I've been given the task to evaluate the integral

$$\int_\Gamma \frac{3z+1}{z^2-z}\,dz$$

over the following two curves

  1. $\Gamma=1+\sqrt{3}e^{it} \,\,\,\,-\pi/2\leq t\leq 0$

  1. $\Gamma=1+\sqrt{3}e^{-it}\,\,\,\, \pi/2\leq t\leq 2\pi$

enter image description here

For the first curve, we use the fundamental theorem of contour integrals. We note that, by partial fractions,

$$\int\frac{3z+1}{z^2-z}\,dz=\int\frac{4}{z-1}-\frac{1}{z}\, dz=4\ln{(z-1)}-\ln z$$

The curve has endpoints $a=1-\sqrt{3}i$ and $b=1+\sqrt{3}$. So, the integral evaluates to


which is the correct answer. However, by the fundamental theorem , the second curve should lead to the same evaluation since it has the same endpoints. This is incorrect. But, if we combine these curves into one closed curve $\Gamma = \Gamma_1-\Gamma_2$, we can also evaluate this integral using cauchy's theorem. We note that the integrand has two singularities, one at $z=0$ and one at $z=1$. By Cauchy's theorem, we then have

$$\int_\Gamma\frac{3z+1}{z^2-z}=2\pi i*(Res[0]+Res[1])=6\pi i$$

We can split the original integral into two pieces to solve for the unknown integral

$$\int_{\Gamma_2}\frac{3z+1}{z^2-z}=-6\pi i+\int_{\Gamma_1}\frac{3z+1}{z^2-z}=-6\pi i + \ln{(\sqrt{3}-1)}+\frac{5\pi}{3}i=\ln{(\sqrt{3}-1)}-\frac{13\pi}{3}i$$

which is the correct answer. My question is, why can't I use the fundamental theorem for the second curve, but I can for the first? What assumption fails?

  • 1
    $\begingroup$ Do you see why $\int_a^b \frac{dz}{z}$ depends on the path $a \to b$ ? In particular $\int_{|z| = 1} \frac{dz}{z} = 2i \pi = \log(e^{2i \pi}) -\log(1)\ne \log(1)-\log(1)$ where $\log(e^{2i \pi})$ is a notation to say "be careful with the branches" $\endgroup$ – reuns Oct 27 '17 at 2:49
  • $\begingroup$ You should state the precise fundamental theorem of contour integration you are using. Other questions here don't seem to have a single-valued version of this theorem. $\endgroup$ – Eric Towers Oct 27 '17 at 2:49
  • $\begingroup$ For the two integrals one needs a different branch of the "function" $4\ln(z-1)-\ln z$. $\endgroup$ – Angina Seng Oct 27 '17 at 2:52

The fundamental theorem of calculus only works if your function is holomorphic along the entire path of integration. Note that

$$ f(z) = \frac{3z+1}{z(z-1)}$$

has two branch points at $z=0$ and $z=1$ and therefore has a principal branch cut along $(-\infty,0)\cup(1,\infty)$. If you use this branch cut to evaluate the antiderivative, it violates the FTOC since the path intersects it.

A way around this is to employ an alternate branch cut on $(0,1)$. I'll leave the computations to you


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.