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$$\int_0^1{{x^n\log x}\over {(1+x)^2}} $$ My Attempt :

The integrand is unbounded for $x=0$. so,

Using $\mu-Test$ , the integral will be convergent if $0<\mu<1$ such than$$\lim_{x -> 0} x^{\mu+n}{{\log x}\over {1+x}}$$ exists

The limit converges for $\mu+n>0$ or $\mu>-n$ This is true for all $n>0$ Hence the given integral must converge for all $n>0$

The answer however given is that the integral is convergent for $n>-1$, How do i deduce that, ans is there another simpler work around ?

Asaf Karagila
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Aman Mittal
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1 Answers1

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If $n>0$ the integrand is bounded at zero, as you can see using L'Hospital rule.

If $-1<\mu<n<0$, then

$$\dfrac{|x^n\ln x|}{1+x^2} < x^\mu$$

for $x$ sufficiently close to 0. Since the integral of $x^\mu$ converges, your integral converges too.

bartgol
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