Does $\int_0^1 \frac{x \ln x}{1+ x^2}dx$ converge?

Question

$\int_0^1 \frac{x \ln x}{1+ x^2}dx$ converges?

Kinda stuck doing this problem. I just need a hint on what to start with. I know that it is an improper integral and I have to use limits but I need to evaluate the integral first.

Thanks in advance for any help.

Answer 1

Yes, it converges. Use these facts:

1. Note that:

$$\frac{|x \ln (x)|}{1+x^2}<|x \ln x|$$

2. Prove that:

$$\lim_{x \to 0}x \ln(x)=0$$

Answer 2

If you let $t=\frac{1}{x}$, so $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$, you get

$\displaystyle\int_0^1\frac{x\ln x}{1+x^2}\;dx=-\int_1^{\infty}\frac{\ln t}{t(t^2+1)}\;dt$.

Then you can use that $\ln t<t\implies\frac{\ln t}{t(t^2+1)}<\frac{1}{t^2+1}<\frac{1}{t^2}$

and the Comparison Test to conclude that the integral converges.