複素積分2

今回はこの積分botさんの積分について解説します。
$\displaystyle\int_0^\frac{\pi}{2}\frac{\log\cos x}{x^2+\log^2\cos x}dx=\frac{\pi}{2}\Big(1-\frac{1}{\log2}\Big)$

$\displaystyle\int_0^\frac{\pi}{2}\frac{\log\cos x}{x^2+\log^2\cos x}dx$
$=\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2}\Big(\frac{1}{\log\cos x-ix}+\frac{1}{\log\cos x+ix}\Big)dx$
$=\displaystyle\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{1}{2\big(\log\cos x+ix\big)}dx$
$=\displaystyle\frac{1}{4}\int_{-\pi}^\pi\frac{1}{\log\cos\frac{\theta}{2}+i\frac{\theta}{2}}d\theta$
$=\displaystyle\frac{1}{4}\oint_{C:z=e^{i\theta}}\frac{1}{\log\frac{\sqrt{z}+{\sqrt{z}}^{-1}}{2}+\log\sqrt{z}}\frac{1}{iz}dz$
$=\displaystyle\frac{1}{4}\oint_C\frac{1}{iz\log\frac{z+1}{2}}dz$
$=\displaystyle\frac{\pi i}{2}\Big(\lim_{z \to 0}\frac{1}{i\log\frac{z+1}{2}}+\lim_{z \to 1}\frac{z-1}{iz\log\frac{z+1}{2}}\Big)$
$=\displaystyle\frac{\pi}{2}\Big(1-\frac{1}{\log2}\Big)$