An integral solves by complex analysis

Question

I try to compute $$ I:=\int_{-\infty}^{\infty}\frac{\arctan^2x}{x^2}\text{d}x $$ by complex analysis.
This is my attempt:
Consider $$ f(z)=\frac{\arctan^2z}{z^2},\frac{\pi }{2}\ge\text{arg}(z)>-\frac{3\pi}{2} $$ and the contour $$\begin{aligned} &\gamma_1:\{t|t:-\infty\rightarrow\infty,t\in\mathbb{R}\}\\ &\gamma_2:\{\infty e^{i\theta}|\theta:0\to\frac{\pi}{2}\}\\ &\gamma_3:\{it|t:\infty\to 1\}\\ &\gamma_4:\{it|t:0\to\infty\}\\ &\gamma_5:\{\infty e^{i\theta}|\theta:\frac{\pi}{2}\to\pi \} \end{aligned}$$ Then $$ I+\int_{i\infty}^{i}\frac{(\arctan t+\pi i/2)^2-(\arctan t-3\pi i/2)^2}{t^2}\text{d}t=0 $$ But obviously note that the equality isn't true.Where are my mistakes?Any help will be appreciated.

Answer 1

$$ I \stackrel{\text{parity}}{=}2\int_{0}^{+\infty}\frac{\arctan^2(x)}{x^2}\,dx \stackrel{x\mapsto \tan\theta}{=}2\int_{0}^{\pi/2}\frac{\theta^2\,d\theta}{\sin^2\theta}\stackrel{\text{IBP}}{=} 4\int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta\stackrel{\text{IBP}}{=}-4\int_{0}^{\pi/2}\log(\sin\theta)\,d\theta$$ and the last integral should be familiar.