Example 2.36. $I=\mathrm{V}$.p. $\int_{-\infty}^{\infty} \operatorname{arctg} x d x$.

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Example 2.36. $I=\mathrm{V}$.p. $\int_{-\infty}^{\infty} \arctan x d x$.
Solution.

$$
\begin{gathered}
I=\lim _{a \rightarrow \infty} \int_{-a}^{a} \operatorname{arctg} x d x=\left|\begin{array}{l}
\left.u=\operatorname{arctg} x . d u=\frac{d x}{1+x^{2}} \right\rvert\,= \\
d v=x, v=x
\end{array}\right|= \\
=\lim _{a \rightarrow \infty}\left(\left.x \operatorname{arctg} x\right|_{-a} ^{a}-\int_{-a}^{a} \frac{x d x}{1+x^{2}}\right)= \\
=\lim _{a \rightarrow \infty}(a \operatorname{arctg} a-(-a) \operatorname{arctg}(-a)- \\
\left.-0.5 \ln \left(1+a^{2}\right)+0.5 \ln \left(1+(-a)^{2}\right)\right)=0
\end{gathered}
$$