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I was reading @Sangchul Lee 's solution to $\displaystyle\int_{0}^{\pi/2} \ln \left(x^{2} + (\ln\cos x)^2 \right) \,\mathrm dx$ on his blog and he makes a statement/an assumption that I don't know how he got to:

Now we know that, choosing the standard branch cut of the logarithm, the function $\log⁡\log⁡ z$ is analytic on the upper half-plane (Indeed, $\log⁡\log⁡ z$ is analytic on $\mathbb C\backslash(−\infty,1]$) and $\color{red}{\log\log(t+0^+i)=\log⁡|\log⁡t|+i\pi}$ for $\color{red}{0<t<1}$.

For computations concerning branch points and branch cuts I'm not sure whether we need to worry about both logs or only the exterior one.

The only intuition on how he could've concluded that would be $$\begin{aligned} \log\log(t+i0^+)&=\log|\log(t+i0^+)|+i\arg(\log(t+i0^+))\\ &=\log|\log(t+i0^+)|+i\arg(\log|t+i0^+|+i\arg(t+i0^+))\\ &=\log|\log(t+i0^+)|+i\arg(\underbrace{\log|t+i0^+|}_{<0,\enspace 0<t<1}+i0^+) \end{aligned}$$ and since the argument of a complex number whose $\mathfrak{Re}<0$ and $\mathfrak{Im}>0$ is $\pi$, and NOT $-\pi$ (if not $\mathfrak{Im}<0$), then, we get $$\log\log(t+i0^+)=\log|\log(t+i0^+)|+i\pi=\log|\log(t)|+i\pi$$ Is this justification correct?

Hug de Roda
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1 Answers1

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This is using the principal branch of the logarithm, where, as stated, the branch cut is the interval $(-\infty, 0]$, and $-\pi < \text{Im}(\log z) < \pi$ for $z$ off the branch cut. Both logs must be considered.

When $0 < t < 1$, $\arg(t+i0+) = 0+$, i.e. as $\epsilon \to 0+$, $\arg(t + i \epsilon) \to 0+$, and $\log(t + i 0+) = \log(t) + i0+$. Since $\log(t) < 0$, $\arg(\log(t+i0+)) \to \pi$, and so $$\log \log(t+i0+) = \log |\log(t)| + i \pi $$

Robert Israel
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