For which values of the parameter $\beta$ is the following integral convergent For which values of the parameter $\beta$ is the following integral convergent
$$\int_0^\infty \frac{3arctan(x)}{(1+x^{\beta-1})(2+cos(x))}\,dx$$
I tried using comparison test with $g(x) = \frac{1}{x^{\beta-1}}$, however I realized that for example for $\beta-1 = \frac{1}{2}$ it won't work and my idea got 'destroyed'
What should be the proper approach to this problem?
 A: Note that the integrand has no singularities on the positive real axis, so your only concern is the behaviour at $+\infty$. As the integrand is non-negative, the integral converges if and only if
$$ f(\beta) \equiv \int \limits_1^\infty \frac{3 \arctan (x)}{(1+x^{\beta-1})(2+\cos(x))} \, \mathrm{d} x < \infty $$
holds. For $x \in (1,\infty)$ we have $ \arctan(x) \in (\frac{\pi}{4},\frac{\pi}{2})$ and $2+\cos(x) \in [1,3]$, so we get the estimate
$$ \frac{\pi}{4} \int \limits_1^\infty \frac{1}{1+x^{\beta-1}} \, \mathrm{d} x\leq f(\beta) \leq \frac{3 \pi}{2} \int \limits_1^\infty \frac{1}{1+x^{\beta-1}} \, \mathrm{d} x \leq \frac{3 \pi}{2} \int \limits_1^\infty x^{1-\beta} \, \mathrm{d} x
$$
for any $\beta \in \mathbb{R}$.
Now if $\beta \leq 1$, we have $x^{\beta-1} \leq 1$ for $x\geq1$, so the estimate from below yields
$$ f(\beta) \geq \frac{\pi}{8} \int \limits_1^\infty 1 \, \mathrm{d} x = \infty$$
and the integral diverges. If $\beta > 1$, we can use $x^{\beta -1} \geq 1$ for $x \geq 1$ to find
$$ \frac{\pi}{8} \int \limits_1^\infty x^{1-\beta} \, \mathrm{d} x\leq f(\beta) \leq \frac{3 \pi}{2} \int \limits_1^\infty x^{1-\beta} \, \mathrm{d} x \, .
$$
