Contour Integration

Put $x=\tan y$ , the integral turns , $\large \displaystyle I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\sec^2 y) dy = -4 \int_{0}^{\frac{\pi}{2}} \ln (\cos y) dy$

Easy result , $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\cos x) dx = -{\frac{\pi}{2}}\ln 2$

So , $\displaystyle I = \color{#3D99F6}{2}\ln \color{#3D99F6}{2}$ .

\newcommand{\LIM}[2]{\lim\limits_{#1\rightarrow #2}} \newcommand{\PLOG}[1]{\text{log}\PAREN{#1}} \newcommand{\diff}{\mathop{}\!\mathrm{d}} \newcommand{\PAREN}[1]{\left(#1\right)} \text{This is for anyone who's interested in the contour integration method.}\\ \\ \text{The branch cut is \$\forall z = ib, -ib, b>1\$.}\\ \text{Since there are no singularities in each closed contour, and the integrands are analytic,}\\ \oint_{C_1} \frac{\PLOG{1-iz}}{1+z^2}\diff z + \oint_{C_2} \frac{\PLOG{1+iz}}{1+z^2}\diff z = 0 \\ \text{For the first integral, by changing the direction of integration in the positive direction,}\\ \oint_{C_1} \frac{\PLOG{1-iz}}{1+z^2}\diff z = \int_{-R}^R + \int_{|z| = R_+} + \int_{|z| = R_-} - \int_{|z|=h}\\ \text{where \$R\_+\$ and \$R\_-\$ represent the semicircle contour with radius \$R\$ on the positive real side and negative real side, }\\ \text{respectively, and \$h\$ is any small arbitrary real number.}\\ \text{Notice that the vertical contours converge to the same contour but in opposite directions,}\\ \text{so the sum of integral on both contours converge to zero.}\\ \text{As \$|z| = R\to \infty\$, the integrand approaches 0 since the order of \$\PLOG{z}\$ is less than \$z\$ and \$z/\PAREN{1+z^2}\$ approaches 0. Hence,}\\ \LIM{R}{\infty} \PAREN{\int_{-R}^R + \int_{|z| = R_+} + \int_{|z| = R_-} - \int_{|z|=h}} = \LIM{R}{\infty} \PAREN{\int_{-R}^R} - \int_{|z|=h}\\ \text{Let \$\epsilon\$ be the distance between the two vertical contours}\\ \text{As \$\epsilon \to 0\$, the contour where \$|z|=h\$ converges to a complete circle. Since the integrand is singular at the center \$i\$, the integral}\\ \oint_{|z|=h}\frac{\PLOG{1-iz}}{1+z^2}\diff z \to 2\pi i Res\PAREN{\frac{\PLOG{1-iz}}{1+z^2}, i} = 2\pi i \cdot \frac{\log(2)}{2i} = \pi \log(2)\\ \text{Therefore,}\\ \oint_{C_1} \frac{\PLOG{1-iz}}{1+z^2}\diff z = \LIM{R}{\infty} \PAREN{\int_{-R}^R \frac{\PLOG{1-ix}}{1+x^2}\diff x} - \pi \log(2)\\ \text{Similarly,}\\ \oint_{C_2} \frac{\PLOG{1+iz}}{1+z^2}\diff z = \LIM{R}{\infty} \PAREN{\int_{-R}^R \frac{\PLOG{1+ix}}{1+x^2}\diff x} - \pi \log(2)\\ \text{Substituting to the first equation,}\\ \LIM{R}{\infty} \PAREN{\int_{-R}^R \frac{\PLOG{1-ix}}{1+x^2}\diff x} - \pi \log(2) + \LIM{R}{\infty} \PAREN{\int_{-R}^R \frac{\PLOG{1+ix}}{1+x^2}\diff x} - \pi \log(2) = 0\\ \LIM{R}{\infty} \PAREN{\int_{-R}^R \frac{\PLOG{1-ix} +\PLOG{1+ix}}{1+x^2}\diff x} = 2\pi \log(2)\\ \int_{-\infty}^\infty \frac{\log(1+x^2)}{1+x^2}\diff x = 2\pi \log(2)\\