residue theorem with logarithmic function

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?

• 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? Nov 16 '15 at 16:01
• thank Dr MV, I try and understand connection of Cauchy Integral Theorem. Please see the changes I make
– dimi
Nov 16 '15 at 16:43

METHOD 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

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

METHOD 2:

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 $(1)$!

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