+x^{2}} \, dx $$ by using a dumbbell/dogbone contour, but I'm having difficulty determining the residue at infinity.

I started by defining $\sqrt{1-z^{2}}=\sqrt{(1-z)(1+z)} $ so that it is a well-defined function if the line segment $[-1,1]$ is omitted.

Similar to an example on Wikipedia, I choose the branches where $0 < \arg(1-z) \le 2 \pi$ and $-\pi < \arg(1+z) \le \pi$.

Then I integrated $ f(z) = \frac{\sqrt{1-z^{2}}}{1+z^{2}}$ clockwise around a dumbbell contour.

Just above the branch cut, $\arg(1-z) = 2 \pi$ and $\arg(1+z) = 0$.

And just below the branch cut, $\arg(1-z) = 0$ and $\arg(1+z) = 0$.

So the integral evaluates to $-I$ both above the cut and below the cut.

Since the integrand is meromorphic outside the contour, I get

$$ - 2I = 2\pi i \left( \operatorname{Res}[f,i]+ \operatorname{Res}[f,-i] + \operatorname{Res}[f,\infty] \right),$$ where

$$\begin{align} \operatorname{Res}[f,i] &=\lim_{z\to i }\frac{\sqrt{|1-z|e^{i\arg(1-z)}\ |1+z|e^{i\arg(1+z)}}}{z+i} =\frac{\sqrt{\sqrt{2}e^{\frac{7\pi i}{4}}\sqrt{2}e^{\frac{\pi i}{4}}}}{2i} \\ &=\frac{\sqrt{2}e^{\pi i}}{2i}=-\frac{\sqrt{2}}{2i}, \end{align}$$

$$ \begin{align}\operatorname{Res}[f,-i] &=\lim_{z\to-i }\frac{\sqrt{|1-z|e^{i\arg(1-z)}\ |1+z|e^{i\arg(1+z)}}}{z-i}=\frac{\sqrt{\sqrt{2}e^{\frac{\pi i}{4}}\sqrt{2}e^{\frac{-\pi i}{4}}}}{-2i} \\ &=\frac{\sqrt{2}e^{ 0\pi i}}{-2i}=-\frac{\sqrt{2}}{2i}, \end{align}$$ and

$$ \begin{align}\operatorname{Res}[f,\infty] &=\operatorname{Res}\left[-\frac{1}{z^{2}}f\left(\frac{1}{z}\right), 0 \right]= \operatorname{Res}\left[-\frac{\sqrt{z^2-1}}{z(1+z^{2})},0\right]\\ &=-\lim_{z\to 0}\frac{\sqrt{z^2-1}}{1+z^{2}} = -(-1)^{\frac{1}{2}}. \end{align}$$

But what is $(-1)^{\frac{1}{2}}$?

I assume it must be $i$, but I don't know how to argue that it can't be $-i$.

EDIT:

There is also the issue that Ted Shifrin mentioned of assuming that $\sqrt{z^{2}}=z$.

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    $\begingroup$ Instead of confused about what $\arg(-1)$ are, why not go back to original integral and observe for $x > 1$ on real axis, $\arg(\sqrt{1-x^2}) = \frac{\pi}{2}$. As a result, $\frac{\sqrt{1-z^2}}{1+z^2} \sim \frac{i}{z}$ for large $z$ and the contribution to the contour integral from $\infty$ is $i (-2\pi i) = 2\pi$. This give us: $$-2I = 2\pi i\left(-\frac{\sqrt{2}}{2i} - \frac{\sqrt{2}}{2i}\right) + 2\pi \implies I = \pi (\sqrt{2} - 1)$$ $\endgroup$ – achille hui Jun 24 '13 at 19:48
  • $\begingroup$ @achillehui Could you post your comment as an answer? I would have accepted it a long time ago had you posted it as an answer. $\endgroup$ – Random Variable Jul 17 '17 at 22:12
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Comment converted (and expanded) into an answer per request.


Instead of confused about what $\arg(−1)$ is, go back to original integral and observe for given choice of argument, $\arg\sqrt{1-x^2} = \frac{\pi}{2}$ for $x > 1$ on real axis. As a result, $\displaystyle\;\frac{\sqrt{1-z^2}}{1+z^2} \sim \frac{i}{z}$ for large $z$.

Deform the clockwise dumbbell contour "continuously" to a clockwise circular contour at infinity. In the middle of process, one pickup two extra counterclockwise circular contours at poles ($\pm i$) of the integrand.

The contribution from the clockwsie circular contour at infinity is controlled by the large $z$ behavior of the integrand. It equals to $$(-2\pi i)i = 2\pi$$

Since $\arg\sqrt{1-z^2}$ is negative on positive imaginary axis and positive on negative imaginary axis, contribution from the two counterclockwise circular contours around the poles is

$$(2\pi i)\left(\frac{-\sqrt{2}}{2i} + \frac{\sqrt{2}}{-2i}\right) = -2\pi\sqrt{2}$$ This leads to

$$-2I = -2\pi\sqrt{2} + 2\pi \quad\implies\quad I = \pi(\sqrt{2}-1)$$

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You really don't want to break the square root up; it's not about separate branch cuts, but about a single branch cut for the square root of the quadratic. Presumably, you want the branch of $g(z)=\sqrt{1-z^2}$ with $g(i)=+\sqrt2$. When you then compute the residue of $f$ at $\infty$, you have to interpret the residue at $0$ of $$\sqrt{1-\big(\frac1u\big)^2} = \frac{\sqrt{u^2-1}}{\sqrt{u^2}},$$ and one has to check that the ambiguities in the two square roots cancel out.

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  • $\begingroup$ I defined the function that way so that it has a single branch cut on $[-1,1]$. That's what I need to evaluate the integral. $\endgroup$ – Random Variable Jun 24 '13 at 18:52
  • $\begingroup$ I found it too confusing. Note in your calculation you blithely assumed $\sqrt{z^2}=z$. $\endgroup$ – Ted Shifrin Jun 24 '13 at 19:09

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