For which real $a$ and $b$ is $I=\int _1^{\infty }\frac{x^a}{1+x^b}dx$ convergent? divergent? 
Let $a, b \in \Bbb R$. Consider the improper integral
$$I=\int _1^{\infty }\frac{x^a}{1+x^b}dx$$
For which values of $a$ and $b$ is $I$ convergent? For which values of $a$ and $b$ is $I$ divergent?
Hint: You might want to consider the two cases $b > 0$ and $b \leq 0$ separately.


What I tried so far:
Maybe translate the questions first:
Let $a,b\in \mathbb{R},I=\int _1^{\infty }\frac{x^a}{1+x^b}dx$
Define $\exists I_1,I_2\subseteq \mathbb{R},P(I_1,I_2)\leftrightarrow \exists I_1,I_2\subseteq \mathbb{R}s.t.\forall a,b \in\mathbb{R},(a\in I_1\wedge b\in I_2\leftrightarrow I \text{ is convergent})$
WTS $\exists I_1,I_2\subseteq \mathbb{R},P(I_1,I_2)\wedge(\forall I_j,I_k\subseteq \mathbb{R},P(I_j,I_k)\rightarrow I_j\subseteq I_1 \wedge I_k\subseteq I_2)$
So this statement answered both questions.
Is it also same to define $\exists I_1,I_2\subseteq \mathbb{R},P(I_1,I_2)\leftrightarrow(a\in I_1\wedge b\in I_2\leftrightarrow I \text{ is divergent})$?
I'm a little confused about this.. I think it might be the same.
Proof.
Let $I_1=(-\infty,b-1),I_2=\mathbb{R}$
First part, $P(I_1,I_2)$
$\rightarrow:$ Assume $a\in I_1\wedge b\in I_2$
$\space \space \space \space \space \space$ Show $I$ is convergent
$$I=\int _1^{\infty }\frac{x^a}{1+x^b}dx=\int _1^{\infty \:}\frac{1}{x^{-a}+x^{b-a}}dx\leq \int _1^{\infty \:}x^{a-b}dx$$
Since $a\in I_1\wedge b\in I_2$ we have:
$$a<b-1\wedge b\in\mathbb{R}\Rightarrow a-b<1$$
Recall that $\int _1^{\infty\:}x^p dx$ is convegent $\leftrightarrow p<-1$
Therefore $\int _1^{\infty \:}x^{a-b}dx$ is convergent
Sicne we have $$0\leq\int _1^{\infty \:}\frac{1}{x^{-a}+x^{b-a}}dx\leq \int _1^{\infty \:}x^{a-b}dx$$
By Basic Comparison Test that
$$\int _1^{\infty \:}x^{a-b}dx\text{ is convergent}\rightarrow \int _1^{\infty \:}\frac{1}{x^{-a}+x^{b-a}}dx\text{ is convergent}$$
Therefore $I$ is convergent
$\leftarrow:$
Assume $I$ is convergent
$\space \space \space \space \space \space $Show $a\in I_1\wedge b\in I_2$
Let's say if it's not the case that $a\in I_1\wedge b\in I_2$, then we have:
$$a\not\in I_1\vee b\not\in I_2$$
Consider 3 cases:
$$(\underbrace{a\not\in I_1\wedge b\not\in I_2}_{Case1}) \vee (\underbrace{a\in I_1\wedge b\not\in I_2}_{Case2}) \vee (\underbrace{a\not\in I_1\wedge b\in I_2}_{Case3})$$
Since $b\in \mathbb{R}$ by assumption, that Case1, Case2 must be false
Case3: $a\not\in I_1\wedge b\in I_2$
That is $$a\geq b+1 \wedge b\in \mathbb{R}$$
$$\Leftrightarrow a\geq b+1 \wedge (b>0 \vee b\leq 0)$$
$$\Leftrightarrow (\underbrace{a\geq b+1\wedge b>0}_{Case3.1}) \vee (\underbrace{a\geq b+1\wedge b\leq0}_{Case3.2})$$
Case3.1: $a\geq b+1\wedge b>0$
$$L=\lim_{x\rightarrow \infty}\frac{x^{a-b}}{\frac{1}{x^{-a}+x^{b-a}}}=\lim_{x\rightarrow \infty}x^{a-b}(x^{-a}+x^{b-a})=\lim_{x\rightarrow \infty}(x^{-b}+1)=1$$
Since the limit exists and $L>0$, by Limit-Comparison Test we have:
$$\int _1^{\infty \:}\frac{1}{x^{-a}+x^{b-a}}dx\text{ and} \int _1^{\infty \:}x^{a-b}dx\text{ are both convergent}$$
However $\int _1^{\infty \:}x^{a-b}dx$ is convergent implies $a-b<-1$ that $a<b+1$ Contradiction.
$$\vdots$$
Second part,$\forall I_j,I_k\subseteq \mathbb{R},P(I_j,I_k)\rightarrow I_j\subseteq I_1 \wedge I_k\subseteq I_2$
$$\vdots$$
This should be the general structure of the proof, and I see what the hint is talking about now.. Is my proof correct so far?
Any help or hint or suggestion would be appreciated.

Updates
Based on @catalogue_number 's answer I rewrite the proof..But there is some disagreement on Case1, is this proof correct?
WTS $$\forall b>0, I\text{ is convergent} \leftrightarrow a<b-1\wedge \forall b\le 0,I \text{ is convergent}\leftrightarrow a<-1$$
Since this is True iff $$\forall b>0, I\text{ is divergent} \leftrightarrow a\ge b-1\wedge \forall b\le 0,I \text{ is divergent}\leftrightarrow a\ge -1$$
Therefore this statement answered both questions.
Proof.
Case 1: $b>0$
Since $I=\int_1^\infty \frac{x^a}{1+x^b}dx = \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx$
And $\forall x \in [0,\infty),x^{-b}\neq 0$
That $\forall x\ge 1, 0 \le x^{-b} \le 1$, then we have: $$1 \le 1+x^{-b} \le 2 \Rightarrow 1 \ge \frac{1}{1+x^{-b}} \ge \frac{1}{2}$$
$$\Rightarrow 0 \le\frac{1}{2}\int_1^\infty x^{a-b}dx \le \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \le \int_1^\infty x^{a-b}dx$$
Recall that $\int _1^{\infty\:}x^p dx$ is convegent $\leftrightarrow p<-1$
By Basic Comparison Test we have:
$$a<b-1 \rightarrow \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \text{ is convergent}$$
And also
$$a\ge b-1 \rightarrow \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \text{ is divergent}$$
$$\Leftrightarrow a<b-1 \leftarrow \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \text{ is convergent}$$
We can conclude that:
$$\int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \text{ is convergent} \leftrightarrow a<b-1$$
$$\Leftrightarrow \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \text{ is divergent} \leftrightarrow a\ge b-1$$
Case 2: $b \le 0$
Since $\forall x \ge 1,0\le x^b \le 1$, then we have $$1 \le 1+x^b \le 2 \Rightarrow 1 \ge \frac{1}{1+x^{b}} \ge \frac{1}{2}$$
$$\Rightarrow \frac{1}{2}\int_1^\infty x^a \le \int_1^\infty \frac{x^a}{1+x^b}dx \le \int_1^\infty x^a$$
Recall that $\int _1^{\infty\:}x^p dx$ is convegent $\leftrightarrow p<-1$
By Basic Comparison Test we have:
$$a<-1 \rightarrow \int_1^\infty \frac{x^a}{1+x^b}dx \text{ is convergent}$$
And also
$$a\ge-1 \rightarrow \int_1^\infty \frac{x^a}{1+x^b}dx \text{ is divergent}$$
$$\Leftrightarrow a<-1 \leftarrow \int_1^\infty \frac{x^a}{1+x^b}dx \text{ is convergent}$$
We can conclude that:
$$\int_1^\infty \frac{x^a}{1+x^b}dx \text{ is convergent} \leftrightarrow a<-1$$
$$\Leftrightarrow \int_1^\infty \frac{x^a}{1+x^b}dx \text{ is divergent} \leftrightarrow a\ge -1$$
Since both case hold we have:
$$\forall b>0, I\text{ is convergent} \leftrightarrow a<b-1\wedge \forall b\le 0,I \text{ is convergent}\leftrightarrow a<-1$$
$$\Leftrightarrow \forall b>0, I\text{ is divergent} \leftrightarrow a\ge b-1\wedge \forall b\le 0,I \text{ is divergent}\leftrightarrow a\ge -1 $$
 A: I'm going to word this quite loosely, since I think the general technique of bounding these bad boys is what's important here.
Like the hint says, I'll treat the two cases separately.

Case 1: $b>0$
Without loss of generality, $\int_1^\infty \frac{x^a}{1+x^b}dx = \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx$ since $x^{-b}$ is never zero at any point on the domain $[0,\infty)$.
Then $x^{-b} \le 1 \forall x\ge 1$, and so $$1 \le 1+x^{-b} \le 2 \Rightarrow 1 \ge \frac{1}{1+x^{-b}} \ge \frac{1}{2}$$.
Then $\frac{1}{2}I \equiv \frac{1}{2}\int_1^\infty x^{a-b}dx \le \int_1^\infty \frac{x^{a-b}}{x^{-b}+1}dx \le \int_1^\infty x^{a-b}dx \equiv I$
This is very important - It says that the original integral converges iff $I$ converges. $I$ can be evaluated, but one must be careful.
$I = \lim_{x\to \infty} \frac{1}{a-b}(x^{a-b+1}-1) $, which converges if $a-b+1 < 0 \Leftrightarrow a -b \lt -1$
If $a \ge b$, $I$ diverges and so the original integral must diverge.
Case 2: $b \le 0$
$x^b \le 1 \forall x \ge 1 $, so $$1 \le 1+x^b \le 2 \Rightarrow 1 \ge \frac{1}{1+x^{b}} \ge \frac{1}{2}$$
Then 
$\frac{1}{2}J \equiv \frac{1}{2}\int_1^\infty x^a \le \int_1^\infty \frac{x^a}{1+x^b}dx \le \int_1^\infty x^a \equiv J$
So the integral converges iff J converges. J converges if $a <-1$.
This exhausts all of the cases.
