In first post of $\pi$ , we see
\[ \pi=\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx \]
If we shorten the interval of integration from $(-\infty,\infty)$ to $[0,1]$ ,then
\[ \int_0^1\frac{1}{1+x^2}dx=\frac{\pi}{4} . \]
Because, put $x=\tan \theta$ . We know
\[ \frac{dx}{d\theta}=\frac{1}{\cos^2\theta} \]
and if $x=0$ , then $\theta=0$, and if $x\rightarrow 1$ , then $\theta\rightarrow \frac{\pi}{4}$ .
Therefore,
\[ \int_0^1\frac{1}{1+x^2}dx=\int_0^{\pi/4}\frac{1}{1+\tan^2\theta}\frac{1}{\cos^2\theta}d\theta=\frac{\pi}{4} . \]
We will get easily
\[ \int_{-1}^1\frac{1}{1+x^2}dx=\frac{\pi}{2} . \]
Thus,
\[ \int_1^{\infty}\frac{1}{1+x^2}dx=\int_{-\infty}^{-1}\frac{1}{1+x^2}dx=\frac{\pi}{4} . \]
