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I want to find the integral: $$\int_0^{\pi}\frac{1}{\ln{(e^{ix}\sin{x})}}dx$$

From $\frac{1}{\ln{(e^{ix}\sin{x})}} = \frac{\ln(\sin{x})}{x^2+\ln^2(\sin{x})}-\frac{x}{x^2+\ln^2(\sin{x})}i$,

therefore $$\int_0^{\pi}\frac{1}{\ln{(e^{ix}\sin{x})}}dx= \int_0^{\pi}\frac{\ln(\sin{x})}{x^2+\ln^2(\sin{x})}dx-i\int_0^{\pi}\frac{x}{x^2+\ln^2(\sin{x})}dx$$.

But it is hard to evaluate these integral.

Are there another method? Thank you.

rack
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  • Don't know if it helps here, but it is often helpful to try to recognize an integrand as, say, the real or imaginary part of a function of a complex variable. For example, you can integrate $\int e^{ax} \cos bx ; dx$ very easily by writing it as $\operatorname{Re}\int e^{(a+bi)x}; dx$ – MPW Dec 08 '16 at 15:12
  • Here is a visually similar integral. However, the methods used might be inapplicable here. – teadawg1337 Dec 08 '16 at 15:44
  • Thanks for information. now, I have changed interval of integral to $(0,\pi)$. – rack Dec 09 '16 at 11:50

1 Answers1

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Notice that the integral can be written as

$$ I = \int_{0}^{\pi} \frac{dx}{f(e^{2ix})} = \frac{1}{2} \int_{0}^{2\pi} \frac{dx}{f(e^{ix})}, $$

where $f(z) = \log\left(\frac{i}{2}(1-z)\right)$. Using the standard branch cut of the complex logarithm, $f$ has the branch cut $\{ 1 - it : t \geq 0 \}$ and the unique zero at $1+2i$. Thus $\frac{1}{f(z)}$ is a well-defined holomorphic function on $\Bbb{D} = \{z \in \Bbb{C} : |z| < 1\}$ and by the mean-value property we have

$$ \frac{1}{2} \int_{0}^{2\pi} \frac{dx}{f(e^{ix})} = \frac{\pi}{f(0)} = \frac{\pi}{\log(i/2)}. $$

Sangchul Lee
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  • do you mean Gauss's mean value theorem? But $\frac{1}{z}$ is not analytic at $z=1$, Why you can use the theorem – Bless Dec 17 '16 at 15:18
  • @Bless, Your objection is correct. To be honest, this follows by the mean value property applied to circles of smaller radii followed by a limiting procedure. – Sangchul Lee Dec 18 '16 at 03:46
  • So we can reduce condition of the theorem? – Bless Dec 18 '16 at 11:00