$$\int_0^1 {x^n\log x\over(1+x)^2} \, dx$$
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.
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You just asked a very similar question; could you give some indication in your question about what you think the potential barriers to integrability are, or some details about the tests you tried? – Eric Tressler Jul 25 '13 at 13:24
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@EricTressler unfortunately they are not similar :(. but i will update the question with whatever i understand :) – Aman Mittal Jul 25 '13 at 13:26
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Hint: Near 0 we have $\frac{x^n \log x}{(1+x)^2} \sim x^n \log x$ (in the sense that their ratio tends to 1, as $x \to 0$) – user1337 Jul 25 '13 at 13:33
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How is this an improper integral? As $x\to0$, the value of the function approaches $0$, so there's no vertical asymptote and no need to take a limit as the lower bound of integration approaches $0$. The function is in fact bounded on the interval over which one integrates, and that interval is bounded. – Michael Hardy Jul 25 '13 at 13:48
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This is not an improper integral. See my answer below. – Michael Hardy Jul 25 '13 at 15:46
4 Answers
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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 $$-\frac{x^n\log x}{(1+x)^2}\sim_0-x^n\log x\geq\frac{1}{x^{-n}}\quad 0<x\leq\frac{1}{e}, \, n<0$$
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Hint: $$\lim_{x\to 0+} x^n\ln x=0,\quad n> 0$$ and $$\left|\int_0^1\ln x\,dx\right|<\infty.$$
TZakrevskiy
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@MhenniBenghorbal What do you mean? If the first limit, then it's true only for $n>0$. If you mean the author's question, then there're no indications for the possible values of $n$. – TZakrevskiy Jul 25 '13 at 14:27
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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?
Mhenni Benghorbal
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if $n$ is negative then $\frac{x^{n+1}\ln(x)}{n+1}\Big|_{x=0}^{x=1}$ isn't zero. – Jul 25 '13 at 13:55
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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.