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First, I know that the identity $\log(z_1z_2)=\log(z_1)+\log(z_2)$ in the title does not work in general.

My question aroused from Showing that $\int_0^1 \log(\sin \pi x)dx=-\log2$ , a problem of Complex Analysis, E. M. Stein. The first line of the best answer of this question is

Consider $$f(z)=\log(1−e^{2πzi})=\log(e^{πzi}(e^{−πzi}−e^{πzi}))=\log(−2i)+πzi+\log(\sin(πz))\\...$$

but I cannot go through.

How can I justify this process?

Any help will be appreciated. Thanks.

Jinmu You
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1 Answers1

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First, notice that we will have problems around $x=0$ and $x=1$ for our integral, so the problem we are trying to solve is the improper integral $$\int_0^1 \log(\sin(\pi x))dx=\lim_{c\to 0^+}\lim_{d\to 1^-} \int_c^d \log(\sin(\pi x))dx.$$ Since we are dealing with a complex logarithm function, you are going to have to define a branch cut, so restrict $\frac{-5\pi}{4}<\arg(z)<\frac{3\pi}{4},$ and $$\log(z)=\log(|z|)+i\arg(z)$$ will be well defined on the domain $\mathbb{C}\sim \{z\in \mathbb{C}|\frac{-5\pi}{4}<\arg(z)<\frac{3\pi}{4}\}.$


Now that we have defined our domain, we will have the property $\log(z_1z_2)=\log(z_1)+\log(z_2)+n\pi$ for $n\in N$ for all $z_1,z_2$ in our domain.
Thus we will see $$\log(\sin(\pi z))=\log\Big(\frac{e^{i\pi z}-e^{-i\pi z}}{2i}\Big)=\log\Big(-e^{i\pi z}+e^{-i\pi z}\Big)-\log(-2i)+n\pi i$$ Now to find what value of $n$ we introduce, consider that $\arg(\frac{e^{i\pi z}-e^{-i\pi z}}{2i})=0$ and $$\arg(e^{i\pi z}-e^{-i\pi z})=\frac{\pi}{2},$$ and $$\arg(-2i)=-\frac{\pi}{2},$$ and thus $$\arg(e^{i\pi z}-e^{-i\pi z})-\arg(-2i)=\pi,$$ so $n=1$. Continuing, we have: $$\log\Big(-e^{i\pi z}+e^{-i\pi z}\Big)-\log(-2i)+\pi i=\log\Big(e^{-i\pi z}(-e^{2i\pi z}+1)\Big)-\log(-2i)+\pi i$$ $$=\log(e^{-i\pi z})+\log(-e^{2i\pi z}+1)-\log(-2i)+(n+1) \pi i.$$ Considering what happens to the argument of these two functions, at $z=.5$, $\arg(e^{-i\pi z})=-\frac{\pi}{2}$ and $\arg(-e^{2i\pi z}+1)=0.$ So the thus, $$\arg(-e^{2i\pi z}+1)+ \arg(e^{-i\pi z})=-\frac{\pi}{2},$$ which is a full $-\pi$ away from what we get from $\arg(e^{i\pi z}-e^{-i\pi z})$ and as $n$ is constant in our domain, $n=-1$. Now, $$\log(e^{-i\pi z})+\log(-e^{2i\pi z}+1)-\log(-2i)=-i\pi z+\log(-e^{2i\pi z}+1)-\log(-2i).$$ Thus we have shown $$\log(\sin(\pi z))=-i\pi z+\log(-e^{2i\pi z}+1)-\log(-2i).$$

Abhirup Basu
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  • Thanks for kind explanation. But I'm sorry that I still cannot go through. For your branch cut, let $z_1=z_2=-1\in\mathbb C-{z\in\mathbb C|Arg(z)=3\pi/4}$, then $\log(z_1)=\log(z_2)=-i\pi$ but $\log(z_1z_2)=\log(1)=0\neq -2i\pi. $I think that when the range of argument is restricted, then the equality cannot be true. Where am I missing? – Jinmu You Apr 23 '17 at 04:52
  • I improved the proof to address the concerns, since $\log(z_1z_2)=\log(z_1)+\log(z_2)+n\pi i$. –  Apr 23 '17 at 06:10
  • And I fixed one last portion, so the argument should be correct now. –  Apr 23 '17 at 15:30
  • Thanks a lot! There is a trivial error exists on your answer : we have to consider $\arg(-e^{i\pi z}+e^{-i\pi z})$, not $\arg(e^{i\pi z}-e^{-i\pi z})$. Then the two $n$s are zero and still the argument is correct. – Jinmu You Apr 24 '17 at 10:04