Answer
$$
\int_{0}^{1} r \ln r d r
$$
We note first that the given integral is improper because $ f(r)= r \ln r $ has the vertical asymptote $r=0 $. The infinite discontinuity occurs at the left endpoint of $[0,1] $,
The given improper integral is convergent and its value is
$$
\int_{0}^{1} r \ln r d r =-\frac{1}{4}.
$$
Work Step by Step
$$
\int_{0}^{1} r \ln r d r
$$
Observe that the given integral is improper because $ f(r)= r \ln r $ has the vertical asymptote $r=0 $. Since the infinite discontinuity occurs at the left
endpoint of $ [0,1] $, we use part (b) of Definition 3:
$$
\int_{0}^{1} r \ln r d r = \lim _{t \rightarrow 0 ^{+}} \int_{t}^{1} r \ln r d r
$$
Integrate by parts with
$$ u=\ln r , \quad dv= r dr \quad \quad \Rightarrow \quad du=\frac{dr}{r} , \quad v=\frac{1}{2} r^{2}, $$
so, we obtain
$$
\begin{split}
\int_{0}^{1} r \ln r d r & = \lim _{t \rightarrow 0 ^{+}} \int_{t}^{1} r \ln r d r \\
& = \lim _{t \rightarrow 0 ^{+}}[\frac{1}{2}r^{2}\ln r]_{t}^{1} -\frac{1}{2}\int_{t}^{1} r d r \\
& = \lim _{t \rightarrow 0 ^{+}}[\frac{1}{2}r^{2}\ln r]_{t}^{1} -\frac{1}{4}[ r^{2}]_{t}^{1} \\
& =\lim _{t \rightarrow 0 ^{+}}[-\frac{1}{2}t^{2}\ln t -\frac{1}{4} (1-t^{2})] \\
& =-\frac{1}{4},
\end{split}
$$
It follows that the integral
$$
\int_{0}^{1} r \ln r d r
$$
is convergent.
