Convergence of integral at different values of $p$

Question

I want to study the following integral.

$$ \int_0^\infty (1+x^2)^{(\log^p(x))}dx$$

For example for $p = 0$ the integral is divergent.

I would love use asymptotic analysis to show that it's similar to $\frac{1}{x^{\alpha(p)}}$ for some $\alpha(p)$ and then impose the known integral convergence theorems.

First of all the problem is only at $x \to \infty$ because at $x=0$ the function goes to 1.

Now, I am struggling on simplifying the following limit:

$$ \lim_{x\to \infty} (1+x^2)^{\log^{p}(x)} $$ any hints?

Answer 1

$\int_0^\infty (1+x^2)^{(\log^p{(x)})}dx = \int_0^1 (1+x^2)^{(\log^p(x))}dx+\int_1^\infty (1+x^2)^{(\log^p(x))}dx$

$(1+x^2)^{(\log^p(x))} \geq 1, \forall p \in \Bbb{R},\forall x \geq 1$ so the second integral diverges and the first integral converges since the limit at zero exists for every $p \in \Bbb{R}$.

So you can extend the integrand to a continuous function on $[0,1]$.

Thus the general integral diverges for every $p \in \Bbb{R}$