Example 2.45. $I=\mathrm{V} . \mathrm{p} . \int_{1 / e}^{e} \frac{d x}{x \ln x}$. Calculate the improper integral of the second kind in the sense of the principal value.
Solution. According to formula (2.18)

$$
\begin{gathered}
I=\lim _{a \rightarrow+0}\left(\int_{1 / e}^{1-a} \frac{d(\ln x)}{\ln x}+\int_{1+a}^{e} \frac{d(\ln x)}{\ln x}\right)= \\
=\lim _{a \rightarrow+0}\left(\ln \left|\ln x\left\|_{1 / e}^{1-a}+\ln \mid \ln x\right\|_{1+a}^{e}\right)=\right. \\
=\lim _{a \rightarrow+0}(\ln |\ln (1-a)|-\ln |\ln 1-\ln e|+\ln |\ln e|-\ln |\ln (1+a)|)= \\
=\lim _{a \rightarrow+0} \ln \left|\frac{\ln (1-a)}{\ln (1+a)}\right|=\ln \left|\lim _{a \rightarrow+0} \frac{(\ln (1-a))^{\prime}}{(\ln (1+a))^{\prime}}\right|= \\
=\ln \left|\lim _{a \rightarrow+0}\left(-\frac{1+a}{1-a}\right)\right|=\ln 1=0 .
\end{gathered}
$$