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。
${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\log \cos x}{x^2+{\log }^2\cos x}dx=\frac{\pi }{2}(1-\frac{1}{\log 2})}$

${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\log \cos x}{x^2+{\log }^2\cos x}dx}$
$={\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{1}{2}(\frac{1}{\log \cos x-ix}+\frac{1}{\log \cos x+ix})dx}$
$={\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2(\log \cos x+ix)}dx}$
$={\displaystyle \frac{1}{4}{\int }_{-\pi }^{\pi }\frac{1}{\log \cos \frac{\theta }{2}+i\frac{\theta }{2}}d\theta }$
$={\displaystyle \frac{1}{4}{\oint }_{C:z=e^{i\theta }}\frac{1}{\log \frac{\sqrt{z}+{\sqrt{z}}^{-1}}{2}+\log \sqrt{z}}\frac{1}{iz}dz}$
$={\displaystyle \frac{1}{4}{\oint }_C\frac{1}{iz\log \frac{z+1}{2}}dz}$
$={\displaystyle \frac{\pi i}{2}(\underset{z\to 0}{lim}\frac{1}{i\log \frac{z+1}{2}}+\underset{z\to 1}{lim}\frac{z-1}{iz\log \frac{z+1}{2}})}$
$={\displaystyle \frac{\pi }{2}(1-\frac{1}{\log 2})}$