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There exists still another way to calculate this integral. Namely, let (the principal values are assumed) $I_1=\int\limits_0^\infty \left (\frac{1}{x^2}-\frac{\cot{x}}{x}\right )dx$ and $I_2=\int\limits_0^\infty \frac{\tan{x}}{x}dx$. Then $$I_1+I_2=\int\limits_0^\infty \left (\frac{1}{x^2}-2\frac{\cot{(2x)}}{x} \right )dx=2I_1.$$ This shows that $I_1=I_2$, so let's concentrate on $I_2$. That follows is somewhat elaborated calculations of Daryl McCullough from http://mathforum.org/kb/message.jspa?messageID=5667982 We have ($\epsilon\to 0$ limit is assumed) $$I_2=\sum\limits_{n=0}^\infty\left [ \int\limits_{n\pi}^{n\pi+\pi/2-\epsilon}\frac{\tan{x}}{x}dx+\int\limits_{n\pi+\pi/2+\epsilon}^{(n+1)\pi}\frac{\tan{x}}{x}dx \right ]=\sum\limits_{n=0}^\infty\left [ \int\limits_0^{\pi/2-\epsilon}\frac{\tan{(x+n\pi)}}{x+n\pi}dx+\int\limits_{\pi/2+\epsilon}^{\pi}\frac{\tan{(x+n\pi)}}{x+n\pi}dx \right ].$$ But for integer $n$, $\tan{(x+n\pi)}=\tan{x}$, and after the substitution $y=\pi-x$ in the second integral, we get $$I_2=\sum\limits_{n=0}^\infty\int\limits_0^{\pi/2-\epsilon}\tan{x}\left[\frac{1}{x+n\pi}+\frac{1}{x-(n+1)\pi}\right ] dx=\int\limits_0^{\pi/2-\epsilon}\frac{\tan{x}}{x}+$$ $$\sum\limits_{n=1}^\infty\int\limits_0^{\pi/2-\epsilon}\tan{x}\left[\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right ] dx.$$ But $\frac{1}{x+n\pi}+\frac{1}{x-n\pi}=\frac{2x}{x^2-n^2\pi^2}$ are all negative for $n=1,2,\ldots$ and $x\in(0,\pi/2)$. Therefore we can use Tonelli's theorem to justify changing the order of the summation and integration in the second term. Hence $$I_2=\int\limits_0^{\pi/2-\epsilon}\tan{x}\left [\frac{1}{x}+\sum\limits_{n=1}^\infty\left (\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)\right ]dx.$$ It remains to recall the Euler's formula for $\cot{x}$: $$\cot(x)=\frac{1}{x}+\sum\limits_{n=1}^\infty\left (\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right),$$ and the result $I_2=\frac{\pi}{2}$ follows immediately (because $\tan{x}\cot{x}=1$).