Question Number 181450 by Socracious last updated on 25/Nov/22
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\infty} {\overset{-\infty} {\int}}\boldsymbol{\mathrm{lnxdx}} \\ $$
Commented by universe last updated on 26/Nov/22
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\infty} {\overset{-\infty} {\int}}\boldsymbol{\mathrm{lnxdx}} \\ $$
Answered by Frix last updated on 25/Nov/22
$$\int\mathrm{ln}\:{x}\:{dx}={x}\mathrm{ln}\:{x}\:−{x} \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{given}\:\mathrm{borders}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{converge}. \\ $$
Answered by mahdipoor last updated on 25/Nov/22
![=[xln∣x∣−x]_(+∞) ^(−∞) =lim_(b→∞) [xln∣x∣−x]_(+b) ^(−b) = lim_(b→∞) [(−blnb+b)−(blnb−b)]=[2b−2blnb] =−∞](https://www.tinkutara.com/question/Q181460.png)
$$=\left[{xln}\mid{x}\mid−{x}\right]_{+\infty} ^{−\infty} =\underset{{b}\rightarrow\infty} {\mathrm{lim}}\left[{xln}\mid{x}\mid−{x}\right]_{+{b}} ^{−{b}} = \\ $$$$\underset{{b}\rightarrow\infty} {\mathrm{lim}}\left[\left(−{blnb}+{b}\right)−\left({blnb}−{b}\right)\right]=\left[\mathrm{2}{b}−\mathrm{2}{blnb}\right] \\ $$$$=−\infty \\ $$