test for convergence of improper integral1

Question

$$\int_0^1 {x^n\log x\over(1+x)^2} \, dx$$

I tried something using practical test, but not much progress.

I see that the integral becomes improper for $x=0$, May be we need to apply the Practical comparison test ? I need some hint please.

Answer 1

It's simple to see that the integral $$\int_0^1|\log x|\, dx$$ is convergent so conclude with the inequalities $$\frac{x^n|\log x|}{(1+x)^2}\leq |\log x|,\quad 0<x\leq1, \, n\geq0$$

and $$-\frac{x^n\log x}{(1+x)^2}\sim_0-x^n\log x\geq\frac{1}{x^{-n}}\quad 0<x\leq\frac{1}{e}, \, n<0$$

Answer 2

Hint: $$\lim_{x\to 0+} x^n\ln x=0,\quad n> 0$$ and $$\left|\int_0^1\ln x\,dx\right|<\infty.$$

Answer 3

A related technique. Note that,

$$ x\sim 0 \implies \frac{x^n\ln(x)}{(1+x^2)}\sim x^n\ln(x).$$

Now, the integral

$$ \int_{0}^{1}x^n\ln(x) dx = \frac{x^{n+1}\ln(x)}{n+1}\Big|_{x=0}^{x=1}-\frac{1}{n+1}\int_{0}^{1} x^{n+1}dx=-\frac{1}{n+1}\int_{0}^{1} x^{n+1}dx. $$

Now, do you know for what $n$ the last integral converge?

Answer 4

This is in no sense an improper integral.

$$ x\log x = \frac{\log x}{1/x}, $$ and by L'Hopital's rule this goes to $0$ as $x\downarrow0$. Therefore $x^n\log x\to0$ as $x\downarrow0$. The denominator stays far from $0$ over the whole interval.

So the function is bounded, and it's integrated over a bounded interval, so there is no "impropriety". No test for convergence is needed.