Improper integral with parameter. I have an integral 
$$\int \limits_{0}^{\infty} \frac{dx}{1+x^{\alpha}}, ~ \alpha \in (1, \infty)$$
I want to find for which $\alpha$ this integral uniformly converges. I proved that it's true on $[t, \infty), ~t>1$ using Weierstrass theorem. However, I don't know how to prove, that on $(1, +\infty)$ integral diverges. My idea was to use corollary from Cauchy criterion:
$$\exists\varepsilon_0 > 0 ~~\forall \xi \in (0, \infty) ~~ \exists b, b' \in [\xi, \infty) ~~ \exists\alpha \in (1, \infty): \int \limits_{b}^{b'} \frac{dx}{1+x^{\alpha}} > \varepsilon_0$$
More specifically, I don't know how to choose $\varepsilon_0, b, b', \alpha$. 
 A: The integral converges for any $\alpha > 1$, but not uniformly on the open interval $(1,\infty)$. 
For any $\xi$ choose $b = n$,$b' = 2n$ where the positive integer $n$ is greater than $\xi$, and $\alpha = 1 + 1/n$.  
Since $(2n)^{1/n} \to 1 $ as $n \to \infty$, we have $(2n)^{1/n} < 2$  for all sufficiently large $n > \xi$, and 
$$\int_n^{2n} \frac{dx}{1 + x^{1 + 1/n}}  > \frac{2n - n}{1 +(2n)(2n)^{1/n}} = \frac{1}{1/n + 2(2n)^{1/n}}  >\frac{1}{1+ 2 \cdot 2} = \frac{1}{5}.$$
With $\epsilon_0 = 1/5$ we have shown that the integral does not converge uniformly on the interval $(1,\infty)$. 
A: Hint:
Consider the integral $$I = \int_{0}^{\infty}\frac{x^{\beta}}{1+x^{\alpha}} \mathrm{d}x$$
The convergence of $I$ when $\alpha>0$, can be found by: $$\text{Integrand} = \begin{cases}x \text{ very large } &\mbox{behaves likes } x^{\beta-\alpha} &\mbox{converges if } \beta-\alpha+1>0\\x\text{ near 0 } &\mbox{behaves like } x^\beta &\mbox{converges if } \beta +1>0\end{cases}$$
This gives us: When $\alpha>0$, integral $I$ converges if, $-1<\beta<\alpha-1$.
Can you take it from here?
