For what value of $p$ does $\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^p}\,dx$ converge?

This integral converges $\iff$ both $\int_{0}^{1} \frac{\ln(1+x^2)}{x^p}\,dx$ and $\int_{1}^{\infty} \frac{\ln(1+x^2)}{x^p}\,dx$ converge.

Using $1<\ln(1+x^2)$ for all $x>x_0$, I get that $\int_{1}^{\infty} \frac{\ln(1+x^2)}{x^p}\,dx$ diverges for $p\leq1$

Using $\ln(1+x^2)<1+x^2$ it is possible to see that $\int_{1}^{\infty} \frac{\ln(1+x^2)}{x^p}\,dx$ converges for $p>3$

From this point I am stuck.

We were not taught inequalities regarding $\ln(x)$ except $\ln(x)<x$, so this is the only comparison allowed regarding to it (and comparing it to numbers of course).


Integrating $\int_{}^{} \frac{\ln(1+x^2)}{x^2}\,dx$ I see that $\int_{1}^{\infty} \frac{\ln(1+x^2)}{x^p}\,dx$ converges for $p\geq2$, and $\int_{0}^{1} \frac{\ln(1+x^2)}{x^p}\,dx$ converges for $p\leq2$

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    $\begingroup$ Please add the slashes for your $\ln(\cdot)$ natural logarithm. $\endgroup$ – Lee David Chung Lin May 14 '19 at 13:01

Note that $$\forall \alpha>0\, \exists M>0\, \forall x>M: \ln(1+x^2) < x^\alpha $$ You can prove it by considering the limit $\lim_{x\rightarrow\infty}\frac{\ln(1+x^2)}{x^\alpha}$.

So as long as there exists $\alpha>0 $ such that integral $\int_1^\infty\frac{x^\alpha}{x^p}dx$ converges, then the integral $\int_1^\infty\frac{\log(1+x^2)}{x^p}dx$ also converges. That will allow you to prove that this integral is convergent for $p>1$.

For the integral $\int_0^1$ note that $$\lim_{x\rightarrow 0} \frac{\ln(1+x^2)}{x^2} = 1 $$ so the integral $\int_0^1\frac{\log(1+x^2)}{x^p}dx = \int_0^1\frac{\log(1+x^2)}{x^2} x^{2-p}dx $ is convergent iff the integral $\int_0^1 x^{2-p} dx$ is convergent, that is, iff $p<3$.


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