Proving that the improper integral is divergent. $\int_0 ^2 x^2 \ln x\,dx$

Question

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.

Answer 1

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.

Answer 2

Hint: $\lim_{x\to0}x^n\ln x=0$, for $n>0$, so this isn't really an improper integral.

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

Answer 3

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.

Answer 4

\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 $0$.