Proving that the improper integral is divergent. $\int_0 ^2 x^2 \ln x\,dx$ The task is "Evaluate the following improper integral or prove that it
diverges" 
$$
\int_0 ^2 x^2 \ln x\,dx
$$
I noticed that we can't evaluate it from $0$ to $2$, so I need to prove that it is divergent, but I can't do it in any way. Please help me.
 A: To prove the divergence of an improper integral, you need to show that the corresponding limit diverges.
Written out, this improper integral is defined as $\lim_{a \to 0^+} \int_a^2 x^2 \ln(x)\,dx$.  Evaluate the integral from $a$ to $2$ as a function of $a$, and then to show divergence you see that this function diverges as $a$ approaches 0 from the right.
A: \begin{align}
& \int(\ln x)\Big( x^2 \, dx\Big) = \int u\,dv=uv - \int v\,du = \frac{x^3}3\ln x - \int \frac{x^3} 3\cdot\frac 1 x \, dx \\[8pt]
= {} & \frac{x^3}3\ln x - \frac 1 3 \int x^2 \, dx \\[8pt]
= {} & \frac{x^3}3\ln x - \frac{x^3} 9 + C.
\end{align}
As $x\downarrow0$, the limit of the first term can be found by L'Hopital's rule applied to $\dfrac{\ln x}{1/x^3}$, and it is $0$.
A: Hint: $\lim_{x\to0}x^n\ln x=0$, for $n>0$, so this isn't really an improper integral.

 $\displaystyle\int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\int\frac{x^2}{3}\,dx$

A: Your integral is improper because $\log x$ is not defined at $x = 0$. To deal with this, we will check if we can integrate over the interval $[\epsilon, 2]$ for all (small) $\epsilon > 0$. That is, we are want to determine whether
$$
I = \lim_{\epsilon \to 0^+} \int_\epsilon ^2 x^2\log x\,dx
$$
is finite.
Integrating by parts,
$$
I = \lim_{\epsilon\to0^+} \frac{1}{3} x^3\log x \Big|_\epsilon^2 - \int_{\epsilon}^2 \frac{x^2}{3}\,dx.
$$
It's clear that last integral is finite so we need only check that 
$$
\lim_{x\to0^+} x^3\log x
$$
exists and is finite. 
