Answer
Convergent $\;,\;$ $\large\frac{-1}{4}$
Work Step by Step
Let \[I=\int_{0}^{1}r\ln r \,dr\]
Since 0 is point of infinite discontinuity of integrand $r\ln r $
\[I=\int_{0}^{1}r\ln r \,dr=\lim_{t\rightarrow 0^+}\int_{t}^{1}r\ln r \,dr\]
\[I=\lim_{t\rightarrow 0^+}\int_{t}^{1}r\ln r \,dr\;\;\;\ldots (1)\]
Let \[I_1=\int r\ln r \,dr\]
Using integration by parts
\[I_1=\ln r\int rdr-\int\left((\ln r)'\int rdr\right)dr\]
\[I_1=\frac{r^2}{2}\ln r-\frac{1}{2}\int\frac{r^2}{r}dr\]
\[I_1=\frac{r^2}{2}\ln r-\frac{r^2}{4}\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow 0^+}\left[\frac{r^2}{2}\ln r-\frac{r^2}{4}\right]_{t}^{1}\]
\[I=\lim_{t\rightarrow 0^+}\left[0-\frac{1}{4}-\frac{t^2}{2}\ln t+\frac{t^2}{4}\right]\]
\[I=-\frac{1}{4}-\frac{1}{2}\lim_{t\rightarrow 0^+}\left(\frac{\ln t}{t^{-2}}\right) \;\;\left(\frac{\infty}{\infty} form\right)\]
Using L' Hopital's rule
\[I=-\frac{1}{4}-\frac{1}{2}\lim_{t\rightarrow 0^+}\left(\frac{\frac{1}{t}}{-2t^{-3}}\right)\]
\[I=-\frac{1}{4}-\frac{1}{2}\lim_{t\rightarrow 0^+}\left(\frac{-t^2}{2}\right)\]
\[I=\frac{-1}{4}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=\frac{-1}{4}$.
