Convergence of $\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$ without Gamma function 
For what values of $a$ does the following integral converge: $$\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$$

We have to study the integral at $0$ and $\infty$:
$$\int_0^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx = \int_0^1 \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx + \int_1^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx$$
I've started checking for $a>0$:
For the second integrand we have equivalence
$$\frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}} \sim \frac{x^{-2a}}{\sqrt{x^a+x^{-a}}} \sim \frac{1}{x^{5a/2}},  \ \ x \to \infty$$
So,
$$\int_1^\infty \frac{\ln(1+x^{-2a})}{\sqrt{x^a+x^{-a}}}dx \sim \int_1^\infty \frac{1}{x^{5a/2}}dx $$ which converges when $5a/2>1$ or $a>2/5$.
But this approach seems to be lots of work to study fully, not to mention that I don't feel I can deal with all of the cases...
Any help is appreciated.
 A: *

*$a > 0$
Since
$$\lim_{x\to 0^{+}} \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} = 0,$$
$\int_0^1 \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges.
By using $\frac{t}{t+1} \le \ln (1+t) \le t, \ \forall t \ge 0$, we have
$$ \frac{1}{4x^{5a/2}}\le \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \le \frac{1}{x^{5a/2}}, \ x \ge 1.$$
Thus, $\int_1^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges
if and only if $a > \frac{2}{5}$.
As a result, $\int_0^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges if and only if $a > \frac{2}{5}$.


*$a < 0$
Since
$$\lim_{x\to 0^{+}} \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} = 0,$$
$\int_0^1 \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges.
It is easy to obtain
$$-a x^{a/2}\ln x \le \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \le -2a x^{a/2}\ln x + x^{a/2}\ln 2, \ x \ge 1.$$
Thus, $\int_1^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges
if and only if $a < -2$.
As a result, $\int_0^\infty \frac{\ln (1 + x^{-2a})}{\sqrt{x^a + x^{-a}}} \mathrm{d}x$ converges if and only if $a < -2$.
