For which values $a$ does the improper integral $\int_0^{\infty}\frac{\ln(1+x^2)}{x^a}dx$ converge

Question

Find the values $a$ s.t. the integral

$$\int_0^{\infty}\frac{\ln(1+x^2)}{x^a}dx$$

converges.

I tried some values of $a$ by programming, it seems that for $a=2$, the integral converges, and for $a=3$, it diverges. But how can we explicitly determine the range of convergence?

Answer 1

Hint: Near $x=0$, $\frac{\log(1+x^2)}{x^a}\sim x^{2-a}$ and as $x\to\infty$, $\frac{\log(1+x^2)}{x^a}\le C_\epsilon x^{\epsilon-\alpha}$ for all $\epsilon\gt0$.

Not that it matters to the question, but $C_\epsilon$ is approximately $\dfrac2{e\epsilon}$ .

Answer 2

The result of the integration is -Pi Sec[a Pi / 2] / (a - 1) under the condition [ 1 < Re[a] < 3 ]. For a less or equal to 1, the integral diverges; the same if a is equal or greater than 3.
The antiderivative of the integrand is
x^(1-a) (-2 + 2 Hypergeometric2F1[1, 1/2-a/2,3/2-a/2,-x^2] - (-1 + a) Log[1 + x^2]) / (a-1)^2 and this explains the above.