I have problem integrating function with logarithm. Problems seems always to be branch cut of $\log$, but here it is different I think.

I have task to integrate $$\oint_{|z| = 1} \! dz \log\left(\frac{z - a}{z - b}\right)$$ given $|a| < 1$ and $|b| < 1$

First I think to check branch points of logarithm. I write $$\log\left(\frac{z - a}{z - b}\right) = \log(z - a) - \log(z - b)$$ Going around $a$ in small circle I pick up term $2\pi i$, going around $b$ I pick up term $-2\pi i$, so going around both I pick up nothing. So I make branch cut $a$ to $b$ and contour does not intersect.

But now I get stuck! I try to find residue of integral at $a$ and $b$ but I cannot get series of at $a,b$, because logarithm always remains inside. How I can evaluate such integrals?

  • $\begingroup$ If your deformed contour did not encircle the branch cuts to infinity of each term, then the contour does not enclose any singularity. What does Cauchy's Integral Theorem guarantee? $\endgroup$
    – Mark Viola
    Nov 16, 2015 at 16:01
  • $\begingroup$ thank Dr MV, I try and understand connection of Cauchy Integral Theorem. Please see the changes I make $\endgroup$
    – dimi
    Nov 16, 2015 at 16:43

1 Answer 1



For the branch points at $z=a$ and $z=b$, we choose branch cuts as straight line contours, that begin at the branch points, intersect the unit circle at $e^{i\phi_a}$ and $e^{i\phi_b}$, then extend to the point at infinity. Note that these branch cuts are not unique.

Now, let $z=e^{it}$. Then, we have

$$\begin{align} \oint_{|z|=1} \log(z-a)\,dz &= \int_0^{\phi_a^-}\log(e^{it})ie^{it}\,dt+ \int_{\phi_a^+}^{2\pi}\log(e^{it})ie^{it}\,dt\\\\ &=\left.\left((e^{it}-a)\left(\log(e^{it}-a)-1\right)\right)\right|_{0}^{\phi_a^-}+\left.\left((e^{it}-a)\left(\log(e^{it}-a)-1\right)\right)\right|_{\phi_a^+}^{2\pi}\\\\ &=2\pi i (e^{i\phi_a}-a) \end{align}$$

Similarly, we have

$$\oint_{|z|=1} \log(z-b)\,dz =2\pi i (e^{i\phi_b}-b)$$

Putting it together, we have

$$\oint_{|z|=1} \log\left(\frac{z-a}{z-b}\right)\,dz =2\pi i (e^{i\phi_a}-e^{i\phi_b}+(b-a))$$

If we choose $\phi_a=\phi_b$, then the integral of interest becomes

$$\oint_{|z|=1} \log\left(\frac{z-a}{z-b}\right)\,dz =2\pi i (b-a) \tag1$$


Here, we note that if the branch cut is chosen to adjoin the branch points, then the integrand is analytic outside $|z|=1$. Therefore, we can evaluate the integral of interest using the Residue Theorem with the Residue at Infinity. To that end, we have

$$\begin{align} \oint_{|z|=1} \log\left(\frac{z-a}{z-b}\right)\,dz &=-2\pi i \text{Res}\left(-\frac{1}{z^2}\log\left(\frac{z^{-1}-a}{z^{-1}-b}\right),z=0\right)\\\\ &=2\pi i\lim_{z\to 0} \left(\frac{1}{z}\log\left(\frac{1-az}{1-bz}\right)\right)\\\\ &=2\pi i (b-a) \end{align}$$

recovering the result in $(1)$!

  • $\begingroup$ Residue at infinity method is very elegant, and I think what example should teach me. Thank you for answer! $\endgroup$
    – dimi
    Nov 17, 2015 at 8:38
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Nov 17, 2015 at 15:17
  • $\begingroup$ @hrodelbert Thank you! Much appreciative. $\endgroup$
    – Mark Viola
    Nov 17, 2015 at 15:18
  • $\begingroup$ Just a small question here. If $f$ is holomorphic in $\mathbb{D}$ how can we evaluate the following integral? $\int_{\partial \mathbb{D}} f(z) \log\left(\frac{z-a}{z-b}\right)dz $ ? $\endgroup$ Sep 29, 2017 at 11:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .