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$$\int_{0}^{\infty} \frac{\ln(x^2+1)}{x^\alpha} dx$$

Now I am a bit confused here, I know that for a very small integer $(x)$, $\ln(x^2+1)$ acts very similar to $x^2$, so I can solve it for $(0,1)$

but checking from $1$ to infinity, I know that $\ln(x^2+1)$ acts like $x^\epsilon$

but I dont really know how to go from there, Ive tried convergence tests but didnt got a defenite limit. treid using $$f(x)=x^\epsilon / x^\alpha $$ but the limit of the quotient is 0.

the answer is $1 < \alpha < 3$

if anyone knows how to solve it :) thank you

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2 Answers

up vote 1 down vote accepted

The criterion(comparison of integrals) you need can be found in the next link: http://www.sosmath.com/calculus/improper/testconv/testconv.html

Since the integral has problems at both ends, you have to split it in two:

  • $\int_0^1 \frac{\ln(x^2+1)}{x^\alpha}dx$ is convergent if and only if $\int_0^1 \frac{1}{x^{\alpha-2}}dx$ is convergent (their ratio has limit 1 as $x \to 0$). The latter one is convergent if and only if $\alpha-2<1 \iff \alpha<3$.

  • $\int_1^\infty \frac{\ln(x^2+1)}{x^\alpha}dx$ is convergent then $\int_1^\infty \frac{1}{x^\alpha}dx$ is convergent, which implies that $\alpha>1$. If $\alpha>1$ then the integral converges, since the logarithm grows slower than any polynomial, so asymptotically, there is a point from where $\ln(1+x^2) \leq x^\beta$ such that $\alpha-\beta>1$.

The details are all in the given link.

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thanks bro!! very helpfull –  YNWA Jan 31 '12 at 14:14
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At $0+$, $\log(1 + x^2) \sim x^2$, so $${\log(x^2 + 1)\over x^\alpha} \sim {x^2\over x^\alpha} = {1\over x^{\alpha - 2}}.$$ The last integrand is integrable at 0 iff $\alpha - 2 < 1$, or if $\alpha < 3.$ Now let's look at what happens at $\infty$. We have $${\log(x^2 + 1)\over x^\alpha}\ge {1\over x^\alpha,}$$ the right-hand side integrates at infinity iff $\alpha > 1$.

This last bit does not seal the deal, but it tells you that $\alpha > 1$ is the best that can be hoped for. However, you know that for any $\epsilon > 0$, $\log(x^2 + 1) < x^\epsilon$ for sufficiently large $n$. You can use this to seal the deal.

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thank you! very helpfull –  YNWA Jan 31 '12 at 14:14
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