2

We want to compue $I=\int_{-1}^1 \frac{dz}{\sqrt{1-z^2}(2+z)}$, so we consider the following countor integral. Since there is a branch cut $(-1,1)$, if we take the contour $C$ (resembling the image below, ignoring the captions) Resembling this icontour

around this cut wrapping around $z=1$ with a small circle of radius $\epsilon$ and extending to the left and closing up in the $z\rightarrow \infty$, we end up in (since the contour does not include the simple pole at $-2$) $$\oint_C \frac{dz}{\sqrt{1-z^2}(2+z)}=2I=0$$ However, $I$ is definitely not zero. What's wrong here? Is it correct to think that $(-\infty, -1)$ part does not cancel due to divergence near $-2$?

Nugi
  • 560
  • But the original interval of integration is $(-1,1)$. Are you sure this integral is equal to the integral around contour $C$? – thedude Oct 15 '21 at 15:06
  • 1
    "wrapping around $z=1$ with a small circle and extending to the left and closing up in the $z\to\infty$". I am not sure I understand what contour you mean here... – thedude Oct 15 '21 at 18:02
  • I too cannot make sense out of your description of the contour, but I think the main problem is your claim it will approach $2I$ as $\epsilon \to 0$, If it is some sort of circuit, about $(-1,1)$, then it will be travelling in opposite directions on each side. So one side will pick up $I$, and the other side, in the opposite direction, will pick up $-I$ for a total of $I - I = 0$. – Paul Sinclair Oct 16 '21 at 01:04
  • I added an image hoping it will aid the description of the contour. Following this way, as @PaulSinclair described and thedude mentioned, it will give $2I$ as it runs in opposite directions, with $-1$ factor arising from the branch cut. – Nugi Oct 16 '21 at 05:22

1 Answers1

1

Indeed, choose the branch of $\sqrt{1-z^2}$ so that its branch cut is $[-1,1]$. For example, we may let

$$\sqrt{1-z^2} = \exp\left(\frac{1}{2}\log(z-1)+\frac{1}{2}\log(z+1) - \frac{i\pi}{2}\right),$$

where $\log(\cdot)$ is the principal complex logarithm. Then for $-1 < x < 1$,

\begin{align*} \lim_{\epsilon \to 0^+} \sqrt{1-(x+i\epsilon)^2} &= \sqrt{1-x^2} \\ \lim_{\epsilon \to 0^+} \sqrt{1-(x-i\epsilon)^2} &= -\sqrt{1-x^2}, \end{align*}

So, the integral of $ f(z) = \frac{1}{(z+2)\sqrt{1-z^2}} $ along the dog bone contour (red contour in the image below)

enter image description here

is $2I$:

$$ \oint_{\color{red}{\text{dogbone}}} f(z) \, \mathrm{d}z = 2I. $$

So, if $R > 2$ and $\epsilon > 0$ is small enough, then

\begin{align*} \oint_{|z| = R} f(z) \, \mathrm{d}z &= \oint_{|z+2| = \epsilon} f(z) \, \mathrm{d}z + \oint_{\color{red}{-\text{dogbone}}} f(z) \, \mathrm{d}z \\ &= 2\pi i \, \underset{z=-2}{\mathrm{Res}} \, f(z) - \oint_{\color{red}{\text{dogbone}}} f(z) \, \mathrm{d}z \\ &= \frac{2\pi}{\sqrt{3}} - 2I \end{align*}

As you see, you first forgot to take account of the pole of $f(z)$ at $z=-2$. When its contribution is properly accounted as in the computation above, then by using that

$$ \oint_{|z| = R} f(z) \, \mathrm{d}z = \lim_{R\to\infty} \oint_{|z| = R} f(z) \, \mathrm{d}z = 0, $$

we end up with

$$ I = \frac{\pi}{\sqrt{3}}. $$

Sangchul Lee
  • 167,468
  • This computation make sense. I also want to know what is wrong with my computation with the original contour. The pole at $z=-2$ is simply outside the contour. – Nugi Oct 16 '21 at 06:00
  • @Nugi, When you make the line segments $M$ and $N$ in your picture closer to $(-\infty,-1)$, they do not cancel out each other. Rather, each of them will pick up half the contribution from the residue at $z=-2$. – Sangchul Lee Oct 16 '21 at 08:40
  • Let aside that you presented the point that elucidates the 'mistake' I was looking for, how can I see half contribution of the residue? – Nugi Oct 16 '21 at 09:17
  • @Nugi, This is essentially because each of $M$ and $N$ will leave an "infinitesimal semicircle" around $z=-2$ in the limit. To make this precise, let $g$ be holomorphic near $z_0$, and let $\gamma_{\epsilon}(t)=z_0+\epsilon e^{i\theta}$ for $\theta_{\epsilon}\leq\theta\leq\theta_{\epsilon}+\alpha$ be a circular arc of radius $\epsilon$ centered at $z_0$ with a central angle of $\alpha$. Then $$\lim_{\epsilon\to0^+}\int_{\gamma_{\epsilon}}\frac{g(z)}{z-z_0},\mathrm{d}z=i\alpha g(z_0).$$ The choice $\alpha=2\pi$ reduces a special case of the residue theorem, and $\alpha=\pi$ covers our example. – Sangchul Lee Oct 16 '21 at 09:30