I want to seek the real value $a$ such that $\int_0^{\infty} \frac{1}{1+x^a} dx$ converges by using comparisons test. Here is my method:
$$\int_0^{\infty} \frac{1}{1+x^a} dx = \int_0^{1} \frac{1}{1+x^a} dx + \int_1^{\infty} \frac{1}{1+x^a} dx$$
The integral $\int_0^1 \frac{1}{1+x^a} dx$ exists for all real number $a$ trivially. Next, since $\frac{1}{1+x^a} < \frac{1}{x^a}$ and $\int_1^{\infty} \frac{1}{x^a} dx$ converges when $a>1$, we conclude $\int_1^{\infty} \frac{1}{1+x^a} dx$ converges as well when $a>1$.
For $a\leq0$, since $$\lim_{x\to\infty}\frac{1}{1+x^a} \not=0 $$ it implies $\int_1^{\infty} \frac{1}{1+x^a} dx$ doesn’t converge. Also, $\int_1^{\infty}\frac{1}{1+x} dx$ doesn’t converge as well.
In the end, I guess that $\int_1^{\infty} \frac{1}{1+x^a} dx$ doesn’t converge when $0<a<1$, but I can’t find a suitable function $g(x)$ such that $g(x)\leq \frac{1}{1+x^a}$ for all $x\geq1$ (or maybe $x$ is large enough) and $\int_1^{\infty} g(x) dx$ doesn’t converge. Can anyone give some hints?