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Let $a\leq b$ and $\Gamma$ be the unit circle (in the complex plane). I found that $\int_\Gamma \frac{\log(b-az)}{z} dz=2\pi i\log(b)$. It seems like Cauchy's integral formula has been used here.

I want to know why the conditions of Cauchy's integral formula are fulfilled. The conditions are that $\log(b-az)$ is holomorphic on some open set $U \in \mathbb{C}$ that contains the closed unit disc.

What can I choose for $U$? Maybe $U = \{z \in \mathbb{C} \mid Re(b-az) >0\}$? Because then I avoid the branch cut and $\log (b-az)$ is holomorphic on $U$. However, then if $a=b$, the closed (!) unit disc is not contained in $U$. And if I change the "$>$" in the definition of $U$ to a "$\geq$", then $\log(b-az)$ is not any longer defined on all of $U$.

What do I do wrong?

Mimimi
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    if $a=b$ apply the Cauchy formula with $\frac{\log(b-brz)}{z}, 0<r<1$ (now $\log(b-brz)$ is analytic on the closed unit disc - ie in a small open neighborhood of it) and use the Lebesgue dominated convergence theorem to take $r \to 1$; if $a<b$ there is no problem – Conrad May 03 '20 at 23:33
  • @Conrad Thank you! Can you maybe give a dominating function in order to use the DCT? – Mimimi May 04 '20 at 12:26
  • show that $|\frac{\log(b-brz)}{z}| \le C, |z-1| \ge .005, |z|=1, |\frac{\log(b-brz)}{z}| \le C_1\log |1-z|, |z-1| \le .005, |z|=1$ and the function $\log|1-z|$ is integrable on the unit circle (in polar coordinates it is about $\log \sin |\theta|$ so about $\log |\theta|$ near zero) since $\log |x|$ is integrable near zero – Conrad May 04 '20 at 13:36

1 Answers1

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Taking account $\ln(b-a z)=\ln b+\ln(1-a z/b)$ and using the usual parameterization of unit circle

$$\int_\Gamma \frac{\log(b-az)}{z} dz=i\,\int_0^{2\pi}\ln b+\ln(1-a\,e^{i\,u}/b)du=2\pi\,i\,\ln b$$

popi
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  • Why is $\int_0^{2 \pi} \ln(1-ae^{iu}/b) du =0$? – Mimimi May 04 '20 at 10:20
  • https://www.integral-calculator.com/#expr=log%281-ae%5E%28ix%29%2Fb%29 ...for example – popi May 04 '20 at 11:57
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    or as Conrad explain before: $a<b$ then $\ln(1+az/b)$ is holomorphic over an closed path, then integral becomes $0$. If $a=b$ you can take llimits. – popi May 04 '20 at 12:15