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

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), so this is the only comparison allowed regarding to it (and comparing it to numbers of course).

EDIT:

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 $$p\leq2$$

• Please add the slashes for your $\ln(\cdot)$ natural logarithm. – 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 $$p<3$$.