Consider integrating $f(z)=\sqrt{1-z^2}$ with a branch cut of $[-1,1]$ around the following contour.

$\gamma_1:[-1,1]\to\mathbb{C}, t\mapsto t+\epsilon i$

$\gamma_2:[-\pi/2,\pi/2]\to\mathbb{C}, t \mapsto 1+\epsilon e^{-it}$

$\gamma_3:[-1,1]\to\mathbb{C}, t\mapsto -t-\epsilon i$

$\gamma_4:[-3\pi/2,-\pi/2]\to\mathbb{C}, t \mapsto -1+\epsilon e^{-it}$

As $\epsilon$ tends to $0$ we have that $\int_{\gamma_2}f(z)dz$ and $\int_{\gamma_4}f(z)dz$ tend to $0$.

Now, $\int_{\gamma_1}f(z)dz$ tends to $\int_{-1}^{1}\sqrt{1-x^2}dx:=I$ and also $\int_{\gamma_3}f(z)dz$ also tends to $I$ (one minus sign from being on the other side of the branch cut, another one from reversing the lower/upper limits).

Now $I=\pi/2$

So the integral around the closed contour $\lim_{\epsilon\to 0}\int_{\gamma_1+\gamma_2+\gamma_3+\gamma_4} f(z) dz=\pi$

Note that $f$ is analytic in $\mathbb{C}\setminus[-1,1]$ with our choice of branch cut and so we can consider the contour integral around $\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4$ as a closed contour around infinity.

i.e. $\int_{\gamma}f(z)dz=-2\pi i Res[f,z=\infty]$ ($-2\pi i$ instead of $2\pi i $ because we are going clockwise around infinity.)

Now, $Res[f,z=\infty]=Res[\sqrt{1-1/z^2},z=0]=\lim_{z\to 0}z\sqrt{1-1/z^2}=\lim_{z\to 0}\sqrt{z^2-1}=i$

So $\int_{\gamma}f(z)dz=-2\pi i (i)=2\pi$

And the two results do not agree. I am not confident my arguments about the pole at infinity. What exactly did I do wrong there?

  • $\begingroup$ You messed with the branches when going from $\sqrt{1-1/z^2}$ to $\sqrt{z^2-1}$, or you didn't define the residue at $\infty$ correctly. For $f$ is analytic on $|z| > 1$ then $\int_{|z|=2} f(z)dz = -\int_{|s|=1/2} \frac{f(1/s)}{-s^2}ds = 2i \pi Res(\frac{f(1/s)}{s^2},0)$ $\endgroup$ – reuns Aug 8 at 22:23

See this answer. You have implicitly used the condition that $f(x + i0) > 0$ for $-1 < x <1$ (otherwise you would get $I = -\pi$). With this condition, $f$ can be written as $$f(z) = -i z \sqrt {1 - \frac 1 {z^2}},$$ where $\sqrt z$ is the principal value of the square root.

$\gamma$ goes around the origin clockwise, therefore $I = +2 \pi i \operatorname{Res}_{z = \infty} f(z)$. By the binomial theorem, the Laurent expansion of $\sqrt {1 - 1/z^2}$ around infinity is $1 - 1/(2 z^2) + O(1/z^3)$, which gives $\operatorname{Res}_{z = \infty} f(z) = 1/(2 i)$.

  • $\begingroup$ I understand that $\sqrt{1-1/z^2}=1-1/(2z^2)+O(1/z^4)$. So, $f(z)=-iz+i/(2z)+O(1/z^3)$ and thus, $f(1/z)=-i/z+O(z)$ which would surely tell us that that the residue at $\infty$ is $-i$ (we aren't looking at the residue at $0$.) $\endgroup$ – daruma Aug 9 at 6:03
  • $\begingroup$ The residue at infinity is $-c_{-1}$ in the expansion around infinity. Or the residue of $-f(1/z)/z^2$ at zero. $\endgroup$ – Maxim Aug 9 at 6:24
  • $\begingroup$ Ah, right. I might have misunderstood that the residue of $f(1/z)$ at $0$ as the residue of $f(z)$ at $\infty$ (where it should be $-f(1/z)/z^2$.) $\endgroup$ – daruma Aug 9 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.